Aim: simplicity amidst compli co mplicatedness. catedness. To understand is to project a known/understood phenomenon onto an unknown/unexplained phenomenon. But understanding is not a rigourous thing. th ing. Whether something is true or not doesn’t care whether you understand it or not. All of the following are (a) known “phenomena”, and (b) non-rigourous, (1) dimensions; (2) scaling; (3) symmetries; symmetries; (4) mathematical tricks
(1.1)
Personal note: ion beam implantation experiment/theory interface. Maslov dug throu gh Danish journal in order to find an important result. Imagine incident part icle with energy E0, and it impinges impinges upon a lattice with atominc binding energy E b. How many atoms are displaced? After 10 pages, N c ( E0 / E b ) . Such a simple simple answer.
Took answer to adviser, and he said “Oh yeah, they just share energy in equal chunks!” By the way, the actual answer is N c E0 / E b , so the above result is wrong. Short story about diffusion: how to describe in very simple, pheno menological way? Well, flux of particles: defined as, diffusion # (1.2) flux 2 ; j Dn; D ; constant L T
This (1.2) should trigger “symmetries” of (1.1), because you bu ilt a vector out of o f a scalar (viz, del operator). Time reversal: the LHS is odd under time reversal: j - j implicity assumed j. But the RHS is not! Why? You implicity dissipation; dissipation; that is, an increase in entropy. e ntropy.
j ; putting this into (1.2) yields, Continuity says n n D 2 n; n n (r, t, D );
(1.3)
How fast does a point-source diffuse to a point where you have, say, 1 standard deviation of, say, a no rmal, 1 distribution ? Use dimensional analysis of (1.1). # L 1 L2 (1.4) [ D ] 3 [ D] ; L2 T LL T Characteristic width of distribution: what is a formula for it? Actually, there’s only one wa y to compute it: D times some characteristic time. Therefore, r 2 ~ Dt , where “ ~ ” means you don’t know the number t . Now let’s use scaling of (1.1). Make a guess about the form of a distribution. distribution. In three dimsneions, the concentration-profile should be spherically-symmetric. You have, g
[unknown function ] g (r /
n (r , t ) r 3 g (r /
Dt )
Dt )
( Dt )3/ 2 f (r /
r 2 r 2 n(r , t )d 3r 0 r 2 n(r , t )4 r 2dr 0 r 2 ( Dt )3/ 2 f ( r / x r /
Dt
4
( Dt ) 3/ 2 Dt
Dt
0
x f ( x)dx Dt 4 2
0
2
x f ( x ) dx
Dt )
The integral is just a number, nu mber, despite it being from [0,infinity]. [0,infinity]. Also, it is of order 1. Now execute some similar machinations with derivatives, derivatives, 1
Actually: Actually: Gaussian may be th e concentration profile here, but I’m not sure.
Dt )
4 r 2 dr
(1.5)
2 1/ 2 n D x n t (D t ) f(
x Dt
)
t ( D t) 1 2
1
1/ 1/ 2
f(
x Dt
) ( Dt )
x n (Dt )1/ 2 f ( x 2 n f (
x Dt
1/ 2
f (
x Dt
)( 21) t
1
x Dt
f
D
( Dt )
3/2
f ( s); s
x Dt
; (1.6)
1/ 2
x
)( D t ) Dt
)( Dt )
3/ 2
Actually, now we have an ODE; look, f
1
f
2
12 sf 0 f 12 sf 12 f f 12 (sf ) f 12 sf 0
f 12 sf 0 f ce
s 2 / 4
n ( x, t )
c Dt
e
x 2 / ( 4 Dt )
;
throw away unbounded-solution; unbounded-solution; want to go to 0 at
n( x, t) dx 1 c
1 2
0 ; n( x, t)
(1.7) 1 2 Dt
x2 / ( 4 Dt )
e
;
Wednesday, January 09, 2013
- going to rearrange times - book by migdal is available. ava ilable. Strongly recommended: Read first chapter of boo k. - gatman and Halpern = hopelessly out of print - tim is making notes; those are reference r eference too. The link to the t he notes is posted on the course-website. - Maslov is trying to make a book. - homework #1 will be posted today; due in 2 weeks. Consider bad/ugly integral,
0
ln(ln x)e x
2
(1 x5/ 2 )2
dx ??
How do you estimate? Constraint: you’re not allowed to sit down and do math. Do an order of magnitude estimate,
0
ln(ln x)e x
2
(1 x5/ 2 ) 2
dx ~ 1
positive-definite for whole who le Proof: what else could it be? QED. The function has no para meters. Integrand is positive-definite range, and is square-integrable thanks to the t he gausian. Better estimate: turns out the integral is about 1.9… the point is that the order o f magnitude estimate is ~ 1. Order of magnitude estimates are perfectly legitimate when you are working wo rking “in theory”. Not all physics questions require precision! 2 Diffusion: how width of diffusion changes. “by units”, t he law is: r ~ Dt in free space.
What happens when you have an interface? How does the law change?
The two concentration profiles would appear as, mathematica: plot Gaussian vs. piecewise-Gaussian for various times
2 1/ 2 n D x n t (D t ) f(
x Dt
)
t ( D t) 1 2
1
1/ 1/ 2
f(
x Dt
) ( Dt )
x n (Dt )1/ 2 f ( x 2 n f (
x Dt
1/ 2
f (
x Dt
)( 21) t
1
x Dt
f
D
( Dt )
3/2
f ( s); s
x Dt
; (1.6)
1/ 2
x
)( D t ) Dt
)( Dt )
3/ 2
Actually, now we have an ODE; look, f
1
f
2
12 sf 0 f 12 sf 12 f f 12 (sf ) f 12 sf 0
f 12 sf 0 f ce
s 2 / 4
n ( x, t )
c Dt
e
x 2 / ( 4 Dt )
;
throw away unbounded-solution; unbounded-solution; want to go to 0 at
n( x, t) dx 1 c
1 2
0 ; n( x, t)
(1.7) 1 2 Dt
x2 / ( 4 Dt )
e
;
Wednesday, January 09, 2013
- going to rearrange times - book by migdal is available. ava ilable. Strongly recommended: Read first chapter of boo k. - gatman and Halpern = hopelessly out of print - tim is making notes; those are reference r eference too. The link to the t he notes is posted on the course-website. - Maslov is trying to make a book. - homework #1 will be posted today; due in 2 weeks. Consider bad/ugly integral,
0
ln(ln x)e x
2
(1 x5/ 2 )2
dx ??
How do you estimate? Constraint: you’re not allowed to sit down and do math. Do an order of magnitude estimate,
0
ln(ln x)e x
2
(1 x5/ 2 ) 2
dx ~ 1
positive-definite for whole who le Proof: what else could it be? QED. The function has no para meters. Integrand is positive-definite range, and is square-integrable thanks to the t he gausian. Better estimate: turns out the integral is about 1.9… the point is that the order o f magnitude estimate is ~ 1. Order of magnitude estimates are perfectly legitimate when you are working wo rking “in theory”. Not all physics questions require precision! 2 Diffusion: how width of diffusion changes. “by units”, t he law is: r ~ Dt in free space.
What happens when you have an interface? How does the law change?
The two concentration profiles would appear as, mathematica: plot Gaussian vs. piecewise-Gaussian for various times
Implicity: Implicity: we introduced the boundary condition
x n |xinterface 0 .
Reminder: do not use math or any sort of calculation.
number-density. The above extrema, vacuum Type III boundary condition: flux of particles proportional to number-density. and impermeable wall, were two extremes e xtremes of an intermediary behavior we’re now go ing to try and capture. The boundary condition is now x n |xinterface n / ; the is a characteristic-length that comes co mes from the derivativeoperator. Question: can you use dimensional analysis to claim that n( x, t ) (Dt ) 1/2 f ( xDt ) ? Can you still use this same 2
ansatz ? Further guess: you will probably have a linear combination of the full Gaussian and the p iecewise solution that vanishes appropriately at the boundary. N.B.: the method of images enters here, too. This is provocative. Limiting cases: classical mechanics is limiting case of quantum theory. S omteims we know higher-order theory, sometimes we know lower order theory. t heory. For diffusion, we know lower-order theory: the Boltzmann equation. Problem with diffusion equation: too little information. Set up system in macrosco pic. What koind of process in real life leads to diffusion? Happens by ballistics and relaxation-events w/probability.
Particle has definite position and velocity. diffusion equation do es not ask about velocity; it only o nly computes concentration. You have averaged all velocity out. Higher order theory keeps information about velocity. Collisions: elastic collisions on event of relaxation. Therefore, energy is conserved. Thus, velocity changes only in direction, not in magnitude.
So: compute function at co nstant energy, f f ( v , r , t ) |v 2 [const]v 2 f ( , x , t ) 0
Boltzmann equation: describes evolution of distribution function f, mean free df f f p f f f 0 f v cos ; ; time t x t x dt
F 0
2
Even so this is a “Guess/ansatz” of a solution, it is highly general; you are only asserting (1) a scale x /
, in which which there r eally isn’t any sort of bad ph ysics or reasoning.
Dt and (2) units
1 L
1 Dt
Collisions Collisions randomize rando mize the angular distribution of velocities. If you sample velocity at two different times, you will observe two different angles, but the same magnitude since you have energy co nservation. Relaxationtime approximation: you have f
( f
f 0 ) / .
tion f f ( , x , t ) in a complete basis. Since you have a ngular distribution, you need to Trick: expand functio expand in basis complete overangels. overange ls. That’s Legendre polynomials of the argument co s(theta), 2 f f ( , x, t ) P (cos ) f ( x, t ) f 0 ( x, t ) 1 cos f1 ( x , t ) O
Derive diffusion equation from this! We actually only ned t wo Legendre polynomials: p0 and p1. Also, let a large number of collisions take place: p lace: t . Truncate the series. Then, you have another statement, n( x , t )
1 2
sin f ( , x , t )d
1
1 1 2
( f 0 pf pf1 )dp
1 2
2 f 0 f 0 n ( x, t ) f 0
There: so the diffusion equation is the leading t erm in the series. The first order correction will tell us about velocity. ut in flux, j
1
1
v0 cos f
1 2
sin d
1
v0 1 p ( f 0 pf1 ) 12 dp 12 v0 f1 f1 2 j / v0 ; f ( , x, t ) n( x, t ) 2 j ( x, t ) cos / v0
Second term disappears when we average over angles! That’s the behavior we want! We want leading term/diffusion-equation term/diffusion-equation beaviour bea viour when we remove angle-dependence by averaging! (Cool). Hm, don’t know why, but, [deviation from equilibrium/Fick's law ] t n cos 2 t j / v0 v0 cos ( x 2 cos x j / v0 ) 2 j cos / (v0 ) (1.8) Multiply whole of (1.8) by cos(theta). Find averag e value of function cos(theta) on both sides. Note the average value of cos2(theta) is ½,and that cos3(theta) has an average value o f 0 due to its odd par ity.
t n 12 (v0 / v0 ) x j 0 t n x j [fick's law, which subsumes matter-conservation ] nd
Extra cosine function kills all terms left in the angle. 2 term vanishes. Now, see if you can obtain from (1.8), the,
t j / v0 12 v0 x n j / (v0 ) j 12 v02 x n t j D x n t j
D
12 v02
You computed diffusion-constant! Diffusion equation is limit when tau 0, or t ≫ tau. Finally: note extra term t j . This imposes constraint on validity of diffusion equation. when t his term is small, fick’s law is good approximation. Thus, current must change slowly in time. Neglect t j : go back to diffusion diffusion equation, then you know j
D x n x ~
Dt .
Thus, x and t are related. Flux should evolve slowly in time.
Number density and flux will not have independent scales. There is no such thing as a scale where flux and number density are not related. We need to look at times t imes t . Derivative: t j ~ j / t , meaning that
t j ~ j / t (and, of course, we mst have t j 12 v0 2 x n ). That scale is the mean free path, x ~
Dt
~ v0 [ mean free path ]
Diffusion equation operates on long time scales compared to length-scales, as relatedby mean free path as related to
Dt .
When you do not have long time scales, the transience is captured by the term t j .
Note: in order to lsoe information about velocity, several collisions must take place. (after first collision, you still should be able to track the trajectory of the particle; not yet randomized). This is why boltzman equation is 1st order in time: when you relate j D x n , you have disparity under time-reversal (c.f., previous lecture),
the paradox. The parado x is resolved in this “information-about-velocity-direction” point of view.
3
Example: strong explosion goes off . How does blast front move in time, given the explosion is strong?
(1.9) What are parameters at work? (1) Energy of the bomb. (2) since explosion is modeled as sp herical front of hot gas effusing (quickly) into cold gas, you have hot temperature and low temperature, forming a pressure gradient, (1.10) parameters R , E0 , 0 , t ; R R (t ) [blast radius] ??? Temperature inside (hot) should be function of E0, so possibly not independent. Temperature outside is independent parameter (in which mass-density of air, spec ific heat, etc., will be subsumed). “Cobble parameters together to make a t hing that you want”: dimensions! Buckingham pi theo rem. 2 3 2 [ E0 ] [ E0 ]( R, t , 0 ) [ M ][ L ] [t ] ; [ 0 ] [ M ][ L ] ;
(1.11)
Note: bomb may be chemical or nuclear. But both are the release of energy, and dimensions doesn’t care which it comes from. Using E 12 mv 2 , and noting E 0 , 0 are material parameters: out of the three units that must be on RHS of result, we can’t make a dimensionless number. Therefore, e ach should enter as a power. Now: the following is a true statement: order of magnitude estimate of R, a b c R ~ E0 0 t ; [surprisingly-constraining statements!]
(1.12)
The (1.12) is by no means rigorous, but it is very informative/helpful. Under what co ndition is it not helpful? When you have scaling function w/o dimensions, a b c (1.13) R ~ E0 0 F ( 0 / t ); [scaling function not captured by dimensional analysis] So: we are linting ourselves to the scope (1.12). Step 2 = find the exponents. Substitute in the units, [ R ] [ E0 ]a [ 0 ]b [t ]c
3
[M ]a b [ L ]2a 3b [t ]c 2a 1 2a 3b; 0 a b; 0 c 2a (a , b , c ) ( 15 , 51 , 25 ) (1.14)
Taken directly from research on the theoretical yield of an atomic bomb.
Hey, that’s 3 equations and 3 unknowns! You get the radius as a function of time,
R R(t ) ~
5
E 0 0
t 2 ;
(1.15)
Shortcomings: you don’t have a character istic distance coming out of (1.15). However, this does tell you that the explosion has a front,
This sacrifice of scales is manifest ; recall that power law functions don’t have scales. You are basically first making a guess; no other units. Look at units, see power laws. See if you’ve found all qualitative behavior: (1) t = 0 behaviour, (2) the expanding front, etc.; basically, units says “you can o nly combine all parameters in one way; mathematically speaking, you can find two solutions: (1) expanding front (@) bell-looking thing; hwoever, neither solution has a scale.= due to power law. Finally: fluid dnamics is nonlinear. This is a hard pro blem (nonlinear fluid dynamics equation, etc.). Before this method, nobody looked at strong explosions, since that was considered a d ifficult problem. This particular solution was most welcome. In the 1940s, propaganda. You saw news reels, not previews. One t ime, there was a news reel with a nuclear explosion. John Archibald Wheeler happened to be in the audience. He o bserved that there were regularlyspaced telephone poles in the explosion. He used the uniform-spacing, and made a computation. His result was so scarily close to the actual result that he got in trouble with the FBI, because his result was scarily close to classified information… More formal stuff – symbols
Go to MM - 001 – terminology.
Wednesday, January 16, 2013 [ D ] L2T 1 R R (t ) L L R (t ) x 2
x2 ;
x 2 n( x, t )d 3 x ( D a ) 22 1 (
x
D a
)2 n ( x , t )4 (
Consider: on the basis of justdimensional analysis,
[dimensional analysis] R (t ) L
t Dt f ( ) ; a
x D a
)2 d (
x D a
)
Now need to determine scaling function. What properties? Well, consider asymptotic t a vs. t a : there’s information in both these asymptotes. You have f (0) ~ 1 and f ( s 1) ~ ? Now: diffusion equation. you have absorption: n D 2 n a n . Assume a solution: n( x, t ) n0 ( x, t ) e t / a as an ansatz to handle the absorption term, scaling function is 1 x2 t / a t / a 2 n D n a n n (x , t ) n0 ( x , t )e exp( )e (1.16) just an exponential 4 Dt 2 Dt Then you have, x 2
x 2 n( x, t )dx e t / x 2 n0 ( x, t )dx et / C tD C Dt e t / a
a
a
(1.17)
Particles mostly at surface. At long time, they most likely forget about the absorption term, So: dimensional analysis only takes you as far as R (t )
Dt f (t / a ) ; most of physics concerns itself with
finding this scaling function. 0.35
0.30
0.25
0.20
0.15
0.10
0.05
1
2
3
4
5
6
Asymptotic accuracy and parametric scaling
We only, only solve a model, which intrinsically neglects factors. All our results are only asymptotically exact. Example: specific heat of Fermi gas: CV
12 2 k B n kBT / EF O 3 ( kE T ) . This is true in the limit where thermal B
F
energy is much smaller than the Fermi energy (as indicated). Our result always is where a parameter is turned to one of its extremes (smallness or largeness). So: Boltzmann statistics: CV 12 d k B n , where d is the 4
dimensionality . What does full behavour look like?
Width of asymptotic
4
Note that Fermi energy is removed as scale.
Width of “crossover” region is about E F / k B
[ K ] [Kelvins] .
example: if a scaling function has a 0 near a point, we can ask if the function has linear, quadratic, or cubic behavior near 0.
Example: a function blows up at the endpoint of an interval
Ratio of heat capacities: thatof Boltzmann and t hat of Fermi,
CV , F CV , B
1 2
k B n (k BT / EF ) 2
1 2
nkB d
2 d
(k BT / EF )
C V , F
1 kBT
CV , B
d 2
EF
kBT EF ;
Parameters you control, numbers you do n’t control (duh). Functions
Dimensionless function f dimensionless argument. Could be two types of functions: (1) normal and (2) pathological.
Function-composition hides the pathology.
Friday, January 18, 2013
(1.18)
Brownian motion: Model as (1) random walk, (2) discrete steps that happen a fter a time (the mean free time). (3) Once a time a elapses (over which time any number of discrete-steps may be taken), the particle
disappears, and gets absorbed,
(2.1) Fick’s second law then has an absorption term added to the time-derivative, n n n / a . We make an ansatz: an exponentially-attenuated concentrated envelope which, if integrated, yields total particle-number at some time (note: normalization is time-dependent for disappearing particles), t / t / 2 (2.2) n D x n ( n / a ); n ( x , t ) n0 ( x, t )e a ; N (t ) n( x, t ) dx e a
Case 1: What if 1 ms and a
1s
(
a
)? Can this model still be used? Yes. The absorption time is
long/absorption doesn't happen that often. diffusive trajectory is formed over many mean-free-times . Diffusion is largely ballistic. Your concentration profile then appears as the usual so lution, 2 g ( x ) f ( x ) d d D x d f Dx x x ; ; n n( x, t ) f ( ) f ( ) f (s ) 2 ; ; s ; 3 2 r dt 2 dt 2 dx x 1 1 d d 2 f ( x ) f ( dtd x ) f d dt f f ( 2 2 D) f 2 D D f ˆ Ln ( D 2 ) D 21 0 sf f 2 f 0; (2.3) dt dx 2 3
f 12 ( sf ) 0
f 12 sf 0 0 d ln f 12 s ds f ce s / 4 n( x, t ) 2
Case 2: Opposite limit:
a
f ( s) / Ce x
2
/(4 Dt )
/;
, in this limit, you have more traps than elastic scatterers. You can consider this
to be a lattice of a high density of traps, and a low density of elastic scatterers. For high density traps (frequent absorption), diffusion equation becomes meaningless, and you have to go back to Boltzmann equation, f F i f (k , r, t ) f 0 1 i I C { f } f k f ; f 0 f 0 (k , T ) ; viri f ; I C { f } (2.4) e k 1 t There are no forces ( F i 0 ), we use the relaxation-time (as indicated in (2.4)), we retain the partial-derivative of the nonequilibrium distribution function, and realize v i ir f v cos x f , and we have,
f f f f 0 f f f 0 f f f f v cos ; v cos ; t x a a t x a
(2.5)
Then: effect an arrangement of ascewnding orders/perturbative-expansion ansatz, regarding isotropy as a leading term, and having a correction 2 j cos / ,
f
f ( , x, t ) n( x, t )
trial solution for f in "almost" isotropic: leading term cos (2.6) boltzmann equation isotropic, plus -dependent correction
2 j
Drawing the trajectory over two separate mean free paths, we have, (2.7) You could also have
f ( , x, 0) ( x) (cos
cos 0 ) g ( x); (cos cos 0 ) ( x vt cos 0 ) e t / ;
f ( , x, t ) g ( x vt cos ) et / a n( x, t )
1
1
a
t / a
(cos cos 0 ) ( x vt cos )e
t / a
d (cos ) ( x vt cos 0 ) e
1 2
Bunch of spikes.
Integrals ,
I
0,
f ( x) dx 0 ; ~ 1
f ( x) dx f x;
Example, f ( x) xe
x
0
0 x ;
1;
f ( x) dx
0
x
A x ; A 1;
vs. support from large region: vs. middle region:
x
xe dx xe |0
x2
x1
xe
x
A
x
x
xe dx xe
well-known dx integral 0 1 x2 2 tan 1 A | A 1
~ 1; 2
A
x A;
dx xe x | A
| x x12
e x |xx
2
1
dx
0
1
e x |0 1;
1 x
1 A
2
0
x
xe dx ~
P Q
0
~ 1;
2 x(1) dx 12
; (x1 , x2 ) ( , A) [interval where integral takes support]
x2
x1
dx
A
1 x
dx 1 x
2
2
tan 1 x | A
tan 1 x | xx
2
1
2
~ 1;
Another example: function with an integrable singularity, 1 1 f ( x ) f ( x ) dx [finite] ?; [well: integrate tiny region] 0 x ( x 1) 2
1.0
0.8
0.6
0.4
0.2
0.5
1.0
1.5
2.0
2.5
1;
e x |A Ae A e A 1;
1;
1;
x
3.0
tan1 A;
0
f ( x )
1
0
x ( x 1) 2
1
dx
0
1
1
2 x (0 1)
dx 2 1
Therefore: the singularity we indicate is still integrable.
a 1 for convergence. The upper bound a < 1 comes from avoiding logarithmic-divergence lim x 0 ln x . This is basically the Convergence theorem from calculus: consider f ( x ) | x 0 ~ x a . We must have
power-law convergence that you are asked to remember.
ln x x
0
x
divergent due to ; lower limit 0
e dx
ln x
0
x
e x dx
ln x
0
x
dx
0
ln xd (ln x )
1 2
(ln x)2 | 0 ;
Question: how do you estimate ln x for x ~ 1 ? Well, you have ln( x ~ 1) ~ 1. Note: for ln( x 1) 0 . Question: of the zillions of special functions out there, how many have independent asymptotic behavior? The point is: asymptotic behavior is indistinguishable from the function whence it came. If you captured asymptotic behavior, you don’t know what special function it came from. E.g., J 0 ( x) | x0 1 , but then again e x | x 0 1 . But
then again it could have come from e x
p
In any event: a Bessel function is J 0 ( x)
| p 0 0 . 2 x
sin( x 4 ) (oscillatory). Special functions have weird behavior,
and we are intererested in their properties at special points, such as 0’s. The function sin(x) goes to x at small x. So: handling asymptotic behavior o f function means massaging it into some form: { x a , ln c x, e x
p
| p 0 } (??).
No class Monday—sleep well.
Wednesday, January 23, 2013
Pool ball example. But now the question is: find the thermal equilibrium,
(2.8) Depends on whether pol table has wa lls or not. If walls, then after finite time all poolballs will be involved in motion. Without friction, the balls certainly will thermalize. Without walls, number of moving balls increases
intime, and at the same time the are of moving bals is also increasing. Ho w does number and/or density of moving balls increase in time? Assumption: spaerse-distribution of balls: d
a . Total number densiy of balls for 2D table is nb ~ d 2 .
Describe such a process in terms of classical cross sect ion. Two balls collide. Each ball has radius o f a. then, the classical cross section is necessarily on order of: ~ a 2 . But total crss section really doesn’t tell you anything… Note: in any process, incoming energy is shared/divided amidst the particles it collides with. a2 energy transfer 2 ~ a ; d d f ( , , a ) d ( ) d ; Ansatz: ~ x a ; 0 a 1; x ; (2.9) ratio The singularity at x = 0 is for forward-scattering. Peaked at small energies. Most probable collisions are such that the particle (1) keeps most of its energy a nd (2) keeps its original direction of motion. Part icle goes through array of stationary atoms. Keeps projectile motion. Motion, itself, is contained in narrow (conical?) region encircling axis of original trajectory. What if 1 a ? Collision leads to rouighly dividing incoming fac Question: whgich of the two scenarios dominates? Not all mot ion which spreads does spread ina Brownian way. Brownian motion is a sub-case of this motion if the original ball (primary) has a mass very different from other particles. Heavy/macroscopic particle is jostled/attacked by molecules which we do not see. Collisions essentially elastic. So: in this scenario, is energy shared even ly, or is there a preference for small energy transfer?
f ( , , a ) . This necessity is realized as, eye-catching yet-integrable c0 ( a 2 / ) ( / ) d a 2 ( x) dx ~ a 2 1 a 2 ~ (2.10) 0 0 singularity, you have small number of these
Note: you must have a normalizable
Picture of equally-shared energy is simple and p lausible. (recall Maslov’s huge excursion into formalism, just to get a laughably-simple result: particles equally-share energy). Estimate number of balls, N ~ 0 / . Look at system as function of time. Consider system when average e nergy is . At the time where you measure the average, you know number of balls that have been struck. Thermalizing: at some point, the things must equilibrate/thermalize. Then, you have Boltzmann distribution function. Energies parameterized by just temperature. Do es system reach thermal equilibrium? st
nd
Estimate mean free path of 1 generation, vs. mean free path o f 2 generaton, ~ a 2 ; n ~ d 2 ; nb3 D ~ d 2 a1 ~ (nb3 D ) 1 ~ d 2 a / a2 d 2 / a ( d / a) d / a 1 d x|x 1 d d; (2.11) Area then increases in proportion to the number of circles,
(2.12) Then you have,
2
R ~ N
2
nm ~
N R 2
~
N R 2 2
~
1
d2
a
; 2
nm nb
~
a2 / d 4 1/ d 2
(
a2
) 1; nb d 2
nm
[#-density of moving-balls]
[number of balls? (per unit vol?)]
Turns out: this does not thermally-equilibrate to some Boltzmann distribution, because it needs to finish its transitory period illustrated in (2.12). When the walls reflect the transferred energy, the n you get thermalizing. When mean free path is increasing function of energy, you have constant mean free path. The radii of the areas of circles will go faster than energies. You t hen have that increase fater trhan the number of particles. Central theorem of dimensional analysis
This is called the “pi theorem”. Already used t his multiple times on homework. Worthless to repeat. S ludge through abstractions, and you see the stuf you used on your homework.
Friday, January 25, 2013 – my birthday! Example: what is resistive force as object moves through fluid?
a 1 1 3 1 1 b a b c d a b 3 c d c d b d L1 M 1T 2 0 0 1 1 1 ..; F R v L M T c 0 1 0 1 2 d See QM 11 - 002 - dimentional analysis for object in fluid and for fine structu re const. NOTE: since both force and velocity are vectors, then force must eb opposite of velocity. thus,w e make the ansatz F v A , where scalar-function, A A(v ) . Symmetry: fixes direction between force and velocity, but doesn’t tell you what A(v) is. Assumption: analyticity: that means F depends on v to the first power, allowing for a Taylor expansion to be linear to lowest-order. G G(
vR
vR ) G( ) G(Re); G(0) ~ 1; V /
Expnanation for behavior of this: at boundary, fluid has same velocity as ball, small Reynold's 2 number limits u(r / R) ~ v / R; F ~ vR; F ~ PR ~ v R; u
large Reynold's need drag-force number limits stuff... January 28, 2013
Estimate the air resistance on your car at 60 mph and assuming “rolling friction” of F RF 0.01N .
(2.13)
c a 14 2b 2 c 3b 2 b 2(1 a) 2(1 a) d a d c b 2b 2c 1 a 14 ca
2
L
T
1 2
K a E b L3b E c K c M d K c a Eb c L3b M d
mv
2
3 2
k BT
vRMS
3k BT 0.3mO2
0.7mN
2
3(0.026 eV) (1.602 1019 ) 27
(0.3 16 0.7 14) 1.7 10
kg
710
m
s
;
2 (2.5 1010 m)2
Wednesday, January 30, 2013
Fractals: something that doesn’t play well with dimensional analysis. It is an arrangement whic h thwarts dimensional analysis. Dimensional analysis (DA) uses SCA: “spherical cow approximation”. This is inappropriate in the coastal line paradox. S olution to coastal line paradox is richardsons’s law, which co ntains a fractional-power of a dimensionless quantity, undetected by SCA/DA. Koch snowflake: 4 2 n 1 P1 3a ( )1 4 a; n 2 P1 3 a ( )2 ; ... n m Pm 3 3 a ln( a / 3) s n n Pn 3aeln(4 n/3) 3aeln ?? 3 3
3 a(
4 3
)m
3 a(
4 3
)m ; P
3 a( ) ;
Topological dimension DT, and dimensionality D (e.g., surface in three dimensional space means DT = 2 and D = 3), R D R D D D D V R T ( ) T R D s T ; DT D DT 1; [line] L R( ) D 1 RD s1 D ; 1 D 2; s s R R S R 2 ( ) D 2 R2 s2 D ; 2 D 3; V R3 ( ) D 3 R D s3 D 3 D 4; s s
Friday, February 01, 2013
What is number of organisms vs. time, given that they breed? Start by thinking about parameters, amount of O2 [ M ] [ M ] R [breathing rate] ; W [weight] [ M ]; [mass density] 3 ; Unit time [T ] [L ] Physicists introduced an additional parameter . You have, amount of oxygen m ; [dimensionality of organism]; tL tL Putting (3.1) and (3.2) together, we have, W R ~ ( ) /3 ; [mysids]
2.4;
[crabs]
2.25;
[humans] 2.4;
(3.1)
(3.2)
(3.3)
Hence, there is some universality between the var ious species. It has to do with the ratio of lung-volume to the rest of the body-mass. A L D [fractal]; 2 D 3; 2 3; (3.4) Fractal dimensionality applied to slight modification of a game of pool
A pool ball hits a cluster, and causes the cluster to eject particles. Those secondary part icles do likewise, and cause a quadratically-larger number of tertiary particles. This sounds artificial, but th is is like what happens in nuclear reactions, and in lightning storms. Should be done with dimensional analysis. Next week: approximating integrals, differential equations, etc. you will be refreshed on any topics in math that are relevant but you have forgotten.
Monday, February 04, 2013
You have integral: approximate it, I ( a )
I ( a)
I (a)
a2 a
0
(e
ax
)
J 0 ( x) a 2 ( ax) 2
ax J ( x) 0
0
e
1 x
J 0 ( x)
0
1 x 2
e
cos xdx ; I ( a 1) ?;
(3.5)
cos x d ( ax)
2
ax
cos xdx
/ a u
J 0 ( ua )
0/ a
1 ( )
e
u 2 a
cos ua d ( ua ) ~
1 a
;
All functions in the integral have a scale o n the order of 1. You can strike out the integral in the region where u/a infinity, because of the decaying-exponential. Similar example, simple, but in more detail. Consider an integral5 where we normally complete the square, I ( a)
ax x 2
0
e
e
dx
0
2
(1 ax 12 (ax ) 2 ...)e x dx; I (a 1) ?; I (a
1) ?;
(3.6)
The series generates a polynomial, whose coefficients are familiar, n the series x x2 1 1 1 1 1 n 1 1 ( n21 ) n 1 x2 ( n 1)/ 2 y ( 0 Cn e dx x xe dx y e dy ) (3.7) n ! n ! n ! 2 2 n ! 2 2 ( n 1) converges 0 0 0
Read up on asymptotic series; this is covered in homework… Example: the exponential vs. the Gaussian e x I
2
1 x 2 12 x 4 ... , y ax ~1 1 1 0 e ax dx 0 e y dy (0 1) a
a
1 a
The main logarithmic approximation
5
Expanding the integral before integrating is actually bad: a mathematician would point out that the integral has a region where
x , so we have no right saying that the series-expansion converges. However: we have a Gaussian. We show the Gaussian causes the series to converge in (3.7).
(3.8)
This is very Example: consider,
I
I (a ) 0
e ax dx 1 x
2
(1 ax ...)dx 1 x
0
2
2
0
( ax ...) dx 1 x
2
2
x dx
1 x 0
2
2
0
dx x
2
ln |0
That’s trouble. But this series does converge. How to prove? Well, cut integral’s first term to being, a 1 a 1 1 1 x dx dy a I ( a) a a [ln(1 ) ln1]; ln1 0; 2 2 1 x 2 2 0 1 y 2 2 a 0
Now: consider
a 1
0
(3.9)
(3.10)
ca 1
dx 0 dx , which yields I (a ) 12 ( a ln(1 ca 1 )) . “Main logarithmic
approximation”: requires that not only a 1
1 but also ln a 1 1 . Then, ln(1 ca 1 ) ~ ln(ca1 ) , and, I ( a ) 12 ( a[ln c ln a ]) 12 ( a ln a ) ;
(3.11)
We sacrificed knowing the coefficient under the log. This is used in theory of superconduct ivity, where powers of a logarithm are,
g g 2 (ln T c1 ) g 3 (ln T c2 )2 ... ~ g g 2 ln T g 3 ln 2 T ...
g 1 g ln T
;
(3.12)
There is a lot of pysics contained in the condition for a series to converge. Theories which “have logs” are knwn as “renormalization group theories”. Bare interaction is then dressed. Watching for logs has become the favourite pastime of theorists; a log means the potential has a chanc e to be nontrivial. Added notes When expanding the integrand doesn’t work
On series-expansion of the integrand, you could wind up with a d ivergent term in the series that is supposed to be small, e.g.,
e ax dx 2 0 1 x
2 a 0
x dx 1 x2
, in which the term leading in a is divergent even so a 1 .
Method 1: you could use a cutoff procedure,
[1 O (ax)]dx 1 dx leading-order 0 0 I (0) 2 2 ax 1 x 1 x 2 term... e dx I I (a ) ; I ( a 1) 2 2 1 x 0 [1 ax O ]dx a x dx divergence from 0 1 x2 infinite-order terms 0 1 x2 2 Taylor series...
(3.13)
remaining integral is dominated by large x. non-analytic series in a: the closed-form integral is not a p ower series as cn a n , but rather cn f n ( a) , where f n f n (a ) does not have a series-expansion for small a.
solution: re-express the upper bound of the integral,
lim a 0 a 1 , but let a be close to zero, but finite, as a
cutoff,,
I
a 1
0 2
x dx 1 x 2
|a a 0
1/ a 2 1 2
0
dy
1 y
|a a 0
1 2
ln(1 a 2 ) |a a 0 ln
1 a
|a a 0 ln a I
a ln a 2
Ambiguity: what if we multiplied the upper limit by ca instead of a ? Turns out this ambiguity is immaterial; consider c c0 100 ~ 100 ; then I ( a 1) 2 a ln ca 2 a(ln a ln c) in which ln a ln c for a 0 .
Recapitulating our treatment of the integral,
Wednesday, February 6, 2013 BCS theory of type-I superconductivity
Convergence vs. divergence of integrals: integral in problem 1 that was
2
ln x x 0
J 0 ( x)dx . Joke problem: once you
see that it’s convergent, you get the integral of order-1. Some said it was divergent. That’s wrong. How do we know it’s not divergent? “Rule of thumb”: logs never change convergence or divergence of integral.
Consider diagram with (p, ) interacting with (p, ) , and emerging as (p, ) and (p, ) (respectively ). 6
You have the Hamiltonian with the energy gap Hˆ
g p ap ap
p 21m p 2 ap† ap g pp ap† a†p a p ap ; H BCS p 21m p 2ap† ap p ap†a †p *a p ap ; (3.14)
Approximate a p ap energy-gap
*
p ap ap h.c. , which is sort of a mean field approximation, and that’s where our
comes from. Replace ap 2
bp , bosons. Then, H BCS p pbp† bp , and the energies are
2
p2m F [ 2pm F ]2 2 . You get energy gap by self-consistency relation. The self7 consistency relation appears as ln( 0 / ) 2 I ( / T ) , in which T , the zero-temperature gap is modified by the split p
1
0 (0) e g and I ( y ) 0 (e x y 1) 1 2
2
dx x
2
y2
. Obviously, ln( 0 / ) 2 I ( / T ) is a transcendental
equation. Limit of large y: take out a factor of y, and expand the exponential, d ( x y ) 1 e y 12 x2 / y e y I ( y ) e dx 2 y ( x / y )2 1 y y ( x / y ) 1 e 1 0 0
2 y 2
2y
y e ;
(3.15)
You can clearly see in (3.15) that as y → 0, you have divergence; specifically, a logarithmic divergence as you get to the lower-end of the integral, x → 0 (e.g., lim
x 0
Meanwhile, use ln realize
6 7
0 0
2 y
e
( x ) 2
y 2 y 1 , which integrates to a log).
0
2 1 ln 1 ln(1 ) ln(1 ) O , and we can then get home free and
y
0 (1
0
0
0
0
0
2 T / 0 e
0
0 / T
0
0
).
It should be explicit by this respectiveness that there’s no spin-flip, for simplicity. The coupling constant g of th e BCS interaction was absorbed into the band-gap equation
Now: the aim is to find the critical temperature. Let the integrand of (3.15) be a function, f(x,y), so you have x 2 y2
f ( x, y ) ( e
1) 1
1 x
2
y2
T C ? . You have y / T (T ) / T C 1 , and,
You have (1) T
f ( x, y) ~
1 y
; f ( y x 1, y) ~
1
1 x
xe
1
; f ( x 1, y) ~
e x x
;
These three regions appear as, 0.25
0.20
0.15
0.10
0.05
1
2
3
4
5
y 2 12 ln( y 1 ) C I I 1 I 1
I
In which
I1
1
20
1
( x
2
y
2
tanh x2
Massaging, A 1 dx I 1 lim( 2 A 0 x 2 y 2
2 x
)dx; I
I1 ( 0
1
1 z
ze
1
1 2z
tanh 2x 2x
)dx
A
1 tanh x2 dx
20
)
Friday, February 8, 2013 Warm up: is the folowiung integral convergent or divergent?
x 0
3/ 2
sin x (ln x)2 dx
Well, at the upper limit infinity, the function’s x-3/2 kills the (ln x) 2 sin x . Also: the ln x function goes to –infinity faster at x → 0 faster than 1/x go es to –infinity.
(3.16)
0.6
0.4
0.2
2
4
6
8
10
0.2
Main logarithmic approximation, Q P0 ln Q ln Q0 ; ln Q0 ~ 1; Q ( P ) P0 ln Q0 Where is log-approximation relevant? Solid state physics; extremely relevant there. Example: Drude model of conductivity has ne 2 / m , in which is computed by the Boltzmann equat ion. at lowest temperatures, what’s main scattering mechanism of atoms? Impurities and defects. (Temperature-dependence o f resistivity is VERY important), T D 0 ln D 0 ln D 0 ln T ln ; T Example: frictional forces; they scale as t he log of the velocity, f fr f 0 ln(v / v0 ) ; this is the friction a series of
OOOOOO atoms exerts on a V point, pointing downward. Example: fluid dyamics, if you have a flow across a surface with friction, you have an x component and y component of velocities, u x and u y respectively. You have u x y0 ln( y / y0 ) .
In theory, all these dependencies are obtained using MLA, or by slugging through some pretty heavy formalism. Cutting stuff off: an integeral-functio n of a t hat has an integrand that goes to 1/x at the lower limit of 0 requieres treatment by cuto ff, a0 1 trouble at a I I (a) f (ax) g (x )dx ( O 2 (x ))dx ~ ln 0 0 a x a lower limit
How to effect a cutoff treatment: consider, a 1
I (a)
e
1 x
ax2 1 vs. ax2 vs. x / a
ax2 2
dx
0
0
1 ax 2 ... 1 x2
dx ~
2
a
a a
x 2dx 1 x2
2
/ a
a
a
dx
2
a
This integral saturates to the value 1 as x infinity. That’s where the divergence comes from. dI/da: When you have exponentials under the integral I
computing
dI da
0
dI da
. You can have,
x 2e ax
2
1 x2
a1
dx
0
x2 e ax
2
1 x2
dx ??
I (a) , it’s a good idea to “play” with the integral by
Well, consider x → infinity. That means the function
x 2 1 x 2
~ 1 . Meanwhile, the Gaussian falls off as 1/ a . The
Gaussian is a “slow” function”. Nothing happens to the Gaussian (it is ~1) until x 1 / a . So basically, you are breaking up the integral into x [0, ] [0, 1a ] [ 1a , ] ; e.g., breaking up the interval in a manner entirely around the properties of the Gaussian. BUT WAIT: that’s not all we could possibly do. Use y 2
dI da
1 1 a a
2
ye
1 0
y 2 2
y a
dx ~
dI 0
da
1
ax2 , and the above intervals are mapped to, a
a y 1
e
y 2
I0 (
dy I
1 a
e
y2
dy )da
y 1
Closing question: what happens when we co nsider a ≫ 1 in the above integral? The Gaussian is much more like 2 1 dI a spike/Dirac delta which samples the function 1 x x 2 close to the origin, at x = 0. Then, da ( a ) 2 (1 2 a ) , and we see this goes to 0 as a → infinity. (this is an interpolative function).
Monday, February 11, 2013
Integrals with oscillatory term inside, times some envelope function. You get these integrals when you have Fourier transforms, I1 ( )
f ( x ) cos x dx; I 2 ( )
0
f ( x ) sin x dx; ;
0
Suppose f ( x ) e x . Then you could have,
I 1 ( ) cos x x Re i x x Re x (1i ) I ( ) dx sin x e dx Im e e dx Im e I ( ) 2 0 0 0 Re 0 1 Re 1 Re 1 i 1 1 2 2 2 ~ 1 ; Im Im Im i i (1 ) 1 1 1 Consider plot of functions e x cos x and e x sin x . The number of oscillations the function undergoes is illdefined for infinite bounds of integration. Integration by parts: consider cosine integral, and integrate by parts,
x
0
x
0
e
sin x dx d
e cos x dx
0
d 1 e x sin x 1
0
1 x
e
0
sin x (e x )dx
cos x ??
1
0
d ( 12 e x cos x ) ... 12 ...
(3.17)
...
And continuing, let f ( x) 11 x 2 ,
I
0
cos x 1 x 2
dx d ( 0
1 sin x 1 x2
)
1
0
sin x
d
(
1
dx 1 x
doesn't have power-law 1 ; 1 x2 ... 2 series... 1 x
)dx 2
Instead, consider the following contour integral with two poles x = + i, as.
1
cos x
dx
1
Re
ei x dx
2 2 1 x 2 2 1 x You put in a non-analytic exponential function,
1 2
Re
e 2 i 2i
2
e
2
exp[
1
]
1/
(3.18)
I
1
e 2
x
cos x dx
The
1 1 x 2
is analytic over the entire real axis. That means real a xis is free of singularities.that means we can do
integral by contour integral. Cauchy theorem just then says that value of integral determined by singularities (specifically, the abveoemntioned x i ), as, As long as you substitute complex umber into oscillatory function, they become exponential functions. Physical example: consider a charge q in 3D. laplace’s equation is
2 4 4 (q 3 (r ) e i ) . Charge
density of ions vs. electrons. Born approximation: ions are rigidly held in place. Then, charge density of electrons is function of position, e e (r ) , resulting in a complicated and nonlinear PDE. But what if charge was small enough to only slightly perturb the background charge? Well, chemical potential governs “how much energy youintroduce per unit charge”. Chemical potential is, (r) F (r ) e (r ) F (r ) e (r ) e e (r ) e ( F (r )) e ( e(r ))
n e e 2 N ( );
series expand e e ( ) e2
n
0
N ( )d ;
n N 0 ( ); e 2 N ( ) e i ;
So then we have Poisson equation that is a sum of a Green function attackable part, with other one, 2 4 4 (q 3 (r ) e2 N ( ) )
Apply Fourier transform, letting image function be (k ) e ik r (r ) d 3 r . We have 2 ( k 4 e N ( )) 4 q 4 q / (k ) , in which we have screening wavevector. Inverting the Fourier transform, 2
(r )
2
2
d 3k
(2 )
3
e
(r)
k 2 dk
(2 )
3
1
2
e
ikr cos
d (cos ) (
1
4 q e ikr cos
1
0
1
In which:
i k r
1 ikr
)( e
2
ikr
k 2 2
eikr ) 2
d (cos );
sin kr ; kr
So we finally have, 4 q 2 1 sin kr 1 screening radius ; r 1k (r ) k dk ; 3 2 2 (2 ) r 0 k
4 e2 N ( ) , the Thomas-Fermi
Now: you already know the answer: you get a Yukawa potential. Note: x sin x . You are able to analytically continue this ito the complex plane,
(r)
1
1 k sin kr dk
2 r 2
k
2
2
e r ;
(r )
q r
e r ;
You see that poles are complex. 1) look at integrand on real axis. (2) look for points of non-analyticity on rea l axis. If there are, can’t go to complex plane. (3) Poles in complex plane substituted into
Use the Cauchy theorem exponential. No Cauchy theorem/non-analyticity power law. Example: consider plane of charge in 3D, and consider z-axis going through normal. you have log behavior. Go back to the Poisson equation, and say 2 4 4 (q 3 (r ) e 2 N ( ) ( z) ) , in which n3 D n2 D ( z ) . The
delta function sends the z-argment to 0, and you have homogeneous equation everywhere else. You basically have ( z ) (r ) ( z ) (z , r ) ( z ) (0, r ) . We must have,
dz (0, r )e
ik z z ik r
e
(0, r )e
2
( z)d r
ik r
2 4 q 4 e N ( )0 ( k ) ( z)d r 0 ( k ); ( k , k z ) ; k z 2 k 2 2
Fourier transform,
0 (k , z ) (k , k z )e 0 (k )
ik z z dk z
2
(k ,0)
2 4 q 4 e N ( )0 ( k ) dk z
k z 2 2
2
0 (k ) 1 2 e k N ( ) 2
z 0
dk z (k , k z ); 2
dk 4 q 4 e N ( ) (k ) 2 k 2
z
0
2 q
2 q 2 q 0 (k ) 1 1 k k 2 D k 2 D
You then have the potential,
q / r ; r 2 D 1; (r ) 3 p / r ; r 2 D 1; Yukawa potential in 2D is a d ipole potential. (interesting). Very important qualitative result. Once a Fourier transform has an absolute value of something, look for power laws. Notice in 3D, you had denominator of 2 2 1 1 k 2 D , wheras in 2D, you have k 2 D .
Wednesday, February 13, 2013
Last time, we thought about screening radius of electronic impurity in charge jellium, ( k )
2 q k 2 D
;
r1 ( r ) 3 r
2 D r 1
; 2 D r 1
Try inverse-transofrming, you hit a snag, 1
F [ ( k )]
2 qeik r d 2 k
k
2 D
(2 )2
q J 0 ( x) x dx r
0
x x0
2
q 2
i kr cos
k e
0 0
dk d
k 2 D
q 2
k dk
2
k 0
e
2D 0
q
I ( x0 ) r
So now, we have to do an asymptotic expansion on the function I ( x0 ) as,
I ( a )
0
J 0 ( p) p dp p a
J1 ( p) p a
|0
0
J1 ( p) dp ( p a) 2
...???
i kr cos
d
q 2
0
k dk 2 J0 ( k r ) k 2 D
“Feynman trick”,
Let: I (a)
pJ ( p) dp e
( p a ) S
0
0
aS
dS e
0
pS
dS J0 ( p) e
0
0
2
S (1 S
S
Se aS dS
(1 S )
2 3/2
p dp
0
)
3/2
1
a
a
Hard cutoff of function, becomes non-analytic, e.g.,
2
1
aS
(1 s )
2 3/ 2
dS ;
0
, from table
1
1 2
e
2
;
f ( k )e ikx dk . Example where this comes up is friedel
oscillations. You have 1D Fermi gas. Finite Fermi energy. Consider solution of single particle SE for o ne electron. Free motion, and then hard wall. Wavefunction should vanish at wa ll, and so the wavefunction should be a sine-function. Look at probability corresponding to this sine function, 2
Mathematica: plot sine vs. sin …
Far away from boundary, you have no trace of the boundary’s effect (free planewaves). Near boundary, you hve sine function. What happens inbetween? Well, look at electron density, dk n f k k ( x) ; f k [equilibrium or non-equilibrium distribution function] 2
Difference between Fermi vs. Boltzmann statistics is felt here. Which wavelength makes w hat contribution to the density? How does wave-interference influence the number density? Etc. The function f k is a box for zero temperatures, so integrate o ver the box only, k F
steady- oscillatory A 2 sin kx dk 2 A n k ( x) dk 2 A part dk state part 2 2 k 0 0 0 sin 2 kF x Normalization of wavefunction vs. 2 A2 sin 2k F x 2 k A k 1 ; F F 2 2 x k F x normalization of e- density A2 k F n 2
dk
kF
2
kF
2
2
1 cos 2kx
k F
2
F
So then you have oscillatory electron density n (x ) n (1 sinc2k F x ) . You can then plot this vs. k Fx and see that both boundary conditions are satisfied. In 2D, this is n( x) 1r sinc 2 k F r and in 3D we have n( x )
1 r 2
sinc 2k F r .
Important effect, “Altshuler-Aronov effect”. Luttinger liquid: 1D wires, quantum hall (integer and fractional), etc. work with system of single-particledescription of electrons.single particle picture. In 2D, more or less ew’re home free and we have simple picture of single particle motion. Fermi liquid theory. FLT sa ys electrons move, and they have free particle motion with renormalized parameters. But in 1D system, you do n’t have this luxury. You can’t just renormalize parameters in 1D. either your way is obstructed, or it isn’t. collective solids take on much diferent para meters. Going from 2D to 1D is going from a traffic jam with alternate routes, and then shutting off alternate ro utes all of a sudden. Potential: study scattering problem in 1D system upont his pot ential,
V ( x ) V0 ( x x) n( x ) dx T (
F ) scattering amplitude ln F
;
Example: liquid helium 3, let it cool down to temperature where it flows without dissipation. BCS t heory for motion.
Liquid he3 is a p-wave superconductor. Most other things are d-wave superconductor. Unconventional superconductivity; we do not suspect pho nons as main culprit of ?? Simple story of Friedel oscillations has far-reaching implications. Can put piece o f copper, place atom on top, and look at distribution of number density about the system.
Friday, February 15, 2013
Physics of the mott problem; we will introduce met hod of steepst descent. Consider mott problem: a bunch o f square wells randomly arranged in space. Ee ach well has 1 bound state. The bound states are at random energy. Question: how do particles move from well to well? Partial anwer: they hop, as quantum particles do. Mechanism: tunneling. The tunneling happens over known distance. Typcal electron will tunnel to potential well closest to it in space. For simplicity, let the electron concentration be low: ne / nwells 1, so we can use Boltzmann statistics, not Fermi/Dirac statistics.
Once you tunnel: you are in enw energy level. New energy levels don’t match. Energy level mismatch betgween old and new positions has to be absorbed. So: let’s say wells couple to infinite heat bath t hat it can borrow energy from. But large energy mismatch means transition-frequency is limited by both (1) space/distance and (2) energy-disparity. So, there must be compromise between these two factors in constituting a hopp ing frequency. Conductivity of this medium is product of two exponentials. Let typical distance between wells be R , and let 8 the extent of localization of the wavefunction be . Meanwhile, in energy-space, let there be energy mis-match of , and let the probability be Boltzmann of this energy scale. then… R exp exp k BT Process of tunneling needs to borrow the energy from the thermal bath. Energy difference large means low probability of hopping. Let there be independence between R and . Then the only way to optimize conductivity is to make R = 0. In real life, you have ( R ) . There is a relation between the energy gap and t hecharacteristic spacing R,
8
Colloquially: the “size” of the wavefunction. For a point-like wavefunction, you have 0 , and no hopping.
N
# of energy levels
energy available
Volume
1 NR
3
exp(
R
1
) exp[ f ] e f ( R ) NR k BT 3
Want to minimize this function, f
R
1 NR k B T
f ( R0 )
R0
1
3
4
f ( R) f
3
1 3
NR0 k B T
4
3
NR0 kB T 1
NkB T
N
(4
0 R04
Nk BT 3
)3
1 kB T
4
3 N kB T T 0 T
R0 4
3
; NkB T
(T ) e
4
T /T 0
;
This is mechanism of conduction in amorphous so lids! Traps and electrons. Only way to go between traps is to hop. Gamma function: is tautology of factorial function. Now: let’s find out where stirling’s approximation comes from using method of steepest descent
( p 1) 0 x e dx 0 e p
x
p ln x x
dx
0
e
f ( x, p )
p 1
dx
???
Plot the function f ( x ) p ln x x Steepest descent: approximate function by Taylor series. Replace f in exponential, noticing that you are at an extreme value, p p 1 p f ( x) p ln x x f (x 0 ) 1 0; f (x0 ) 2 2 0 x0 x0 p p (3.19)
f ( x) f ( x0 ) 12 f ( x0 )(x x0 ) 2 O3 ; So then, p ln p p
( p 1) e
1 ( x x0 ) 2 2 p
0
e
y x x0
dx
1 y2 2p
e p ln p p 2 0 e
dy e
p ln p p
2 2p
2
e p ln p p 2 p p p e p 2 p (3.20)
You can clearly see stiring’s approximation if you take the log of both sides of (3.20). Assumptions: we assumed integral came from narrow region about maximum. We see that
x x0 x0
~
p p
~
1 p
x x0 y ~
p
1/2
; p 1 p
p; z
y 2p
~ 1
1
2p
2
e z dz; y ~
p;
0
Steepest decent for complex integrals: suppose you have integral of a smooth function g ( z ) in complex plane. Let it be a contour between points z1 and z2. Let f(z) and g(z) be analytic functions. (note: this excludes
the innocent-looking
2 1 z ). Then,
I
I ( z1 , z2 ) [ z ,z ] g ( z )e f ( z ) dz; f ( z) | z z 0; 1
0
2
f ( x, y) u( x, y) i v( x, y);
f ( z1 ) f ( z )
f ( z); z1 z Recalling a bit of complex analysis, thining abo ut derivatives, z x f xu i x v; z i y; 2u 0 2 v;
f
lim
z1 z
u ( x, y dy) u( x, y) i v( x, y dy) i v( x, y) idy
y v i yu;
xu y v x 2 u x y v; y u x v y 2 u y x v 2 u 0 2 v Electrostatics: Earnshaw’s theorem! Can’t form a st able system. 1 u ( z0 ) u d 2 R
That is the ultimate freedom of complex analysis. Idea: deform contour such that (1) it passes through saddle point and (2) in direction of steepest decent. Then integral becomes integral over real line. Do same thing as with gama function: expand thunctionnear saddle point. And do Gaussian integral for small deviations. Example: Bessel, or any cylindrical function. All are given by contour-integral representation. Integral is from 0 to pi band in complex plane. Trick: textbook has particular choice of axis. He’d ask you to draw contour of integration. But deformed contour is equally valid. Bessel function: saddle point on this interval jut happens to be Pi/2.
All large-value formulas from mathematical physics, from Bessel functions to whatever, Legendre p olynomials, hermite polynomials, they all come from the same interval.
Monday, February 18, 2013
Gotta use tim’s notes…
Wednesday, February 20, 2013
Consider sums. Diamagnetism of 2D electron gas. Yo u have k BT F and C
F . You have C eB and c
C (n 12 ) . Since you have A B , the Hamiltonian decouples in this uniform vector potential, P ec A . You have A must be lienarin the potential. On squaring the ( P ec A ) 2 , you have a term the spectrum n
2
that is proportional to A , an this one is … simple. The cross term is an anticommutator wh ich is a shift (?). recall: p ( / i) , and you have eip x/ being the generator of translations (?) by x . Free energy: compute intensive quantity free energy per unit area; this appears as, F A
F 2 B
2
k BT 2 B
2
ln(1 e n 0
n
); B
c ; n n ; eB
Sum over positions of circles, in which you have localized states. You have to estimate the sum.
(3.21)
Discrete energy levels at fixed chemical potential. Change field, stretch the levels. When you stretch levels, some “pass through” the chemical potential. If you measure M B F , you will see oscillations in the oscillations. Anytime the chem potential crosses a level, you ha ve some undulation, but the next level is equivalent (note: n 1 n C ), and you get oscillation. Lab people: put graphene in magnetic field. Measure current. Current starts to oscillate, because the resistance is oscillating as a function o f the field. Effect when a quantity oscillates as a funct ion of the field: de Haas/von-Alphen oscilltaions. (See Ashcroft and Mermin). Tells you the shape of the Fermi surface. This is the reason why we built the high field lab in talhassee, FL: we put materials in a magnetic field and we watch for oscilations. That’s the purpose. To see good oscillations, we need strong magnetic fields. Two things (1) estimate sum and (2) expect sum to be oscillatory function of B. (not e: B-dependence is buried inside of the n n ( C ) and C C ( B ) . Look at, 1
n
n
2
1
coth 3.15 ~ 1
Rules for sum-convergence is same as rules for integral-convergence, because integral is a sum.
n
n
1 2
a2
Coth[a ] a
1
2
a
3
2
4a 2 45
n
O[a] ; 3
dn 2
a2
a
;
Note that in the integral, dependence on coth[a] is lost. In general, f n 1 f n f n example: y
n a
y ~1 n ~ a f
a 1
n
n
1 2
a2
f (n ) 2n
(n
1
2
a
a2 1
2
f (n )
1;
2
a
2 2
)
2 a2
4
1; ~
a a
4
...
1 a2
1 3
a
; n ~ a;
f ( n ) f ( n)
1
1
2
a2
2 ... ~
~
1 a
;
But replacing sumby integral may kill some important physics: e.g. , quantum effects. In that regard, you can’t do the integral-approximation. Poisson formula: wonderful invention. consider identity where you su m infinite number of delta functions. This formula appears like our formulae for densities of states,
( x n) e
n
2 ikx
; n ;
n
Exercise: compute, m
m
[ ( x n)]dx ?; [ e
2 ikx
m n
Anyway, consider e2 ikm
m n
e2 i 1
] dx ?;
Poisson formula says that the sum of the function is equal to the sum of Fourier transforms. K plays the role of the wavenumber.
x m
f (x) ( x n)dx f (x )[ e
]dx;
k
2 ikx
f (n) f (x)e
n
2 ikx
dx
k
Thing you gained: it is often easier to find the asmptoi value of the sm of transforms, rather than the original sum itself. In the original example, 1 1 f ( n) 2 ; f (x) 2 2 n a x a2 Fourier transform of this function,
e2 ikx dx
x
k
2
a2
x
dx
a2
2
2 k 1
cos2 kx dx x2 a 2
2Re a
k 1
e2 ikx dx
x2 a 2
Now do contour integration: you have contour along real axis. K is positive. Therefore, make integrand vanish when circle of pole goes to infinity. Carefully choose the orientation of each contour so that the exponential 2 ka vanishes at infinity. residues are I 2 i e 2 ia ( / a )e 2 ka . Therefore,
Re
e2 ikx dx
x k 1
2
a2
e a
2 ka
k 1
e2 ikx dx
x
k
2
a2
a
2 a
e
2 a
1 coth a ... 1 2 2 a a 1 e a 2
In the opposite limit, when we go to 1/a, you get behavior that goes as 1/a . We got both of these limits in our estimation! The icing on the cake is when we plug in a = 1, and get what we got before… Returning to the sum over landau levels:
S
ln(1 e ( ) ) n
n0
Problem: sum is asymmetric. So: consider a limit: C
F . Large population between occupied levels.
Occupy large number of levels semiclassical approximation regime. This allows us to use a t rick on this semi-infinite sum. Go back to the Poisson-formula,
n
( x n )
k
e 2 ikx . Multiply both sides by f(x)
and integrate over x from x [ a, ] . Imagine sum of delta functions vs. x (e.g, the “Dirac comb” you t reated in the very first solid state physics problem last semester). Ignore the interval below x = a,
f (n) e n0
2 ikx
f ( x) dx
k a
Homework questions: integrate by parts on problem 2a. / a 1 a 2 d 1 x 1 1 1 a I ( a ) dx 2 2 2 2 2 1 x a 1 x a 1 a 0 0/ a a
rescale the variable;
Friday, February 22, 2013
( 0
d 2
1)( 1a )
Electrons in quantized magnetic fields. You have n
C (n 12 ) . Add kinetic energy of z-axis motion is just
constant shift tohamiltonian, so not interesting physics here. Regime where po pulated by landau-levels: ratio ofchemical potentials to spacing between landau levels is large,
C
1 . Also consider k BT and
k BT C
being
arbitrary. Let free energy be, and we have the su
F
F0 ln(1 e( )/ k T ); n
F 0
B
n 0
k BT 2 B
There is a cut to confine the sum to be finite. The exponential of n
;
2
(3.22)
0 very rapidly becomes small. That
invites, not a step-function treatment, but rather a Po isson formula treatment,
a
k 1 a
f (n) f ( x) dx 2 f ( x) cos(2 kx) dx F F
c
n 0
Foscl ; 1 a 0;
(3.23)
Problem difficult to formula mathematically: not interested in leading/largest term, but rather oscillatory term. Each time you cross chemical potential, you have oscilation. Experiment: manifestation of magnetization. We don’t know how mathematicians split into oscillatory vs. non-o scillatory. So how do we split? First term: although large, is non-oscillatory, on physical grounds. This co ntribution to free energy just is a constant shift to F: it is likefree energy of gas in absence of magnetic field, and we’re not interested in that. Argument: integral term is actually classical contribution to freenergy which neglects quant ization due to landau levels. Sum keeps discrete nature of spectrum. Sum consists of two parts as indicated in (3.23). first term is a smearing out term (?). if you were asked to find free energy of classical gas in B field, you should see first term. Bohr-von Lewween theorem: “one o f the few useful theorems” in theoretical physics. There is no diamagnetism in classical systems. (Actually: there is no magnetism in a classical system, period). Magnetism is a pure ly quantum/relativistic effect. Any induced magnetism is due to an external field. Spin is “hopelessly excluded” by classical mechanics; it is purely relativistic. The effect of the spin is due to an expansion in 1/c, where c is the speed of light. The leading ter m is the Zeeman term. If you didn’t have the Dirac equation, we would have no reason to even suspect spin existed if it weren’t for experiment (e.g., the structure o f the periodic table). Hamiltonian for such a system, (p ce A) 2 H ( p, r ) 2m
2 2m
; Z
e
H ( p ,r
) /( k BT )
H ( p ,r )/(k BT )
d rd p e 3
3
d r d ; 3
3
(3.24)
Now: in quantum mechanics, position and momentum are non-commuting operators. However, in classical mechanics, position and momentum do comute, and p p ec A and r r r is just a canonical transform. One then integrates over canonical variables, and you “cannot tell about magnetism”, period . Magnetic effects just enter in as a shift in momentu m, and that’s just a Galilean transform which measurement cannot detect. Therefore: magnetism is a purely quantum and/or relativistic effect. Hence…the first term in (3.23) does not affect the energy levels! The second term is the o scillatory part of trhe kx free energy! Integrate by parts using cos(2 kx ) dx d ( sin2 ) , and you get, 2 k
Fosc
F0 ln(1 e k 1 a
( x )/ kBT
sin 2 kx e ( )/ k T C ) cos(2 kx)dx F0 d dx ( )/ k T 2 k 1 e k T k 1 B a x
B
x
B
Introduce dimensionless energy y ( x ) / kBT ; this is the measurement of the width of the energy distribution where we have occupied states. We have
(3.25)
x
y
k BT
C (a 12 ) kBT
~
C k BT
k BT
we lose information about "a", and (3.26) consequently about few lowest landau-levels
You are losing information about 0, 12 C , 23 C ,... . These levels play no role; they are too deep. So: in the integral (3.25), we have y, Fosc We have
2 C
k BT C
C k BT
F0
Im[ e k 1
2 ik /(2 C )
e
2 iky
e y 1 ey
dy]
F0 Im[ e
2 ik / (2 C )
k 1
e2 iky
e
y
1
dy] ;
(3.27)
1 . Quantum mechanics is lost in two ways: e ither (1) you heat the system up (you still have
kBT ≪ mu, but you spread out the thermal spread such that you encompass more landau levels; window becomes large, and you lose oscillations, and you do so in an exponential way; see the pole ), or you (2) decrease the field. At zero temperature, a weak field going to 0 still makes you lose oscillations in the free energy. For reference, 1 Tesla produces 1 k BT of energy/frequency (think hbar = 1). Caution: midterm on Wednesday. Review session on Mo nday at 6 PM.
2 k 2 Ry 2m 2
E
8 k 2kea0 4 3
a0
3
6
ke a0
2
Cosine is analytic over the real. Problem number 2: Taylor series expansion! 1 series expansion... x3 1 Problem number 3: 2 y dy I1 ( 1) J 0 ( y ) 2 2 2 0 y
Must put 0 . Two ways to do this (1) strike out delta2 in denominator. BUT…subsequently integrate by 2 parts and evaluate J0(y) at a small argument, approximating it as 1. Upper limit of integration can be kept as infinity! OR…less formal way (2) you could strike out the upper limit of integration and replace it with 1. Ok to “destroy information’about delta. Going to put de lta in lower limit anyway. Two ways to proceed.
Monday, March 11, 2013 – “A Robinson Curusoe’s approach to special functions” Final stretch to end of semester. APS meeting will interfere. Class will be canceled on Wednesday, march 20, 2013. First make up class shall be Wednesday, march 13, 2013, 6 PM… Current topic: differential equations. Estimate regions of relevant behavior, matching boundar y conditions. Consider problem from QM: you have hard-disk of radius a in plane. Potential well. Electrons move in the plane, so k k x xˆ k y yˆ . You have potential U ( r a ) U 0 and U (r a ) 0 . Get fundamental solutions of Schrödinger equation Bessel functions. But…pretend you don’t know anything about Bessel functions.
1 k
Simplifying assumptions: s-wave scattering. That means
~ a (e.g., long-wavelength scattering). We also
assume isotropic scattering. Height of potential well is U0; let it be that U 0
E 2mk . Small but very-high potential-barrier. Another note: 2 2
consider thin, high potential; knee-jerk reaction is to model it as a delta function! Legitimate re placement, but you’ll never be able to investigate physics of finite radius a, or barrier-height U0. Isotropic scattering: wavefunction only depends on radial distance. Two Schrödinger equations: one inside disk, one outside disk, 2 ( r a); ( r a); 2m 2
U E ;
r 1 2 k 2 ; r 1 0 k 2 ; Now let’s get scattering cross section of the disk,
2
2mU0 / 2 ;
k
2
2 mE / 2 ;
(3.28)
( , a,U 0 ) . Simplification: k inside the disk. Each
nd
of these is a 2 order differential equation. pretend we don’t know what Bessel functions are. Get properties of solutions. Decide which fundamental solutions to keep. 1 1 2 {1,2} 2 {1,2} (3.29) r k ; r ;
( r ) a , and let x r , x 1 ; a( a 1) xa 2 2 axa 1 xa ; axa 1 xa ; a( a 1) xa 2 2 axa 1 xa axa 2 xa 1 xa
Inconvenience: having both 1st and 2nd derivatives. So introduce ansatz
a 1/2
x 3/ 2 x 1/ 2 21 x 3/ 2
1 1 2
2
x2 ( x2
x1/ 2
14 )
(3.30)
Thus, the ansatz helped us make this into something easy to integrate,
x ;
12 ;
x;
x a /
x
x 1;
x/
(3.31)
We arrived at 1 for the solution if we neglected the x2 on the RHS of (3.30). Now, rewrite the derivative and the factors as an operator,
x 2
x
2
14 x 2
0 1 x
x 2
2 x
14
x
x 2
x
x 2
2 x
14 x5/ 2 x 2 x5/ 2 0 (3.32)
Perturbation theory: must discard terms of higher order in a consistent manner. Use power-law ansatz Ax . That gives you, 5/2; A 14 1 5/ 2 5/ 2 A ( 1) x 14 Ax x x 14 x (3.33) ( x 1) 1 14 x2 ... ; x
This gives you solution behaiour near the origin. Other solution: should not behave well. Well-behavedness at the origin is orthogonal to divergence at orgin. You need a log-function for that, 1 1 1 (2) (2) (1) (2 ) (3.34) x ~ x12 x x ~ ln x (???); ( x ) ~ ln x; A1 A2 ; Let y kr , and write,
( y 1 ) ; Our solution is then,
x ;
1 2
4 y1
2
0;
(3.35)
A sin y B cos y
C sin( y ) C (sin y cos cos y sin ) C (sin y cot cos y ) C y (sin y cot cos y )
Now, play the matching game, S ( x);
G ( y ) cot F ( y ) | y kr ;
(3.36)
|r a |r a ; |r a |r a ;
(3.37)
Turn this into a matching game of their log-derivatives. This is convenient, because normalization as an unknown drops out, and we can focus on finding the unknown phase shift, S ( a) G (ka ) cot F (ka ) case-1 ka 1; a 1; case-3 a 1; ; k case-2 a 1; case-4 a ~ 1; S ( a ) G (ka ) cot F (ka )
(3.38)
The radius of the disk enters everywhere. Thus, we definitely sacrifice pysics by using a de lta function.
Wednesday, March 13, 2013
Summarily, we had, for dimensionless 2 and y
2mU 0 / 2 and k 2 2mE / 2 , and variables x r S ( x) C ( x)
kr , the eigenfunctions,
1 14 x 2 ... r ; ln x 1
r 1
x0
a
2
2 mU 0 / a ~
k 2
1
cot sin y cos y ; (3.39) y Wavefunction should be localized in wavefunction of the disk, but there is zero-point motion happening inside the disk, 2
2
ma U 0 /
(3.40)
Mapping: don’t want to solve the problem again. Compare two solutions inside and o utside the disk. Infer how solution of outside behaves in same region. Not e the two equations in (3.39): they differ only in sign. Map one equation onto another by switching two dimesionless distances. Consider replacement: r ir . Both the
r 1 drd change sign under this switch. What happens whe n we effect nd this switch in (3.39)? map the 2 expansion (i x ) ~ 1 14 x 2 , and (i x ) ~ ln ix ln x ln i ~ ln x . operators
d2 dr 2
i 2 drd drd and 2
2
2
2
r
1 d
dr
Now, note also that the transform x x leaves the differential equation f x 1 f f unchanged. This differential equation is also homogeneous, which means you can multiply through by a constant and leave the solution unchanged. Multiply through by i , x 1 2a 2a 1 ax 2 a 41 ; (3.41) 2 1 1 2 0 ax 1 4 x ; 2a; x 2a ; Go back to (3.39) and concern ourselves with the phase-shifts, and get, cot 2ka ka1 0 ka1 S S ( a) cot G( ka ) F ( ka) | a |a k ~k ~ k S S ( a) cot G( ka) G( ka) cot 1 ln ka cot ln ka
cot ln ka
S ( a) aS ( a)
~ ln ka 1; cot
1
big 0 cot(dx)
1 dx
O;
1 ln ka1
(3.42)
1 ;
This is actually a very weird solution. Recall this is 2D scattering off a disk, not a sphere. In 3D, you have a phase shift which is also small, but it is small as a power-law, ka 1 . The log-divergence is an oddity of 2D quantum mechanics. What’s 3D scattering cross section? Must estimate! Consider the wavenumber and phase-shift: t hen, (k , ) . Cross section and phase shift tell you everything. The k is the initial incident energy, and the phase shift is the change due to the interaction. 1
In two dimensions: note that the units of are L expression is very weird.
4k 1 sin 2 . Know that k
L21 L D 1 . So,
1 2
1
~ k ~ k / (ln
1 2 ka
) . The exact
E . So, small particle becomes large. 2D quantum mechanics is
Example: graphene. If you measure the drudian collision-frequency 1/ vs. electron-number density n, and you imagine a renormalized-Drude-conductivity ne2 / m* , you get 1 quantum mechanics is bizzare, but real and observable.
1 (n) ~
2 n / ln
n . Awesome. 2D
Opposite limit: consider a 1 ; we need S ( x 1) . This is like making our disk-radius really huge compared to the wavelength. Use mapping again or you could start from scratch. Recourse to the original differential equation 1 x 2 (3.43) S x S S ~ S 0 S e | x1 ; S 1 14 x ... | x 1 ; Let’s imagine that S is a power-series, S
n 0 a2 n x 2n . Put this back into the first equation S x 1S S of
(3.43), and we get,
a (2 n 2)(2 n 1) x 2 n n 0 2n 2
1
n 0 a2n x 2 n a2n 2 (2n 2)(2n 1) a2n a2n 22 n (n !)2 ;
(3.44)
Opposite limit: you could have tiny disk, a 1 , so S ( x ) 1 14 x 2 , and S ( a ) 1 . Youthen give yourself the derivative S ( a) 12 ( a) , and you get,
cot ln
1 a
2 ( a) 2
1; S
two regimes: (ka ~ 1) ~
1 ln ka1 ( 2a ) 2
( a ) k
; ~
1 k
2
; ( ka 1) ~
ln ka1
2 ( a ) 2
2
; (3.45)
ln
2 1 ka
k
Thomas Fermi model: should be part of standard graduate QM. You say “how big is the atom?”, f (r / a B ) Getting the Bohr radius without heavy formalism: should scale as t he Bohr radius, a B
(3.46)
me , ratio of quantum2
2
force to Coulomb-force. You have [e 2 / a B ] J m / m J . Zero point moton kinetic energy is T0 ~ ( / a B ) 2 / m ; then, U ~ e 2 / a and T U a ~ 2 / ( me 2 ) 0.522 A . 0
B
0
0
B
Now consider Z-proton atom with one electron. It must be a much-smaller atom. You have a B ~ 0.525 A / Z . Watch out: you don’t have any dimensional control over the pure-number Z. ( z ) (3.47) a B aB f ( z ); Z 1;
Many nodes: for large-Z, you have many nodes in the wavefunction. That means you’re far away from the zero point energy. That means you can actually use semiclassical approximation. So: consider orbiting electrons as a Fermi gas. The gas species are near ly-free particles, 3 4 N 2deg 3 k F ( r ) ; T 0 E E F ( r ); n( r) V (2 )3 (3.48) 1/3 2/3 1 1 k F ( r ) 3 2 n( r ) ; F kF 2 3 2 n( r) ; 2 2 Chemical potential, (3.49) F ( r ) (r ) 0 const 0 F (r ) (r ); Now, Poisson equation is 2 4 n Z (3) (r ) . Zeroth order solution (r 0) ( r ) Z / r . You have the density related to potential in a nonlinear way,
2
8 2 3
3/ 2
Z (3) (r );
(3.50)
Um,
r 2 c 3/ 2 ;
|r 0 ~ Z / r ; |r 0;
(3.51)
Weak localization: If you shine a laser at a gas-cloud, the probability of backscatt ering is four times the classical value. This was discovered in the 60s (when the laser was invented), but it was not paid attention to. Shine light on a medium, and the medium, itself, seems to “trap” the light. 5
In a seemingly-separate vein: they observed the Bloch resistivity vs. temperature, T law. At low temperatures, 5 resistivity deviated from this T law in thin metallic films, and the deviation was extremely-strong in 1D systems. (Iron whiskers: pretty much one-dimensional). Essentially, you have, P AB
C1 C2
2
C1
2
2
C2 2 Re C1C2*
C1
2
2
C 2 2 C1
C 2 cos(1 2 )
(3.52)
And, P ~ (Dt )3/ 2 ; R ~ 1
Dt ;
A ~ 2
Dt; V ~ 3 (D t )3/ 2 ;
(3.53)
Probability is divergent: T T 2 2 v dt v dt divergent in certain v dt T v dt P3 D ; P2 D ln ; P1D T ; (3.54) 3/ 2 3/ 2 ( Dt ) ( t ) dimensionalities t D D t D 0
VERY successful theory in solid state physics, even more so than BCS theory.
Monday, march 18, 2013
You missed first 20 minutes. Consider continuous symmetries in classical mechanics: symmetry under (1) translation in space (2) translation in time (3) rotation (4) gauge invariance. Let’s look at gauge invariance. You have, A A A ;
f ;
(3.55)
Four vectors, x
(ct , r ) x ( ct, r ); j ( c , j); A A A x ;
(3.56)
Now, think about Lagrangian densities, L
L0 Lint L d 3 x;
1c
j A 1c ( j A);
Lint
(3.57)
Subject the interaction-part of the Lagrangian to a gauge transform, Lint
Lint Lint 1c ( x ) j Lint 1c x (j ) x j ;
(3.58)
Electrodynamics of superconductors: natural consequence o f gauge invariance. Wavefunctions are not invariant under gauge-transform: you get a relative phase, e ); A A A ; exp( (3.59) i c To prove this, consider what happens to the canonical momentm, p ec A i ec A i ec ( A )
(3.60)
Aharanov bohm effect; persistent current. Wavefunction of cooper pairs will similarly suffer this phase, F
ˆ (r )ˆ (r )
F e i exp(
2e i c
);
(r );
(3.61)
Supercurrent of cooper pairs related to gradient ofF in the same way that number current is related to the gradient of the wavefunction j 2mi ( * * ) , and this appears as, jS
2e 2 m 2i
( F F * F *F )
ie
2m
nS e i ( nS e i ) nS e i ( nS e i )
e
m n S
(3.62)
Make a gauge-invariant combination, jS
e m
nS
2 e c
A jS
e m
nS
2 e c
A
e m
nS
2e
c
B
2e 2 / mc
nS B
(3.63)
Meanwhile: maxwell’s equations appear as,
B
4 c
jS ;
B 0;
2
B
1
B; 2
2
mc2 4 nS e 2
plasma ; P frequency ; P 2 c2
(3.64)
Monday, April 08, 2013
Assignment 4 posted. Due in 2 weeks. Last homework of semester. Nearing end. Let’s talk about qualitative methods in high e nergy physics/quantum field theory. Consider decay of K-mesons. Tells us symmetry: e.g., parity, charge-conjugation, time-reversal (PCT). Four K-mesons: K , K , K0 , K o . You have ~ 10 10 s . You have quantum number “strangeness”: ( K , K0 ) S 1 , and ( K , K0 ) S 1 . You could have following processes,
0 ; K 0 ; K 0 0 0; K 0 ; K0 ; K 0 ; K 0 ;
K 0
(3.65)
Hamiltonian to capture decay-physics (3.65) then app ears as, H
g1 † † K 0 g 2 † † K 0 g3 † † † K 0 g 4 † † † K 0 h.c.
(3.66)
K 0 and K 0 are energy-degenerate eigenstates. There are two superpositions: the short-living superposition S and the long-living superposition L. you have, K S†
N ( g1 , g2 ) g1 K0 g2 K0 ; KL† N( g1 , g2 ) g2 K0 g1 K0 ; N N( g1 , g 2 )
1 g1
2
g 2
2
;
(3.67)
Omitting the 3-roder terms, we can write the ket s K S † 0 and K L† 0 , and thereby get, H L
HK L† 0 g1 g2 † † K0 K0† 0 g1 g2 † † K0 K 0† 0 O ( † † † )
(3.68)
g1 g2 † † (1) 0 g1 g2 † † (1) 0 1 1 g1 g 2 † † 0 0 In contrast to (3.68), we have HK S † 0 would have
L (2 ) L (3 )
0 ; we have L (3 ) S (2 ) . In experiment, we have K L . We
~ 10 3 .
Parity to inversion: you obviously have the following properties, r r; (r ) ( r); P (r) ( r ); P 2 1; P[e iqr ] e iqr
q q;
(3.69)
Book: Lipkin, “quantum mechanics
Wednesday, April 10, 2013
Could you have the reaction 0
? I do not think so. It is not time-reversal invariant.
and K 0 . The Hamiltonian for these decays was (3.66), e.g., H g1 † † K 0 g 2 † † K 0 g3 † † † K 0 g 4 † † † K 0 h.c.
Recall kaon decays K 0
(3.70)
Parities: the parity of the interaction is -1. Then, PK0† P K 0† and P P † . Time-reversal, i t (t ) H (t ) i t (t ) H ( t )
Suppose you have H
*
i t * ( t ) H * * ( t );
H * ; this is ordinary circumstance of quantum mechanics. Then you have, T (t ) * ( t ); Teiqr e iqr ei(q ) r ; TAT 1 A* ;
(3.71)
(3.72)
How does time-reversal affect the kaon? Consider t he action of time-reversal as described by (3.72) up n the Fourier-decomposition of the state of the kaon,
K (q ) K (r )e iq r
TK 0 (q)T 1 K 0 (q);
Similarly: you can time-reverse a 180-degree rot ation, T D180 (q) q T D180 K0 (q)T 1D180
K0 (q);
(3.73)
(3.74)
What if we had H H ? Would be a more general interaction…might not conserve probability density, which could represent the transmutation of one particle into another (?). *
Charge conjugation: if this is symmetry of Hamiltonian, consider transform, CAC 1 C ( A); g1C ( K0 ) g1 K0 ; g2 C( K0 ) g2 K 0 ;
(3.75)
Bring back long-living and short-living combinations, (3.67), but rewrite as,
N g1K 0 g2 K0 Ng1 K0 K 0 ; 1 N N ( g , g ) ; 1 2 g g 2 2 † g1 g 2 K L N g2 K0 g1K 0 Ng1 K0 K 0 ; K
† S
g1 g 2
1
(3.76)
2
Action of charge-conjugation (3.75) upon (3.76) appears as, C ( K S ) K S ; C( K L ) KL ;
(3.77)
How does PC apply to the decays? Parity and charge conjugation, PC ( K S ) K S ; PC ( K L ) KL ; PC( ) | KS ; PC( ) | K S
;
(3.78)
This is why you have “short living” and “long living” combinations. That was terminology before 1964. Then, decays were discovered. After 1964, we learned that K L . That means that CP is not conserved. CPT: Luden-Pauli theorem is that T is broken. No CPT. CP at T interactions CPT.
Friday, April 12, 2013
Last time, looked at time-reversal. Transformation appears as T (t ) * ( t ) . But suppose Hamiltonian is not † ˆ Hermitian, H H . This happens in the Zeeman e ffect, where H 1 p 2 I B , in which, 2m
0 B 1
1 B x 0
B
Bx iBy B z 0 i 1 0 B B ; y 0 1 z B iB B x y z i 0
(3.79)
Spin orbit interaction: couples spin to orbital angular momentum. Can only rotate the two together. Yyou do not have axis-freedom. Must consider 3 components of spin. Contributions, B SO
ie 2 E v E ; H SO B B SO (E ) i H 2 c c m 2mc 1
1
p
H † , and we have, † H SO i x Hx y H y z H z ;
Here, we have a Hermitian
(3.80)
H
*
SH S
1
S ( H * * ) ES * 1
S( H * S 1 S * );
ES ; H SH S ; HS ES ; *
*
*
*
(3.81)
That means the thing solves the same Schrödinger equation dn thus gives the same physics. Explicit form of Hamiltonian, using S x 1 S 1 x and S y S 1 y and S z S 1 z , is as follows, H
i x Hx y H y z H z
(3.82)
This is a linear system. A lot of algebra could be done to show the following result: S y . Note, also, that, y x y 1
y x y y i z i i x x ; y y y 1 y ; y z y 1 z ;
(3.83)
Spin-orbit interaction, T
y * y K
Complex conjugation operator, K [complex-conjugaton operator]; Ka a* ; T
(3.84)
y K ; THT 1 H ;
(3.85)
This means: spin in Hamiltonian will cause intricacies under time-reversal symmetry. Exercise: go back to rashba Hamiltonian, and check time-reversal symmetry. It fits into the above parad igm of the Hamiltonian, but it doesn’t contain certain terms. Discrete symmetries
Bibliography: QM 06, ch 12; Tinkham’s book Group Theory and Quantum Mechanics, Dresselhaus’s Group Theory: Applications to the Physics of Condensed Matter , and Ramond’s book Group Theory: A Physicist’s Survey. Group properties: closure, associativity, ident ity, inverse.
Monday, April 15, 2013
Symmetry operations. Irreducible representations of D3. You aso have D4. Example: CO2 molecule. Symmetries of it are important. Has D2 symmetry.
Group theory doesn’t solve the moton. It specifies transformation criteria. If one was to transform, one must retain isomorphism. That is a very physical concept. Group theory is analytical analogy. Important theorem: The trace of a matrix is basis-independent. Orthogonality of characters: some exercises. You find out that characters play the role of orthogonal vectors. Thus, there is more than one way to represent a physical reality. CO2: go back to dumbbell molecule. What wavefunctions can be about the molecule? Well, potential is an even function of x if x is the axis along the line, and you define 0 to be at carbon atom.
1 x y z
D2
I
C x2
C y2
C z 2
A1 A2 A3 Ay
1 1 1 1
1 1 -1 -1
1 -1 1 -1
1 -1 -1 -1
G of the group is 4. You basically have the cayley t able of transforms of the CO2 molecule.
“Van Vleck” orthogonality theorem for “irrep” (“irreducible representation”)...
“decomposition theorem”: which elements of a symmetry gro up will “survive”. Contains elements of group, plus elements of perturebation. E.g., an electric field along main body diagonal of cube Example: lifting of cubic degeneracy. A strain could be along the main diagonal. There are degenerate levels, and they are split by this straining.
Gamma15: pesky notation for the group properties of a square, which is a huge t opic in and of itself. “lifting of 3fold degeneracy”. Again, pesky square-notation, A, E, F, etc. The F2(x3) splits into A1 and E2 doubly degenerate state. Notes updated: chapter on continuous symmetries. Big pdf file. Look at content. Is the same. Doesn’t have chapter 10. But at end of chapter 9, you have separately-inserted .pdf.
SCl6, etc., there are actually molecules which feature symmetry. Lifting of 3-fold degeneracy: you just add up the numbers. Apply perturbation that breaks degeneracy. Smaller table is symmetries of equilateral triangle. Big table; all representations/classes of smaller table. That means you can deco mpose the big table into a representation that the smaller one serves as a basis vecto r for. In general, to split reducible into irreducible, you need to know basis vectors. Symmetries that are responsible for energy levels of electrons. Molecular vibrations: caveat; you have rotational, translational, and vibrational DOF. Must eliminate some. That’s to come later. Lattice symmetries: reminder: you build lattices out of blocks, and o nly certain types of blocks by the Crystallographic Restriction Theorem. Pretty pictures from solid state 1: review this. Ohm’s law on a lattice: how do es ohm’s law transform under rotations? Should be invariant if you have angularly-homogeneous material (duh). Ohm’s law on a lattice (cont’d): ohm’s law also is necessarily symmetric under time reversal. rhombohedral vs. orthorhombic Nontrivial statement: there is no difference, fro m a symmetry point of view, between a 90 degree rotated parallelogram and that which is not. D2 = C2V group: you know that 180 rotation about x-axis imposes no constraints. D2 group don’t learn anything new. Focus on various symmetry elements. Rotate about one of the da shed lines in rhombohedral vs. orthorhombic (the slike w/the nontr ivial statement).
Thinking about the hopping that may happen along a rectanglular-lattice. That’s why you might have anisotropic conduction. Next symmetry element: tetragonal lattice. Diagonal compoentn of conductivity. D4 = C4v: tetragonal. Symeries of D2 must be taken care of. Symmetries of conductivity tensor. Relation between sigmaxx and sigmayy. Chose element of C4v, instead (why not?). reflection in diagonal vertices. Square, cut by diagonal. Reflection in plane. Any tensor describing physical properties…: you look at tensor. Matrices of elastic model. Tensor of effective masses. All have same symmetries as we found. We did a very general calculation. Once you have gro up with particular symmetry, you know what tensor should look like. Do the exercise (final exam‼!). Vibrational modes of H2O molecules: bent molecules. E.g., water. What are symmetries of normal vibrational modes? Count up degrees of freedom. Picture of H2O: in no way is the H2O molecule a rhombus or rectangle, but it does have same symmetries. C2v group symmetries very simple. 2
“Wonderful formula from lecture on Monday”: sumirreps (dim (irrep)) = g. N.B., cyclic gropu is special case of more general group called Abelian group. E.g., 1D rotations commute amidst one another. Add extra dimensions and you get non-commutivity. How do group elements act on coordinates? …group, cyclic group, Abelian group; 1D, so no degeneracy. Must now figure out and in which wa y the table is telling us about possible normal-mode vibrations. Tonight (make up class): figure out how to get rid of extra degrees o f freedom.
Wednesday, April 17, 2013, 18:00 9
Water molecule. A and B in character table . We are not solving motion problem; we are considering basisfunctions. Equivalence: inversion
g , u ;
rotation A, B ; e.g., things like Geq , eq , a.s . . Possibly, Geq
2 A1 B1 .
Translation and rotation: vector of displacement is transformed. Just like any ot her vector. Corresponding character will be 1 + 2 + 3? Subst “group theory eliminates the need for differential equations and integrals”. E.g., you can Interesting project: is there a Mathematica program for splitting up a reduc ible group into an irreducible one? Transition to chalkboard
“selection rules”; “identical representation”. f* H i dq 0 or finite .
9
Old notation; Ag means “gerade”, and Au means “ungerage”.
Certain energy could correspond to more than one wavefunction (good ol’ quantum degeneracy). Must rotate to space with non-degenerate (“good”) eigenstates.
( ) i
dq
c0
(3.86)
1
Under direct product operation,
( ) ( ) ( ) 1 ( ) ( ) ( ) 1
(3.87)
H eEx
(3.89)
Linearfield,
(3.88)
Linear molecule, xi
xi ; xi x, y, z ;
(3.90)
And,
funny business ( ) ( ) ( ) 1
x dx 0; * x
0
* x
(3.91)
x 0 dx 0;
(3.92)
“the universe is nothing more than a direct product of represetnations of symmetry groups”. The C_2v (of water, H2O). C_2V Z, A_1 Xy, A_2 Oc, B1 Y, B2
I 1 1 1 1
C_2 1 1 -1 -1
The D3, D_3 1, A_1 Z, A2 (x,y) E
Sigma_v 1 -1 1 -1
I 1 1 2
Sigma’_v 1 -1 -1 1
2C_3 1 1 -1
3C_2 1 -1 0
Consider, representation-index i( ) basis-number, i = 1,2,...,f ; f [dimension of representation];
( ) i
dq 0;
( ) 1 identical representation, symmetric ;
Proof, sum over the group elements,
( ) i
dq
group
( ) i
dq
G
ˆ
( ) i
dq
G
ˆ
( ) ki
i( ) dq;
(3.93) (3.94)
(3.95)
G
And,
g i dq dq ( )
( ) ki
( )
k
0
(3.96)
G
Orthogonality theorem,
G
Matrix elements,
( ) ik
(j ) g ;
( 1 ) j 1
G
( ) ik
g 1 ;
(3.97)
H
H 0 H ; H 0 G; H [lower-symmetry group];
(3.98)
Direct product of represntations of a group, C A B; D AB C; Aij Bk ; 1 s m, p; 1 r na; A n n; B m n; C nm nm;
(3.99)
You can have,
k( ) H i( ) dq
(3.100)
Direct product of representations itself. Example of directproduct operator,
b11 a b 11 21 b13 b31 b23 b b33 11 a21 b21 b31
b11 b12 a11 a12 A B b21 b22 a a 21 22 b b 31 32
b12 b22 b32 b12 b22 b32
b13
b33 b13 b23 b33 b23
b11 a12 b21 b31 b11 a22 b21 b31
b12 b22 b32 b12 b22 b32
b13
b33 a11bij a21bij (3.101) b13 a21bij a22 bij b23 b33 b23
This is in contrast to matrix-multiplication, which is an array of dot products/convolutions (or, more generally, inner-products). (1) direct products are non-commutative. (2) direct products a re associative. Note: the diagonal elements of (3.101) are composed entirely of diagonal elements of A, B. that allows us to say stuff about the trace, Tr A B a11 ( b11 b22 b33 ) a22 ( b11 b22 b33 ) ( a11 a22 )( b11 b22 b33 ) Tr ATr B; (3.102) Two repesentatiosn two matrices. You want to think about
ij( ) and (k ) , as ( ) ( ) . Go back to the water molecule. Regard H as
perturbation. What’s the symmetry of that perturbation? 1; A1 B1 B2 ;
A2 E ;
(3.103)
You have, ( j )* H i( ) Theorem:
( ) ( ) ( ) ( ) 1 ...
(3.104)
( ) ( ) a11 22 ... ; then, the proof is, a1
See if product
( ) ( )
1 g
r r * ( r 1 )
G
1 g
G
1 g
2
G
has overlap. Watch for occurrence of
( )
g g
1;
(3.105)
. Could be there or not there.
Now we write down the Hamiltonian of something in the microwave: H
2
21m P ec A 21m P 2 mce A P O 2 ( A) H 0 H ;
Important to distinguish H0 and H’ each as po lar vectors. Matrix elements and the group-gamma, A , j | P | , i ; A1 B1 B2 ; A1 ( A1 B1 B2 ) A1 A1 B1 B2 ;
(3.106)
(3.107)
Conclusion: All normal modes of the molecule are coupled to each other. Problem: consider equilateral triangle. C3 axis. You have e lectron going around the molecule. Wavefunctions of electron must transform as members of D3 group. ???…Need to multiply characters of table at this point.
Wednesday, April 24, 2013 – the last day of classes
“Analytic” character of physical law. Build a Dru de-force-balance of an electron’s conductivity in an isot ropic 2 e sample. You get EOM appearing as p eE mc p B 1 p . Resulting conductivity is xx 0 (1 [C ] ) , in which 0
eB nem and C mc . There is a hole in the argument. That’s because you should, actually, have a non2
analytic function. Deadline for final: Tuesday at 10 AM.
Let’s use dimensional analysis again. This time, we w ill solve the problem of Brownian motion: what causes it? You have a particle of mass M, immersed in a gas, and it jostles about. Estimate “mobility”. E.g., for a given force that you pull on the part icle, what’s the velocity? Parameters: mass of gas-particles m, density n, cross-section A. In order to have motion at constant velocity, you need two forces. Other force is frictional force. That frictional force comes from the cross section. u F; F R au; (3.108) Retarding force must be a function as, F F ( a, m, M , n, A, u ) a( m, M , n, A) u
(3.109)
Dimensions, [ a]
[ F ] [u ]
[ M ][ v] / [ t] [ v]
[ M ] [t ]
(3.110)
Large mass: fixed temperature will have big particle-velocity. same t emperature, light molecules move faster. Collision-rate should be determined by relative velocity between particles. Large mass should disappear. It moves slowly compared to little atoms. Slightly more elabourate: recall collision of light cart and heavy c art from baby-physics. You could have little cart with velocity v and mass m, and large cart with velocity u and mass M. estimate, F ~ p / . Balance between two processes: (1) velocities u and v parallel and (2) veloicities u and v anti-parallel will determine motion. Write down momentum and energ y balances, as usual, mv Mu m( v) Mu; 12 mv2 12 Mu 2 12 mv2 12 Mu2 ; (3.111) Then, substitute them into each other,
m(v v)( v v) M ( u u)( u u) p2 u; m(v v)2v M ( u u)2 u; m(v v)2u M ( u u)2 u p2 u;
(3.112)
(v v)v 2 vu v v 2u
p m(v v) m( v v v v) m(2 v 2 u) 2 m( v u) To leading order, you have the momentum shift being just twice o f … something,
p p p 4mu p ~ mu
(3.113)
Now recall that 1 is the collision-rate. It should be speed divided by mean free path. We should use “the little particles’ velocity-magnitude” as the speed. But what do we use as the mean free path? Note that the mean free 2 path has units of L, and the cross section has units of L . Their product has units of inverse-number-density. But which density should we use? Large part icle or small paricle density? Use small-particle-density, 1 small-particle-speed v ~ (3.114) mean free path nA Then,
F ~ (mvnA)u; F
1
1
a
mvnA
au; a ~ mvnA; ~
;
(3.115)
Another step that Einstein did: compute diffusion coefficient. The particle M is undergoing “random-walk” motion. Find proportionality between , k BT , D . Recall, [ D ]
L2 T
; [ ]
T
; [ k BT ] M
M
L2 T 2
;
(3.116)
What will the proportionality be? Write, [ D]
T ML2 M T 2
L2 T
D ~ k B T
(3.117)
Alternate method: recall the diffusion equation, but rewrite it as a flux balance that tallies to 0, N N v U 1 j D ; N ( x) Ne U ; ; N F 0; F x x x k B T N
(3.118)
Finally: let’s do an integral,
I
e ax dx
I (a )
I ( a 1)
I ( a 1)
(1 x 2 ) 2 0
1
e y
a 0 (1 a 2 y 2 ) 2 1
e
dy ~
y
a (1 a y ) 2
0
; I ( a 1) ?; I ( a 1) 1?;
2 2
dy ~
1
e y
a 1/ a (0 a2 y2 )2 1
e a 0
y
dy
1 a
dy
1 a5
a 1
(3.119)
e y y4
dy ; (3.120)
