This essay will discuss the above quotation (from Giannetti) regarding formalism and support the outlined elements of formalism with Eisenstein’s theories and film form in general. In addition to t...
Descrição: A cheat sheet for the basic Japanese particles ga, wa, no, mo, so, wo, ni, he, de, ka, to, ya, yo, and ne. Originally this was shared on tofugu.com but it's no longer available there.
Descripción: JLPT N5 Particles
The official score/sheet music to one of the must-have songs featured in Twilight: Eclipse as the "unofficial theme" to kickstart the Saga mid-way.
Collision between identical particles Collisions between identical particles are particularly interesting as a direct illustration of the fundamental fundamental differences differences between between classical classical and quantum mechanics. mechanics. We shall examine first the elastic scattering of two identical spinless bosons and then analyze elastic collision between two identical spin-
1 2
fermions.
Scattering of two identical spinless bosons Let us consider the elastic elastic scattering scattering of two identical identical bosons of mass m. n the centre of mass frame! the time-independent "chrodinger equation is
2 2 − ∇ + ( ) ψ ( r ) = E ψ ( r ) ############### V r 2 µ #$1% wher wheree µ =
m 2
is the reduced reduced mass mass and r = r 1
− r 2 is
the relati&e position &ector of the two
colli collidin ding g parti particle cles. s. 'he situat situation ion in the centre centre-of -of-m -mass ass system system is illust illustrat rated ed in (ig. (ig. 1. 'wo identi identical cal particl particles es 1 and 2 approac approach h one another another!! mo&ing mo&ing parallel parallel to the z-axis z-axis in opposit oppositee directions. )fter an elastic collision the &elocity of each particle is changed in direction but remains unchanged in magnitude. ) detector counts the particles scattered into the direction characterized by the polar angles (θ ! φ ) . "ince the particles 1 and 2 are identical! there is no way of deciding whether a particle recorded by the detector results from a collision e&ent in which the particle 1 is scattered in the direction (θ ! φ ) $see (ig. 1$a%%! or from a collision process in which the particle 2 is scattered in that direction! so that particle 1 is scattered in the opposite direction ( π − θ ! φ + π ) $see (ig. 1$b%%.
n classical mechanics the differential cross-section for scattering in the direction (θ ! φ ) would simply be the sum of the differential cross-sections for obser&ation of particle 1 and particle 2 in that that dire direct ctio ion. n. f the the same same were were to be true true in quan quantum tum mechan mechanic ics! s! we woul would d obta obtain in the the differential cross-section the *classical+ result result d σ classical d Ω
= f (θ ! φ ) 2 + f ( π − θ ! π + φ ) 2 ###########
#$2% where f (θ ! φ ) ! the center of mass amplitude for scattering in the direction (θ ! φ ) ! is related to the asymptotic beha&ior of the solution ψ ( r ) satisfying the usual boundary condition
ψ ( r )
exp( ikr ) θ φ ########### ( ) ( ) + f ! → r r →∞
exp ik .r
##..$,% owe&er! we shall now show that the expression $2% for the differential cross-section is incorrect. ndeed! we now that wa&e functions describing systems of identical particles must be properly symmetrised with respect to permutations of identical particles. n particular! a wa&e function describing a system of identical bosons must be completely symmetric. 'hus! in the case of two identical spinless particles! the wa&e function must be symmetric under the interchange of the spatial coordinates of the two particles. /ow the interchange r 1 ↔ r 2 corresponds to replacing the relati&e position &ector r by − r ! which in polar coordinates corresponds to ( r ! θ ! φ ) being replaced by ( r ! π − θ ! π + φ ) . 'he wa&e function ψ ( r ) satisfying the boundary condition $,% does not ha&e the required symmetry! but the symmetric combination
ψ + ( r ) = ψ ( r ) + ψ ( − r ) ################ #..$0% is also a solution of the "chrodinger equation $1% and does ha&e the required symmetry ψ + ( − r ) = ψ + ( r ) sing $,%! the asymptotic form of ψ + ( r ) is seen to be r →∞
#$3% 'he amplitude of the spherically outgoing wa &e is the symmetric amplitude f + (θ ! φ ) = f (θ ! φ ) + f ( π − θ ! π + φ ) #############.
.$4% so that the differential cross-section is d σ d Ω
= f (θ ! φ ) + f ( π − θ ! π + φ ) 2 #############
#.$5% a result which we can write in the form d σ d Ω
$7%
= f (θ ! φ ) 2 + f ( π − θ ! π + φ ) 2 + 2 6e[ f (θ ! φ ) f ∗ ( π − θ ! π + φ ) ] ########.
t is important to note that this formula differs from the *classical+ result $2% by the presence of the third term on the right! which arises from the interference between the amplitudes f (θ ! φ ) and f ( π − θ ! π + φ ) . We also remar that the total cross-section is σ tot
= ∫ f (θ ! φ ) + f ( π − θ ! π + φ ) 2 d Ω ############
$8% is equal to twice the number of particles remo&ed from the incident beam per unit time and unit incident flux.
n the simple case for which the interaction potential is central! the scattering amplitude is independent of the azimuthal angle φ . 'he differential cross-section then reduces to
d σ d Ω
= f (θ ) + f ( π − θ ) 2 ################
$19% or d σ d Ω
= f (θ ) 2 + f ( π − θ ) 2 + 2 6e[ f (θ ) f ∗ ( π − θ ) ] ########..
$11% so that the scattering is symmetric about the angle θ =
π 2
in the centre-of-mass system. "ince
P l [ cos( π − θ ) ] = ( − 1) P l ( cos θ ) ############# l
$12% it is clear that the partial wa&e expansion of the symmetrised scattering amplitude π f + (θ ) = f (θ ) + f ( π − θ ) contains only e&en &alues of l . :oreo&er! at θ = we note that the 2
quantum mechanical differential cross-section is equal to
d σ θ =
π
2
d Ω
2
π #############. = 0 f θ = 2
$1,% and hence is four times as big as if the two colliding particles are distinguishable! and twice as big as the *classical+ result
d Ω
d σ classica θ =
$10%
π
2
2
π = 2 f θ = #############.. 2
(urthermore! if there is only s-wa&e $ l = 9 % scattering! so that the scattering amplitude f is isotropic! we see from $19% that two colliding spinless bosons ha&e differential cross-section four times as big as if they were distinguishable particles! and twice as big as the classical result.
Coulomb scattering off two identical spinless bosons Let us consider two identical spinless bosons of charge Ze interacting only through Coulomb forces. 'his is the case for example in the scattering of two identical spinless nuclei $e.g. the scattering of two alpha particles or two 12C nuclei% at colliding energies which are low enough so that the nuclear forces between the two colliding particles can be neglected due to the presence of the coulomb barrier. 'herefore! the differential cross-section d σ d Ω
= f c (θ ) + f c ( π − θ ) 2 ##############
$1% where f c (θ ) is the Coulomb scattering amplitude! f c (θ )
( Ze ) 2 where γ = . ( 0πε 9 ) v 'his is called the :ott formula for the Coulomb scattering of two identical spinless bosons. Problem:
;roton-proton scattering at sufficiently large energies or scattering angles can get contributions not only from the electrostatic potential but also from the strong nuclear force. (or moderate energies! we need only consider the l = 9 partial wa&e in a phase shift analysis so that the scattering amplitude is gi&en by f (θ )
where F INT ( θ ! δ 9 ) gi&es the interference between the Coulomb and nuclear amplitudes.
Scattering of two identical spin-
1 2
fermions
'he scattering of identical fermions is more difficult to analyze than that of spinless bosons because of complications due to the spin. (or simplicity! we shall only consider the case of two identical spin-
1 2
fermions interacting through central forces. "ince the interaction is in general
different in the singlet $ S = 9 % and triplet $ S = 1 % spin states of the two fermions! we shall start from two $unsymmetrised% scattering amplitudes f S ( θ ) and f t (θ ) corresponding respecti&ely to the singlet and triplet cases. 'he full wa&e function describing a system of two identical spin-
1 2
fermions must be
antisymmetric in the interchange of the two particles! i.e. when all their coordinates $spatial and spin% are interchanged. /ow! if the system is in a singlet spin state $ S = 9 %! the spin part of the wa&e function is gi&en by χ 9! 9
=
1 2
[α (1) β ( 2 ) − β (1)α ( 2 ) ] ##########..
$1% ence the corresponding spatial part of the wa&e function must be symmetric in the interchange of the position &ectors r 1 and r 2 of the two particles. )s a result! the symmetrised singlet scattering amplitude
f S +
= f S (θ ) +
f S ( π − θ ) ############.$2%
and the differential cross-section in the singlet spin state is d σ
= f S (θ ) +
d Ω
2
f S ( π − θ ) #############
$,% f! on the other hand! the two spin-
1 2
fermions are in a triplet spin state $ S = 1 % the
corresponding three spin functions are symmetric in the interchange of the spin coordinates of the two particles
χ 1!1
= α (1)α ( 2)
χ 1! 9
=
χ 1! −1
=
1
[α (1) β ( 2 ) 2 β (1) β ( 2)
+ β (1)α ( 2) ] #############
#.$0%
'he spatial part of the wa&e function must therefore be antisymmetric in the interchange of the position &ectors r 1 and r 2 ! so that the symmetrised triplet scattering amplitude is gi&en by d σ t d Ω
=
2
f t ( θ ) − f t ( π − θ ) #############
#..$3% f the *incident+ and *target+ particles are unpolarized $i.e. their spins are randomly orientated%! the probability of obtaining triplet states is three times that of singlet states! so that the differential cross-section is gi&en by!
d σ d Ω
=
1 d σ s 0 d Ω
+
, d σ t 0 d Ω
=
1 0
f S (θ ) + f S ( π − θ )
2
+
, 0
2
f t ( θ ) − f t ( π − θ ) #########$4%
(or the particular case of spin-independent central interactions! where f s (θ )
$5%
= f t (θ ) = f (θ ) ################
We find from! $4% that! d σ d Ω 1
=
=
1 0
f ( θ ) + f ( π − θ )
[ f (θ ) 0
2
2
+
, 0
f (θ ) − f ( π − θ )
2
+ f ( π − θ ) + 2 6e f (θ ) f ∗ (π − θ ) ] +
[ ( )
1 d σ 2 = f θ 0 d Ω 1 , = + f θ 0 0
2
[ f (θ ) 0 ,
2
+ f ( π − θ ) − 2 6e f (θ ) f ∗ (π − θ ) ]
+ f ( π − θ ) + 2 6e f (θ ) f ∗ ( π − θ ) ] + 2
2
[ f (θ ) 0 ,
2
+
(
f π
− θ ) − 2 6e f (θ ) f ∗ ( π − θ ) ] 2
1 , 2 1 , + + f ( π − θ ) + − 6e f (θ ) f ∗ ( π − θ ) 0 0 2 2 0 2 2 0 − 2 6e f (θ ) f ∗ ( π − θ ) = f (θ ) + f ( π − θ ) + 0 0 2 2 2 = f (θ ) + f ( π − θ ) − 6e f (θ ) f ∗ ( π − θ ).....................................................................................$7%
( )
2
We note that this formula differs from the *classical+ result by the presence of the third term on π the right! which again is an interference term. We also remar that at θ = the quantum 2
mechanical differential cross-section $7% is gi&en b y!
d σ θ =
π
2
π ##############$8% = f θ = d Ω 2 π d σ classical θ = 2 π . and hence is equal to one-half of the classical result 2 = 2 f θ = d Ω 2 2
(urthermore! if there is only s-wa&e ( l = 9) scattering! the differential cross-section $7% is four times smaller than the corresponding one for the scattering of two identical spinless bosons.
