Quantum Mechanics - Homework Assignment 5 Alejandro G´ omez omez Espinosa∗ October 16, 2012
Shankar Shankar Ex. 5.2.1 A particle is in the ground state of a box of length L. Suddenly the box expands (symmetrically) to twice its size, leaving the wave function undisturbed. Show that the probability of finding the particle in the ground state of the new box is (8/ (8/3π )2 ? A particle in a box of length L, in the ground state (n=1), is in the state:
2
ψ(x) =
L
sin
πx L
0
|x| ≤ L |x| ≥ L 2 2
if we double symmetrically the length of the box, the new system would be: φ(x) =
1 2L
sin
πx 2L
0
|x| ≤ L |x| ≥ L
Then, the probability of finding the particle in the ground state of the new box is 1 : P = φ ψ
| | |
2
√ √ √ L/2
=
L/2
−
= =
2 L
8 3π
2 πx πx sin sin L L 2L
2 2L 3π
2
dx
2
2
[email protected] Int Integrat gratio ion n made made in Mathe athema mattica. ica. Wolfr olfram am Alpha Alpha LLC LLC. http://www.wolframalpha.com (access October 15, 2012). ∗
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Shankar Shankar Ex. 5.2.4 Consider a particle of mass m in the state n of a box of length L. Find the force F = ∂E/∂L encountered when the walls are slowly pushed in, assuming assuming the particl particlee remains remains in the nth state of the box as its size changes. changes. Consider Consider a classical particle of energy E energy E n in this box. Find its velocity, the frequency of collision on a given wall, the momentum transfer per collision, and hence the average force. Compare it to ∂E/∂L compute above.
|
−
−
The energy of a particle of mass m in a box of length L is: 2
π 2 n2 , 2mL2
E =
n = 1, 2, 3,...
Next, let’s calculate the force: F=
−
∂E = ∂L
−
∂ ∂L
2
π 2 n2 2mL2
2
=
π 2 n2 2E = mL3 L
On the other hand, classically, we know that the energy is:
√
p2 2mE n E n = p = 2mE n v= 2m m In addition, the time that one particle needs to collide again on a given wall, i.e., its frequency is: 2L 2Lm t= = v 2mE n The change in momentum, in an elastic collision, when the particle hits the wall would be: ∆P = p ( p) p) = 2 p = 2 2mE n
⇒
⇒
√
−−
Finally, using all the previous computed results, the average force of the particle is: F =
∆P 2(2mE 2(2mE n ) 2E n = = t 2mL L
Shankar Shankar Ex. 5.2.2 Hint: The spirit of this problem is to see how the kinetic and potential energies scale as α 0 (i.e., for a very wide Gaussian) and show that the (positive) kinetic energy drops to zero faster than the (negative) potential energy.
→
(a) Show that for any normalized ψ , ψ H ψ E 0 , where E 0 is the lowest-energy eigenvalue. (Hint: Expand ψ in the eigenbasis of H.) Let’s define define n as the orthonormal eigenbasis of H, i.e., H n = E n n and n n n = 1. Then, we can compute:
| |
| | | ≥ |
|
ψ|H |ψ
=
|
| | | | | | | | | | | | ψ m m H n n ψ
m,n
=
E n ψ m m n n ψ
m,n
=
E n ψ m n ψ δ m,n m,n
m,n
=
E n n ψ
2
n
= E 0 0 ψ E 0
≥
2
| | | 2
+ E 1 1 ψ
| | |
2
+ E 2 2 ψ
| | |
2
+ ...
|
(b) Prove the following theorem: Every attractive potential in one dimension has at least least one bound state. state. Hint: Hint: Sinc Since V is attract attractive ive,, if we define define V ( V ( ) = 0, it follows that V ( V (x) = V ( V (x) for all x. To show show that there there exists exists a bound state state with E < 0, consider a 1/4 −ax /2 ψa (x) = e π and calculate
|
∞
|
2
2
E (α) = ψα H ψα ,
H =
| |
−
d2 2m dx2
− |V ( V (x)|
Show that E (α) can be made negative by suitable choice of α of α. The desired result follows from the application of the theorem theorem proved proved above. Let’s calculate the value of E (a): ∞
E (a) = ψa H ψa =
| |
a π
−∞
1/2
−
e
ax
2
2
−
2
d2 2m dx2
|
V (x) − |V (
e−
ax
2
2
dx
Let’s calculate separately the value of H:
−
2
d2 2m dx2
|
V (x) − |V (
−
e
2
2
ax
2
=
−
=
−
d2 2m dx2 2 d 2m dx
2
=
− −
−| | − − | e−
ax
axe−
a2 x2 e−
2m
2
V ( V (x) e−
2
ax
2
2
−
− axe
2
= Returning to the value of E (a): ∞
E (a) =
= =
a π
∞
a π a π
1/2
−∞
2
a 2m
a 2 2m π
−
e
2
ax
2
2m
2
a −ax e dx 2m
π a
a π
2
2
a x ∞
2
2
− ax 2 2
− | | V ( V (x)
a 2 −ax x e dx 2m
−∞
−| | e−
− ax − |V ( V (x)
2
− − − − | √ − | | −∞
=
2m
a2 x2
|
2
ax
2
2
∞
2
2
V ( V (x) e−
2
ax
2
ax
e
−
ax
2
ax
2
V ( V (x) e−
|
ax
2
ax
2
2
2
2
2
dx
∞
−
−∞
ax
−
|V ( V (x)|e
2
dx
V ( V (x) e−ax dx
0
|
−∞
∞
2
V ( V (x) e−ax dx
−∞
≥ E
0
from (a)
where here the first term corresponds to the kinetic energy and the second to the potential energy. In addition, we can see that when a 0 the kinetic term decrease faster than the second term due to the exponent of a in each term. The theorem can be prove using part (a), where we can see that the expression for E (a) is greater or equal than the lowest-e lowest-energy nergy eigenv eigenvalue. Therefore, Therefore, in a one dimensional problem there is at least one bound state.
→
3
Shankar Shankar Ex. 5.2.3 Consider V Consider V ((x) = aV 0 δ (x). Show that it admits a bound state of en2 2 ergy E ergy E = ma V 0 /2 2 . Are there any other bound state?. Hint: Solve Schr¨ odinger’s equation outside the potential E < 0, and keep only the solution that has the right behavior at infinity and is continuous at x at x = 0. Draw the wave function and see how there is a cusp, or a discontinuous change of slope at x = 0. Calculate the change in slope and equate it to + d2 ψ dx dx2 −
−
−
(where is infinitesimal) determined from Schr¨ odinger’s equation. Let’s calculate the Schr¨odinger odinger equation with the potential V ( V (x): 2
−
d2 ψ + aV 0 δ (x)ψ = Eψ 2m dx2
d2 ψ 2m + 2 (E dx2
⇒
))ψ = 0 − aV δ (x))ψ 0
Now, let’s integrate this relation between an infinitesimal change: +
d2 ψ dx dx2 − + d dψ dx dx dx − dψ dx − dψ( dψ() dψ( dψ( ) dx dx
−
−
+
=
− −
−
+
=
2maV 0 2
2maV 0 2
(E
))ψ − aV δ (x))ψ
=
2
2mE dx + 2
−
= 0+
2m
0
+
−
2m 2
dx
aV 0 δ (x))ψdx ))ψdx
ψ (0)
ψ(0)
Figure Figure 1: Diagram Diagram of a particle moving moving in a delta function potential, potential, according to Shankar, Shankar, Exercise 5.2.3. On the other hand, the equation of motion of a particle moving in a V = 0 potential is: 2 2 d ψ = Eψ 2m dx2
−
4
where if we separate the cases for in this exercise, shown in Figure 1, the wavefunctions of a particle are: ψI (x) = Aekx + Be −kx = Aekx ψII (x) = Ce kx + De −kx = Dekx where k 2 = 2mE/ and, B and C vanish due to boundary conditions at both infinities. Applying the boundary conditions when x = 0:
−
ψI (0) = ψII
A=D
⇒
Finally, let’s calculate the discontinuous change of slope at x = 0 using the previous results: dψII (0) dx
− dψdxI (0) −kD − kA −2kD
2maV 0
=
2
2maV 0
=
2
2maV 0
=
2
taking ψ(0) = ψII (0)
4m2 a2 V 02 4
4m2 a2 V 02
=
2
D
2
k2 =
−
ψ (0)
2maV 0
k =
2mE
ψ (0)
4 2
2 0
− ma2 V
E =
2
In this case, there is no other bound state because E and the wavefunction do not depend on any integer. 5) In class we derived Shankar (5.2.23) k tan ka = κ for the even solutions in a finite well of depth V 0 . Repe Repeat at for the odd solutions solutions and show that satisfy: k cot ka = κ
−
We can separate this problem problem in three parts, described in Figure Figure 2, according to the value of the potential. In the case when V = V 0 : 2
−
d2 ψ + V 0 ψ = Eψ 2m dx2
where ψ = ∂ 2 /∂x 2 , and k2 = 2
2m 2
(V 0
⇒
ψ (x) = k2 ψ
− E ).). When V = 0:
2
− 2 m ddxψ = Eψ ⇒ 2
5
ψ (x) = α2 ψ
Figure 2: Sketch of the square well potential. where α = 2mE . Therefore, Therefore, following following the notation ψI as the wavefunction of the part (I) in Figure 2, we found: 2
ψI (x) = Aekx + Be −kx = Aekx ψII (x) = C cos( C cos(αx αx)) + D sin(αx sin(αx)) ψII I (x) = Ee kx + F e−kx = F e−kx Here, we can quickly said that B and E must vanish due to conditions in very large and small values of x. Next, we need to satisfy boundary conditions: ψI ( a) = ψII ( a)
−
−
ka
= C cos( C cos( αx) αx) + D sin( αa) αa)
ka
= C cos( C cos(αx αx))
−
Ae
−
Ae
−
−
sin(αa)) − D sin(αa
ψII (a) = ψII I (a) F e−ka = C cos( C cos(αx αx)) + D sin(αa sin(αa))
(1)
(2)
ψI ( a) = ψII ( a)
−
−
Ake−ka =
αx) + Dα cos(−αa) αa) −C α sin(−αx)
Akeika = Cα sin(αx sin(αx)) + Dα cos(αa cos(αa))
(3)
ψII (a) = ψII I (a)
−F ke
ka
−
=
sin(αx)) + Dα cos(αa cos(αa)) −C α sin(αx
(4)
sin(αa)) − F ) F )e ka = −2D sin(αa (A − F ) F )e ka = 2Dα cos(αa cos(αa)) 1 tan(αa tan(αa)) =− k α cot(αa)) −α cot(αa −
(5)
−
(6)
Solving Solving this equations: equations: (1)
− (2) ⇒ (3) + (4 ( 4) ⇒ (5)/(6) ⇒ k
=
(A
6
6) a) Show that
j =
(ψ ∗ ψ
∗
∇ − ψ∇ψ )
2mi for the current density can be rewritten rewritten
j =
m
ξ 2 φ
∇
where ξ (r) and φ(r) are the magnitude and phase of ψ(r), resp respec ectivel tively, y, i.e., iφ ψ = ξe with ξ and φ real. Defining ψ = ξe iφ , then ψ ∗ = ξe −iφ , and the expression inside the brakets is: ψ
∗
∇ψ − ψ∇ψ
∗
= ξe
iφ
−
∇ − ∇ ξe iφ
= ξ 2 e−iφ ieiφ
ξe iφ
ξe −iφ
∂φ ∂φ + ξ 2 eiφ ie−iφ ∂ r ∂ r
= 2ξ 2 i φ
∇
Replacing in the expression above: j =
2mi
2ξ 2 i φ =
∇
m
ξ 2 φ
∇
b) Use the result of (a) to do Shankar Ex. 5.3.2.: Convince yourself that if ψ = cφ˜, where c is constant (real or complex) and φ˜ is real, the corresponding j corresponding j vanishes. Using the result of (a): j = = = = =
(ψ ∗ ψ
∗
∇ − ψ∇ψ ) 2mi (c φ˜∇ cφ˜ − cφ˜∇ c φ˜ 2mi (c cφ˜∇φ˜ − cc φ˜∇φ˜) 2mi |c| (φ˜∇φ˜ − φ˜∇φ˜) 2mi 0 if φ ˜ ∈ R ∗
∗
∗
∗
)
2
c) Use the result of (a) to show that the probability current density of a plane wave ψ(r) = ψ0 eik·r is what you would expect, based on its probability density and the momentum (and thus velocity) of the particles. Using the result of (a) with ψ (r) = ψ0 eik·r : j =
m
ψ02 (k r) =
∇ ·
7
m
ψ02 k =
p 2 ψ m 0
