Quantum Mechanics - Homework Assignment 7 Alejandro G´ omez omez Espinosa∗ November 14, 2012
1) Show that the potential V ( V (x, y) = + for x > L/2 L/2 and V ( V (x, y) = 12 mω 2 y2 for x < L/2 L/2 (”harmonic oscillator in a strip”) is separable, and write the ground state wavefunction ψ (x, y). What relati relation on between L and ω must hold if the first two excited states are to be degenerate?
∞
||
||
This potential can be separable in the x
− y components, like this: 0 |x| < L2 V 1 (x) = ∞ |x| > L2
1 mω2 y2 2
V 2 (y) =
where the total product V ( V (x, y ) = V 1 (x)V 2 (x) is the potential for an harmonic oscillator in a strip. From this potential we can see that we split this problem in two: in x we have a particle in a infinite square potential and, in y we have the usual harmonic oscillator, where the general solutions are: ψn (x) = ψn (y) =
√ 2 nπx nπx cos + sin L L L 1 mw 1/4 − e H n 2n n! π mwy
2
2
mw
y
where the general state wavefunction is the product ψnm (x, y) = ψn (x)ψm (y ). Therefore, Therefore, the ground ground state is given by: 2 πx mw 1/4 − ψ10 (x, y) = cos e L L π In addition, the total energy for the system is the sum of the energy in each case described above:
E nk nk =
2 2 2 π n 2mL2
+ k+
1 2
mwy
2
2
w
The first two excited degenerate states are when n = 2 k = 0 and, n = 1 k = 1. Replac Replacing ing this this values in the energy and match them: 4 2 π2 1 + w = 2mL2 2 4 2 π 2 + wmL wmL2 =
2 2 π 2mL2
3 w 2 2 2 wmL2 π + 3 wmL 3 π2 w = 2mL2
∗
[email protected]
1
+
2) a) Show that if operators C (2) and D(2) commute in space
V 2, then [A(1) ⊗ C (2) , B (1) ⊗ D(2) ] = [A(1) , B (1) ] ⊗
[A
(1)
(2)
⊗ C
,B
(1)
⊗D
(2)
] = = = =
A
(1)
(2)
(1)
(2)
C (2) D(2)
−
(1)
(2)
V 2:
V
| | | | | | | |⊗ | | ⊗| | | | | | | | | ai i
t =
i
x =
(s
x )( t
i
c j j j
y =
d j j j
j
ai∗ c j∗ i j
y )=
bi d j i j j
ij
=
ij
ai∗ bi i i
st xy =
V 2. (Let’s use V 2 are |α, so
bi i
j
Then:
V | | V | | ⊗ | | ⊗ | | |
|s =
and in
(2)
B ⊗D A ⊗ C ⊗ C B ⊗ D A(1) B (1) ⊗ C (2) D(2) − B (1) A(1) ⊗ D(2) C (2) A(1) B (1) − B (1) A(1) ⊗ C (2) D(2) ⇐⇒ C (2) , D(2) commute [A(1) , B (1) ] ⊗ C (2) D(2)
b) Let s and t be arbitrary vectors in 1 , and x and y be arbitrary vectors in the notation developed in class that the basis vectors of 1 are i and those of that s = i ci i , etc. Show that ( s x )( t y )= st xy . Let’s define the vectors in 1 :
| | | |
(1)
(1)
ij
c j∗ d j j j j
i
ai∗ c j∗ bi d j
ai∗ bi c j∗ d j
=
j
(2)
ij
Since (1 (1) is equal to (2 (2), the theorem is proved.
Shankar Shankar Ex. 10.1.2 Imagine a fictitious world in which the single-particle Hilbert space is two-dimensional. Let us denote the basis vectors by + and . Let
|
(1) σ1
|−
a b = c d
and
(2) σ2
=
e f g h
| (1)|− | ⊗ | | ⊗|− |−⊗| |−⊗|−
be operator in V1 and V2 , respectively (the signs label the basis vectors. Thus b = + σ1 etc.) The space V1 V2 is spanned by four vectors + +, + , +, . Show (using the method of images or otherwise) that:
±
⊗
(1) (1)⊗(2)
σ1
(1)
= σ1
2
⊗ I (2)
a 0 = c 0
0 a 0 c
b 0 d 0
0 b 0 d
(Recall (Recall that α (1)⊗(2)
σ1
= =
=
| ⊗ β | is the bra corresponding to |α ⊗ |β .) (1) σ1 ⊗ I (2) + + |σ1 ⊗ I | + + + + |σ1 ⊗ I | + − + + |σ1 ⊗ I | − + + + |σ1 ⊗ I |−− | −− + − |σ1 ⊗ I | + + + − |σ1 ⊗ I | + − + − |σ1 ⊗ I | − + + − |σ1 ⊗ I |−− | −− − + |σ1 ⊗ I | + + − + |σ1 ⊗ I | + − − + |σ1 ⊗ I | − + − + |σ1 ⊗ I |−− | −− −−|σ1 ⊗ I | + + − − |σ1 ⊗ I | + − −−|σ1 ⊗ I | − + − − |σ1 ⊗ I |−− | −− +|σ1|++|I |+ +|σ1|++|I |− | − +|σ1|−+|I |+ +|σ1|−+|I |− |− +|σ1|+−|I |+ +|σ1|+−|I |− | − +|σ1|−−|I |+ +|σ1|−−|I |− |− −|σ1|++|I |+ −|σ1|++|I |− |− −| −|σ1|−+|I |+ −|σ1|−+|I |− |− −|σ1|+−|I |+ −|σ1|+−|I |− |− −| −|σ1|−−|I |+ −|σ1|−−|I |− |−
1 0 1 c 0
0 1 0 1
a 0 c 0
b 0 d 0
a
=
=
0 a 0 c
1 0 1 d 0 b
0 1 0 1
0 b 0 d
(2) (1)⊗(2)
σ2
Following the results of part (a): (1)⊗(2)
σ2
e g = 0 0
f h 0 0
0 0 e g
0 0 f h
⊗ σ2(2) + + |I ⊗ σ2| + + + + |I ⊗ σ2| + − + + |I ⊗ σ2| − + + + |I ⊗ σ2|−− + − |I ⊗ σ2| + + + − |I ⊗ σ2| + − + − |I ⊗ σ2| − + + − |I ⊗ σ2|−− − + |I ⊗ σ2| + + − + |I ⊗ σ2| + − − + |I ⊗ σ2| − + − + |I ⊗ σ2|−− −−|I ⊗ σ2| + + − − |I ⊗ σ2| + − −−|I ⊗ σ2| − + − − |I ⊗ σ2|−− +|I |++|σ2|+ +|I |++|σ2|− +|I |− |−+|σ2|+ +|I |− |−+|σ2|− +|I |+−|σ2|+ +|I |+−|σ2|− +|I |−−| |−−|σ2|+ +|I |−−| |−−|σ2|− −|I |++|σ2|+ −|I |++|σ2|− |− −| −|I |− |−+|σ2|+ −|I |− |−+|σ2|− −|I |+−|σ2|+ −|I |+−|σ2|− |− −| −|I |−−| |−−|σ2|+ −|I |−−| |−−|σ2|−
= I (1) =
=
e g e 0 g
f h f h
e g 0 0
0 0 e g
1
=
=
f h 0 0
e g e 1 g 0
f h f h
0 0 f h
3
(3) (1)
(σ1 σ2 )(1)⊗(2) = σ1
⊗ σ2(2)
ae ag = ce cg
af ah cf ch
be bg de dg
(1)⊗(2)
Do part (3) in two ways, by taking the matrix product of σ1 (1) (2) computing the matrix elements σ1 σ2 .
bf bh df dh
(1)⊗(2)
and σ2
and by directly
⊗
(1)
⊗ σ2(2) + + |σ1 ⊗ σ2| + + + + |σ1 ⊗ σ2| + − + + |σ1 ⊗ σ2| − + + + |σ1 ⊗ σ2|−− + − |σ1 ⊗ σ2| + + + − |σ1 ⊗ σ2| + − + − |σ1 ⊗ σ2| − + + − |σ1 ⊗ σ2|−− − + |σ1 ⊗ σ2| + + − + |σ1 ⊗ σ2| + − − + |σ1 ⊗ σ2| − + − + |σ1 ⊗ σ2|−− −−|σ1 ⊗ σ2| + + − − |σ1 ⊗ σ2| + − −−|σ1 ⊗ σ2| − + − − |σ1 ⊗ σ2|−− +|σ1|++|σ2|+ +|σ1|++|σ2|− +|σ1|−+|σ2|+ +|σ1|−+|σ2|− +|σ1|+−|σ2|+ +|σ1|+−|σ2|− +|σ1|−−|σ2|+ +|σ1|−−|σ2|− −|σ1|++|σ2|+ −|σ1|++|σ2|− |− −| −|σ1|−+|σ2|+ −|σ1|−+|σ2|− −|σ1|+−|σ2|+ −|σ1|+−|σ2|− |− −| −|σ1|−−|σ2|+ −|σ1|−−|σ2|−
σ1 σ2 = (σ1 σ2 )(1)⊗(2) = σ1 =
=
e g e c g
f h f h
e g 0 0
0 0 e g
a
=
=
=
ae ag ce cg
f h 0 0
af ah cf ch
e g e d g b
f h f h
0 0 f h
be bf bg bh de df dg dh
We can also multiply the results of part (a) and (b): σ1
⊗ σ2
=
=
=
(1) σ1
a 0 c 0 ae ag ce cg
4
(2)
⊗ I 0 a 0 c
b 0 d 0
af ah cf ch
⊗ I (1)
0 b 0 d
e g 0 0
be bf bg bh de df dg dh
(2)
σ2
f h 0 0
0 0 e g
0 0 f h
4) For the infinite infinite square-well square-well potentia potential, l, let γ = 2 π2 /2mL2 so that the energies are E n = γn 2 for n = 1, 2,... Now suppose two indistinguishable, non-interacting particles are placed in this potential. The total energy for the two indistinguishable particles in this potential is: E n
1 ,n2
=
2 2 π 2mL2
n21 + n22
(3)
a) Find the energies of the ground state and of the first three excited states in the case that the two particles are bosons. Using (3 (3): E 11 11 =
2 2 π 2mL2
E 12 12 =
2 2 π 2mL2
E 22 22 =
2 2 π 2mL2
E 13 13 =
2 2 π 2mL2
2
1 +1
2
=
2 2 π mL2
5 2 π2 2mL2 4 2 π2 = mL2 5 2 π2 = mL2
12 + 22 = 22 + 22 12 + 32
b) Find the energies of the ground state and of the first three excited states in the case that the two particles are fermions. Using (3 (3): E 21 21 =
2 2 π 2mL2
E 12 12 =
2 2 π 2mL2
E 23 23 =
2 2 π 2mL2
E 13 13 =
2 2 π 2mL2
2
2 +1
2
=
12 + 22 = 22 + 32 = 12 + 32 =
5 2 π2 2mL2 5 2 π2 2mL2 13 2 π2 2mL2 5 2 π2 mL2
Shankar 10.3.2 When an energy measurement is made on a system of three bosons in a box, the n values obtained were 3,3, and 4. Write down a symmetrized, normalized state operator.
|ψ = √ 13 (|334 + |343 + |433)
5
Shankar 10.3.6 Consid Consider er a composi omposite te obje object such such as the hydro hydrogen atom. Will it behave ehave as a boson oson or fermion? Argue Argue in general that objects containing an even/odd number of fermions will behave as bosons/fermions. Consid Consider er carefu arefully lly the case of two hydr hydrogen atoms. atoms. Suppose Suppose i is a basis of one-electron states and I is a basis basis of one-proton one-proton states. Argue Argue that the basis states for the system of two hydrog hydrogen en atoms (that is, for the system of two electrons and two protons) consists of all states of the form ij,A IJ,A , and discuss that the eigenvalue of this state is under the permutation operator that interchanges the two atoms (that is, that interchanges both the electron states and the proton states). Do H atoms behave behave like fermions or bosons? Argue Argue in general general (some ”hand-waving” ”hand-waving” is OK) that objects containing even/odd number of fermions will behave as bosons/fermions.
| | ⊗|
|
A composite object as the hydrogen atom is made of two fermions, an electron i and a proton I which which besides besides being being fermio fermions, ns, the com combin binatio ation n of both create createss a boson. boson. This This result resultss is based on the permutation permutation operator acting in both states. states. Let’s consider consider the case of two two hydrogen hydrogen atoms. A basis state has the form ij,A IJ,A . If we permute permute the first state we will have have a minus minus sign and we found another minus minus sign from the permutation permutation of the second term. Since both particles are distinguishable and the permutation does not change the state, the combination of this two hydrogen atoms will behave as a boson.
|
|
⊗|
|
From this result we can conclude that composite objects with an even number of fermions will behave as a bosons while, objects with an odd number of fermions will remain behaving as fermions.
6