Quantum Mechanics - Homework Assignment 8 Alejandro G´ omez omez Espinosa∗ November 26, 2012
1) If ψ is a simultaneous eigenvector of Hermitian operators A and B that anticommute, AB + BA = 0, what can you say about the eigenvalues a and b of ψ ? As an exmaple, discuss the case case where where the operators are Π and P (parity and momentum) for a particle in 1D.
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If A and B anticommute, we know that AB + BA = 0. Applyi Applying ng this to the eigenv eigenvect ector or of both operators: AB ψ + BA ψ = bA ψ + aB ψ = 0
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| then, we can know that A|ψ is an eigenvector of B and B |ψ is an eigenvector of A.
But, But, if this this
expression vanishes, a or b must be zero.
In the case of Π and P for a particle in 1D: ΠP x + P Π P Π x = pΠ x + P
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thus, p = 0. 2) An ammonia molecule has its nitrogen nucleus fixed at the origin, and three hydrogen nuclei fixed at (a, 0, c), ( a/2 a/2, 3a/2 a/2, c), and ( a/2 a/2, 3a/2 a/2, c), wher where constatn onstatnss a and c are are appr appropriat opriately ely chosen.
−
√
−
−√
a) Argue, in general terms, that C 3† H C 3 = H , where H is the Hamiltonian for the electrons in this system, and C 3 is the unitary operator corresponding to a rotation of 120◦ about the z axis. If C If C 3 is the unitary operator corresponding to a rotation of 120◦ about the z axis, we can define it as: C 3 = U [R(120)] = e− 120iL 120iLz
then, C 3† H C 3 = e
120iL 120iLz
H e−
120iL 120iLz
= H
that is clear for a system to rotate 120† and then turn back the same angle. b) What are the allowed eigenvalues of C 3? C 3 can be written as the matrix:
C 3 =
∗
cos cos 120 120 sin 12 1 20 0
− sin 120 cos 12 1 20 0
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1
−√
0 0 1
=
1 2 3 2
0
√
− −
3 2 1 2
0
0 0 1
Then, to calculate the eigenvalues:
− − − √ √ − − − − − − − − − − √ √ − − − − − √ − √ − 1 2
3 2
λ
det
3 2
1 2
0
0
λ
(1
1
λ) +
2
1 2
1 2
λ
2
1 2
0 0
λ
+
3 4
λ+
i 3 2
1 2
i 3 1 2
−λ
−λ − i
λ
3 (1 4
(1
λ) = 0
λ) = 0
i 3 2
λ
=0
3 1 2
(1
(1
− λ) = 0
− λ) = 0
Thus, the allowed eigenvalues of C of C 3 are: λ=
−
√ −
√ −
i 3 1 i 3 1 , 1, 2 2
c) If ψ is a non-degenerate eigenvector of H, what can you say about C 3 ψ If C 3† H C 3 = H , we know that [C [C 3 , H ] = 0, i.e., i.e., C 3 commute with H , then:
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H C 3 ψ = C 3 H ψ = EC 3 ψ
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where E is the energy associated to H . Therefore, we can say that C 3 ψ is an eigenvector of H with eigenvalue E .
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3) a) Let W be a symmetry of the Hamiltonian, i.e., W † H W = H . Show that if ψ is a non-degenerate eigens eigenstat tatee of H, then then ψ A ψ = ψ W † AW ψ . Why is the cave caveat about about non-de non-degen gener eracy acy needed? Due to W is a symmetry of the Hamiltonian and ψ is a non-degenerate eigenstate of H, then W ψ is also an eigenvector of H , as explained in part (2c). Therefore:
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ψ|A|ψ = ψ|W † AW |ψ The requeriment of the non-degeneracy state is needed because otherwise W ψ = ψ . b) Show that if W † AW = A, then ψ A ψ = 0 for any non-degenerate eigenstate of H. If W † AW = A, then W A + AW = 0, i.e., A and W anticommute. Then, using the result of part (1), if A if A ψ = a ψ and W ψ = w ψ , a or w must be zero. Therefore:
− |
−
|
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| |
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| ψ|A|ψ = 0
c) Use these results to argue that the CO 2 molecule cannot have a dipole moment in its ground state. state. (More (More presic presicely, ely, assuming that the electr electronic onic ground state is non-degener non-degenerate ate and that the nuclei are fixed according to the summetric structure of the C the C O2 , show that a measurement of the dipole moment operator d = eX is certain to give a zero value.) Due to the structure of the CO 2 molecule, the position of the oxigen atoms have parity simmetry. Thus, Π† d Π = d . Therefore, any measurement of the dipole moment operator in one of the oxigen atoms will give us the value of the other oxigen atom, i.e., the value will be zero.
−
−
2
Shank Shankar Ex. 12.2.1 Provide the steps linking U [R] x, y = x
| | − y , x z
z
+y
to
x, y|U [R]|ψ = ψ(x + y , y − x ) z
z
(Hint: (Hint: Rec Recall the derivation derivation of equation quation (11.2.8) (11.2.8) from from equation equation (11.2.6)). (11.2.6)). Recalling the derivation of equation (11.2.6):
|ψ U
= U [ U [R] ψ
|
= U [ U [R]
∞ |
x, y x, y ψ dx dy
|
−∞
=
∞ | − x
y z , xz + y x, y ψ dx dy
|
−∞
Making a change in the variables: x = x
y = xz + y
− y , z
we can found the values of the old variables respect to the new ones: x + y z x= , 1 + 2z
y x z y= 1 + 2z
−
where where the denomina denominator tor became became one for an infinit infinitesi esimal mal approa approach ch.. Replac Replacing ing this this values alues in the previous relation: U [R] ψ
|
=
x , y|U [R]|ψ
= =
x , y|U [R]|ψ
∞ | − | −∞∞ | − | −∞ − | − x ,y
x + y z , y
x ,y x ,y
x + y z , y
x + y z , y
Lz
x z ψ dx dy
x z ψ
= ψ x + y z , y
Shank Shankar Ex. 12.2.3 Derive
x z ψ dx dy
x z
→ − ∂φ∂
by doing a coordinate transformation on eq (12.2.10), and also by direct method mentioned above. Equation (12.2.10) shown the relation: Lz = x
− − − ∂ i ∂y
∂ i ∂x
y
=
−i
∂ x ∂x
−
∂ y ∂x
(1)
Express the coordinates as x = r cos φ and y = r sin φ and, found the partial derivative respect to φ: ∂ ∂φ
=
∂x ∂ ∂y ∂ + ∂φ ∂x ∂φ ∂y ∂ ∂ r sin φ + r cos φ ∂x ∂y ∂ ∂ y +x ∂x ∂y
=
−
=
−
3
Replacing it in (1 (1):
−i ∂φ∂
Lz = Next, by direct method:
− 1
i
ψ(r, φ)
δφ
Lz ψ(r, φ) = ψ (r, φ
δφ) − δφ)
− i δφ L ψ(r, φ)
= ψ (r, φ)
−i δφ L ψ(r, φ)
=
−δφ ∂ψ ∂φ
Lz ψ(r, φ) =
−i ∂ψ ∂φ
z
z
+ O[(δφ [(δφ)) ] − δφ ∂ψ ∂φ 2
5) a) Review the discussion on p.311 and fill the missing details leading from (12.2.25) to the commutation relations [P x , Lz ] = i P Py and [P y , Lz ] = i P Py .
−
i
I + z Lz
= =
i
i
i
i
I
= I + I + = I + I +
x P x
x z P x Lz 2
x z 2
− L P z x
z
2
i
I
I + I + x P x + y P y
i i I + x P x + y P y
× −
− − − − − × − −
I + I + (x P x + y P y )
I + z Lz
i
x
z x
y
x z
z
x
2
[P x , Lz ] +
y z 2
+
i
I
z Lz
z
i
x
I
2
2
− L P z
y
2
[P y , Lz ]
where comparing with the RHS of equation (12.2.25), we found that: [P x , Lz ] =
−i P P ,
[P y , Lz ] = i P Py
y
b) Carry out a corresponding derivation to show that [Lx , Ly ] = i Lz . From Shankar (12.4.1), we know that: Lx = Y P z Ly = Lz =
4
− Z P ZP − X P X P − Y P
y
x
z
y
x
x P x
z y Lz P y
z Lz
y z
i
2
y z P y Lz
(x P x + y P y ) =
z y Lz P y
i + z Lz
2
− L P
I
− i P − L P y
z Lz
i
i
− P y
y
then, [Lx , Ly ] = Lx Ly = = = =
−L L (Y P − ZP )(Z )(Z P − X P ) − (ZP − X P )(Y )(Y P − Z P ) Y P Z P − Y P X P − Z P P + Z P X P − Z P Y P + Z P P + X P Y P − X P Z P Y P (P Z − ZP ) + XP (Z P − P Z ) − Z [P , P ] + X Y [ Y [P , P ] (Z P − P Z )(XP )(XP − Y P ) y
z
z
x
x
y
x
x
z
z
z
z
z
z
z
x
2
y
y
y
z
x
y
z
z
z
x
2
z
y
x
z
2
y
x
z
y
z
z
z
z
x
= [Z, P z ]Lz = i Lz
6) Working in 2D, consider the family of rotations by angle θ about the point (a,0). Find the expression for the operator operator G that is the generator for this family of operations. operations. The unitary operator that represents the rotation can be written as: U [R(θ)] = e−
i
Gθ
where G is the genera generator tor operator operator for this family of operator operators. s. Let us write write down the rotation rotation matrix in 2D: cos θ sin θ R(θ) = sin θ cos θ
−
and the coordinates of the rotation about (a, (a, 0) are: x = (x + a)cos θ + y sin θ and y = y cos θ (x + a)sin θ. Now, the unitary operator acting on the state r is given by:
|
| − | −
ψ (r) = r U [R(θ)] ψ = r
|
But also:
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i
I
Gdθ
∂ψ( ∂ψ (r) ψ (r) = ψ(r ) − dθ = ψ(r) ∂θ
where ψ(r ) = ψ(x, y ) = ψ(x cos θ
ψ = ψ(r )
∂ψ( ∂ψ (r ) ∂x ∂x ∂θ
−
− i Gdθψ( Gdθψ (r)
∂ψ( ∂ψ (r ) ∂y ∂y ∂θ
− y sin θ − a, x sin θ + y cos θ), then:
∂ψ( ∂ψ (r ) ∂ψ (r ) ψ (r) = ψ(r ) − y dθ − x ∂ψ( ∂x
∂y
Finally, comparing (2 (2) with (3 (3) we found that: G = i
∂ψ( ∂ψ (r) y ∂x
5
−
∂ψ( ∂ψ (r ) x ∂y
−
(2)
dθ
(3)
y