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L.J. Sham 02/04/
Physics 130C Mid-Term Examination Solutions
- Short questions. Each answer must be accompanied by a brief explanation.
(a) Use Fig. 1 to give a description of how a photon can enable the transformation of an
unentangled state of two two-state atoms into an entangled state.
u〉
l 〉
u〉
g〉 →e〉
g〉 →e〉
BS
Figure 1: The colored arrows denote the photon states.
Solution – The key effect of the photon-atom interaction is to excite the atom from the
ground state to the higher energy state by absorbing a photon. With both atoms initially
in the ground state, the relevant transformation consists of,
|u〉 ⊗ |g, g〉 → | 0 〉 ⊗ |e, g〉, (1)
|ℓ〉 ⊗ |g, g〉 → | 0 〉 ⊗ |g, e〉. (2)
Passing a photon through a beam splitter yields a linear combination of two photon
states, which by a linear combination of the two processes in Eq. (2),
(|u〉 + |ℓ〉) ⊗ |g, g〉 =
| 0 〉 ⊗ (|e, g〉 + |g, e〉), (3)
achieving an entanglement of the two atom states as well as leaving the photon in the
zero state.
(b) (i) Show that any operator on the composite space H
AB can be written as a sum of
outer product of product operators, e.g., |n, k〉〈n, k|. (ii) Show that the Hamiltonian of
the hydrogen atom can be put in this form.
Solution –
i. To show the form of an operator
D on H
(AB) , we first define a basis set {|n〉} in
H
(A) and a basis set {|k〉} in H
(B)
. Then in terms of the product basis {|n, k〉 =
|n〉 ⊗ |k〉} in the composite Hilbert space H
(AB)
, we can use the completeness
relation to find the operator
D,
D =
I
(AB) ˆ D
I
(AB)
n,k
|n, k〉〈n, k|
D
m,ℓ
|m, ℓ〉〈m, ℓ|
n,k;m,ℓ
D
n,k;m,ℓ
(|n〉〈m|) ⊗ (|k〉〈ℓ|), (4)
which is a sum of the product operators with coefficients D n,k;m,ℓ
ii. Let H
(R) be the Hilbert space of the relative motion states |n〉 and H
(C) be the
Hilbert space of the center of mass motion states |K〉. Denote the corresponding
eigenenergies of the |n〉 by E n
and of the |K〉 by E K
. Then, the Hamiltonian
H
(RC)
=
n
d
3
K
(2π)
3
|n, K〉〈n, K|(E n
+ E
K
[
n
|n〉〈n|E n
]
I
(C)
I
(R)
⊗
[∫
d
3 K
(2π)
3
|K〉〈K|E
K
]
(c) A composite state of qubits A and B is given by,
|Ψ〉 = | 0 , 0 〉 + | 0 , 1 〉β + | 1 , 0 〉γ + | 1 , 1 〉δ. (7)
Find the conditional states of qubit B on the states | 0
(A) 〉, | 1
(A) 〉 respectively, and find
the relations for the constant coefficients β, γ, δ in order that |Ψ〉 be a product state.
Solution – The conditional states of qubit B on the states | 0
(A) 〉, | 1
(A) 〉, are, respec-
tively,
|ψ
(B)
0
(A)
|Ψ〉 = | 0
(B)
〉 + | 1
(B)
〉β, (8)
|ψ
(B)
1
(A)
|Ψ〉 = | 0
(B)
〉γ + | 1
(B)
〉δ. (9)
If |ψ
(B)
1
〉 = λ|ψ
(B)
0
〉, then,
(A)
〉 + | 1
(A)
〉λ)|ψ
(B)
0
is a product state. The condition on the coefficients are,
γ = λ, and δ = λβ, (11)
or δ = γβ. (12)
i. Prove that they are all entangled states.
ii. They are members of a basis set. Prove that the set is orthonormal and complete.
Solution –
i. Prove that the Bell states are all entangled states. By definition, it is equivalent to
showing that none of them can be put in the general product form, i.e., for any of
them
(AB)
〉 6 = |ψ
(A)
〉 ⊗ |φ
(B)
〉. (19)
In the |z±〉 basis set for each spin,
|ψ
(A) 〉 = |z+〉α + |z−〉β, |φ
(b) 〉 = |z+〉γ + |z−〉δ, (20)
so, |ψ
(A)
〉 ⊗ |φ
(B)
〉 = |z+, z+〉αγ + |z+, z−〉αδ
+|z−, z+〉βγ + |z−, z−〉βδ. (21)
To equate the product state to the Bell state, we need either
|αγ| =
= |βδ|, and αδ = 0 = βγ, (22)
or αγ = 0 = βδ, and |αδ| =
= |βγ|. (23)
Neither set of conditions can be simultaneously satisfied.
ii. Prove that the set is orthonormal and complete. Use the product basis set, {|n〉, n =
1 − 4 } ≡ |z±, z±〉. The Bell set is related to the product basis by a transformation,
(AB)
(AB)
(AB)
−
(AB)
−
U (|z+, z+〉, |z+, z−〉, |z−, z+〉, |z−, z−〉)
or equivalently, |Ψ n
U |n〉, (24)
with the matrix representation given by Eq. (18):
U =
Thus, the transformation
U is unitary. Then,
〈Ψm|Ψn〉 = 〈m|
U
† ˆ U |n〉 = 〈m|
I|n〉 = δm,n, (26)
shows orthonormality.
n
|Ψn〉〈Ψn| =
n
U|n〉〈n|
U
†
=
U
n
|n〉〈n|
U
†
=
U
U
†
=
I, (27)
shows completeness.
(b) Prove that by means of the single spin Hamiltonian with the magnetic field in the z
direction within an appropriate time interval,
H
(AB)
=
H
(A)
z
I
(B)
, where
H
(A)
z
= ℏλ
Z, (28)
it is possible to transform |Φ
(AB)
〉 to |Φ
(AB)
−
〉 and |Ψ
(AB)
〉 to |Ψ
(AB)
−
〉 and vice versa.
Solution – Actually we could transform every Bell state to every other Bell state by a
local spin operation by adding the Hamiltonian
H
(A)
x =^ ℏλ^
X. Note that for λt = π/ 2 ,
e
−i
ˆ H
(A)
z
t/ℏ
|z±〉 = ∓i|z±〉, (29)
e
−i
ˆ H
(A)
x t/ℏ |z±〉 =
I cos(λt) − i
X sin(λt) = −i
X|z±〉 = −i|z∓〉. (30)
Hence, using them on the Bell states,
e
−i
ˆ H
(A)
z
t/ℏ
|Φ
(AB)
〉 = −i|Φ
(AB)
−
e
−i Hˆ
(A)
z
t/ℏ
|Ψ
(AB)
〉 = −i|Ψ
(AB)
−
e
−i
ˆ H
(A)
x t/ℏ |Φ
(AB)
〉 = −i|Ψ
(AB)
e
−i
ˆ H
(A)
x t/ℏ |Φ
(AB)
−
〉 = i|Ψ
(AB)
−
These together with the inverse transforms will link any Bell state to another.
- You are given a single mode of the electromagnetic field, with the Hamiltonian given by
Eq. (13) and the electric field operator at a given position and time,
E
x
= Ei(ˆa − ˆa
†
), where E =
ℏω j
2 ǫ 0
L
3
Choose either part (a) or part (b) but not both :
(a) (i) Find the mean and root-mean-square (or fluctuation or uncertainty) of the electric
field for the vacuum state | 0 〉. (ii) Find also their time dependence.
Solution –
(i) The mean electric field,
Ex| 0 〉 = Ei〈 0 |(ˆa − aˆ
†
)| 0 〉 = 0, (36)
since aˆ| 0 〉 = 0, 〈 0 |ˆa
†
= 0.
For the variance,
(ˆa − ˆa
†
)| 0 〉 = −| 1 〉, 〈 0 |(ˆa − ˆa
†
)
†
= −〈 1 | (37)
〈 0 |(ˆa − ˆa
†
)
2
| 0 〉 = 〈 0 |(ˆa − ˆa
†
)(ˆa − ˆa
†
)| 0 〉
= 〈 0 |(−1)(ˆa − ˆa
† )
† (ˆa − ˆa
† )| 0 〉
