22.51 Quantum Theory of Radiation Interactions
Mid-Term Exam
October 27, 2010
Problem 1:
Solution
Electron Spin: Magnetization
20 points
~ = Bz ~z at zero temperature. The spin is
Consider an isolated electron with spin- 21 , placed in a large magnetic f eld B
in the state
|ψ i = c0 |↑i + c1 |↓i
where |↑i ≡ |Sz = + ~2 i (|↓i ≡ |Sz = − ~2 i) represents the spin state aligned (anti-aligned) with the vertical z-axis.
~ = γ ~ σ~ of this single spin (with γ the gyromagnetic
We now assume that we can measure the magnetic dipole ~µ = γS
2
ratio of the electronic spin):
a) What is the probability of f nding an outcome µz > 0? What is the spin state immediately after the measurement?
Solution:
The eigenstates of µz are |↑i and |1i with eigenvalues ±γ ~2 . Assuming γ > 0, the probability of f nding µz > 0 in a
measurement is simply p(µz > 0) = |c0 |2 . The spin state is projected into the corresponding eigenstate |ψi′ = |↑i.
(For γ < 0 it would have been p(µz > 0) = |c1 |2 and |ψi′ = |↓i)
b) What is the average magnetization hµz i in the z direction?
Solution:
We need to calculate hψ|µz |ψi = (c∗0 h↑| + c∗1 h↓|)µz (c0 |↑i + c1 |↓i). Since µz is already diagonal in the basis |↑i, |↓i
this is simply hµz i = γ ~2 (|c0 |2 − |c1 |2 ) = γ ~(|c0 |2 − 21 ).
~ = Bz ~z and the spin is in its ground state, what are c0 and c1 ?
c) If the magnetic f eld is aligned with the z-axis, B
What is now hµz i?
[Assume that the only interaction is the Zeeman interaction, HZ = ~γBz σz /2]
Solution:
The energy levels of the Zeeman interaction are simply the eigenvalues of HZ , ±γBz ~2 corresponding to the eigenvectors |↑i, |↓i. Assuming Bz > 0 and γ > 0 (or more generally γBz > 0), the eigenvector |↓i has thus the lowest
energy. Then the state is simply |ψi = |↓i, that is |c1 | = 1 (or c1 = 1 up to an unimportant phase factor) and c0 = 0.
The average magnetization is then hµz i = −γ ~2 .
If γBz < 0 the ground state is instead |↑i and hµz i = γ ~2 .
Now assume that we cannot achieve zero temperature, but only a temperature T (as provided e.g. by liquid Nitrogen),
~ = Bz ~z.
so that the spin is at thermal equilibrium in the f eld B
d) What is the state of the spin?
Solution:
At f nite temperature, we expect
to have a mixed state as given by the canonical ensemble. Thus the state is given by
ρ = e−βHZ /Z, where Z = Tr e−βHZ . For this simple Hamiltonian we can calculate the state explicitly:
ρ=
e−βγBz ~/2 |↑ih↑| + eβγBz ~/2 |↓ih↓|
e−βγBz ~/2 + eβγBz ~/2
1
Notice that the partition function Z = 2 cosh (βγBz ~/2). We can also write the state as:
ρ = e−βHZ /Z = [cosh (βγBz ~/2)11 − sinh (βγBz ~/2)σz ]/Z =
1
[11 − tanh (βγBz ~/2)σz ]
2
e) What is the average magnetization? How many spins would you need in order to achieve the same magnetization
as in question c?
[Hint: you can assume a temperature T = 77K which corresponds to ≈ 1600GHz and a magnetic f eld B ≈ 6 Tesla
which gives a Zeeman energy ~γB ≈ 160GHz. ]
Solution:
For a mixed state, the expectation value of an observable is hOi = Tr {ρO}. Thus the magnetization is hµz i =
Tr {ρµz } or:
−βγBz ~/2
e
|↑ih↑| + eβγBz ~/2 |↓ih↓| γ~
hµz i = Tr
(
|↑ih↑|
−
|↓ih↓|
)
2
e−β γBz ~/2 + eβγBz ~/2
γ~ e−βγBz ~/2 − eβγBz ~/2
γ~
=−
tanh(βγBz ~/2)
2 e−βγBz ~/2 + eβγBz ~/2
2
This result could have been found also by remembering that all average properties of a system can be found from the
ln Z
. The magnetization µz is related to the internal
partition function, in particular the internal energy is hEi = − ∂ ∂β
energy by the simple relation Hz = µz Bz . Then
=
hµz i = −
1 ∂ ln Z
1 ∂Z
γ~Bz 2 sinh (βγBz ~/2)
γ~
=−
=−
=−
tanh(βγBz ~/2)
Bz ∂β
Bz Z ∂β
2 Bz 2 cosh (βγBz ~/2)
2
160GHz
1
At the temperature given, T ≈ 77K, 1/β = kb T ≈ 1600GHz. Then βγBz ~/2 ≈ 21 1600GHz
= 20
. To f rst order
γ~ 1
tanh (0.05) ≈ 0.05. Then the magnetization of one spin at 77K is hµz i = − 2 20 and we need about 20 spins to have
the same amount of magnetization as one spin at zero temperature.
Problem 2:
Electronic Spin: Dynamics
30 points
We consider again an electronic spin-1/2 subjected to an external f eld Bz via the Zeeman interaction.
a) We consider two cases, where the initial state is either what you found in Problem 1, question c or in Problem 1,
question d. The initial state is rotated by 90◦ by the operator Uy = eiπ/4σy , to be aligned with the x-axis and it then
evolves under the Zeeman interaction.
By choosing the most eff cient “picture” (Schrödinger , Heisenberg or interaction picture), calculate hµx (t)i for the
1
two initial states. [Hint: i) What is U eA U † for U unitary? ii) eA Be−A = B + [A, B] + . . . n!
[A, [A, [. . . , B]]].]
Solution:
Since we want to calculate the evolution of an observable (and for different initial states) it is more convenient to adopt
the Heisenberg picture, in which the observables are time-dependent and the states are constant.
First we calculate µx (t) in the Heisenberg picture under the action of the the Zeeman Hamiltonian HZ .
~γ
†
iHz t/~
µx (t) = U (t)µ)x(0)U (t) = e
σx e−iHz t/~
2
Now HZ = ~γBz σz /2 and we can call ω = γBz . Also remember that
eiωtσz /2 σx e−iωtσz /2 = σx cos(ωt) − σy sin(ωt).
1
This could have also been calculated from the formula above, eA Be−A = B + [A, B] + . . . n!
[A, [A, [. . . , B]]], with
A = iωt/2σz and B = σx and the usual commutation relationships of the Pauli matrices. Then
µz (t) =
~γ
(σx cos(ωt) − σy sin(ωt))
2
2
and hµx (t)i = ~γ
2 (hσx i cos(ωt) + sin(ωt) hσy i). We thus need to calculate hσx i and hσy i with respect to the two
initial states (pure state at zero temperature and mixed state). The initial states after the rotation Uy are :
1
Uy |↓i = |+i = √ (|↑i + |↓i),
2
Uy e−βHZ /Z Uy† =
eβ~γBz σx /2
2 cosh (βγBz ~/2)
†
which follows from U eA U † = eUAU and Uy σz Uy† = −σx . Also notice that this state can be written as:
ρx =
eβ~γBz σx /2
1
= [11 + tanh (βγBz ~/2)σx ]
2 cosh (βγBz ~/2)
2
From these states we can calculate hσx i = h+|σx |+i = 1 and Tr {ρx σx } = tanh(βγBz ~/2), while hσy i = 0 for
both states. Notice that these results could have been obtained even by simply noting that after rotating the states,
the expectation value of σx should have been equal to the expectation value of σz (calculate in Problem 1) before the
rotation.
Finally, we have hµx (t)i = γ~
2 cos(γBz t) for the ground state at zero temperature and
γ~
hµx (t)i = 2 tanh(βγBz ~/2) cos(γBz t) for the thermal state.
b) We now want to describe the thermalization process that gives rise to the state you found in Problem 1.d. When
we raise the temperature from T = 0 to T ≈ 77K, the electronic spin will undergo an evolution to reach a new
thermal state under the action of the Zeeman Hamiltonian
with a reservoir. We can represent this
√
√ and the coupling
thermalization process by the Lindblad operators L1 = ασ− , L2 = 1 − ασ+ where σ+ = |0ih1| (σ− = |1ih0|)
and α = 12 [1 − tanh(β~γBz /2)]. Write out the Lindblad equation describing the total evolution of the system.
Solution:
The Lindblad equation is the differential equation:
ρ̇(t) = −i[H, ρ(t)] +
X
Lk ρ(t)L†k
k
1 †
†
− (Lk Lk ρ(t) + ρ(t)Lk Lk )
2
In the specif c case of the problem this becomes:
1
1
1
ρ̇ = −i ~ω[σz , ρ] + α h1|ρ|1i|0ih0| − (|1ih1|ρ + ρ|1ih1|) + (1 − α) h0|ρ|0i|1ih1| − (|0ih0|ρ + ρ|0ih0|)
2
2
2
†
where we used the fact that (σ− )† σ− = (|1ih0|)(|0ih1|) = |1ih1| and σ+
σ+ = |0ih0|.
c) What is ρ0̇0 , where ρ00 = h0|ρ|0i? What is ρ̇11 ? Take the steady-state (SS) limit of the system of equations you
SS
found and calculate ρSS
00 and ρ11 . Compare the result with the state you found in Problem 1.d.
Solution:
We need to project out h0|ρ̇|0i:
1
1
ρ0̇0 = h0| −i ~ω[σz , ρ] |0i + αh0| ρ11 |0ih0| − (|1ih1|ρ + ρ|1ih1|) |0i
2
2
1
+(1 − α)h0| ρ00 |1ih1| − (|0ih0|ρ + ρ|0ih0|) |0i
2
The f rst term is zero, since σz = |0ih0| − |1ih1| yielding h0|[σz , ρ]|0i = h0|(σz ρ − ρσz )|0i = h0|ρ|0i − h0|ρ|0i. The
second term yields αρ11 and the third term −(1 − α)ρ00 . Thus we have:
ρ0̇0 = αρ11 − (1 − α)ρ00
We can calculate ρ̇11 in a similar way, or remember that ρ11 + ρ00 = 1 so that ρ̇11 = −ρ̇00 :
ρ̇11 = −αρ11 + (1 − α)ρ00
We then have the system of equations:
ρ0̇0 = α − ρ00
ρ1̇1 = (1 − α) − ρ11
3
At the steady state, ρ̇ = 0 we obtain ρ00 = α =
1
2 [1
eβ~γBz /2
− tanh(β~γBz /2)] =
e−β~γBz /2
2 cosh(β ~γBz /2)
and ρ11 =
1
2 [1
+
tanh(β ~γBz /2)] = 2 cosh(β ~γBz /2) . These are the same values as for the thermal state in Problem 1. If we can prove
that at the steady state ρ10 = ρ∗01 = 0 then we have recovered the thermal state.
Notice that in the Exam I had written α = 21 [1 − tanh(β~γBz /2)] which would have given a result off by a factor 2
d) Now calculate as well ρ0̇1 and the relative steady-state ρSS
01 to prove that the steady-state is indeed the thermal
state found in Problem 1.d.
Solution:
By taking the projection h0|ρ̇|1i we obtain :
1
1
ρ̇01 = −i~ωρ01 − αρ01 − (1 − α)ρ01 = −(i~ω + 1)ρ01
2
2
Then, at the steady-state ρ01 = 0. Thus the equilibrium state (such that ρ̇ = 0) reached under the action of this
Lindbladian process is indeed the thermal equilibrium.
Problem 3:
Neutron interferometer
20 points
Consider a Mach-Zehnder neutron interferometer such as the one seen in class and in the problem sets. The possible
states of the neutrons are described by its momentum, either |U i for neutrons moving upward or |Di for neutrons
moving downward.
Phase
BS
z
Mirror
BS
U-detector
|ψ〉0=|U〉
|D〉
collimator
D-detector
|ψ〉1
|ψ〉2
|ψ〉3
|ψ〉4
The interferometer components act on the neutrons passing through with the following unitary operators:
iϕ
1
1 1
0 1
e U 0
UBS = √
Umirror =
Uphase =
1 0
0 eiϕD
2 1 −1
where BS stands for beam-splitter. Uphase represents a phase shift that neutrons acquire when passing through a material
sample: the phase is ϕU for neutrons crossing the material while going upward and ϕD for neutrons going downward.
a) We send in a beam of neutrons moving upward, |ψ0 i = |U i. What is the neutron state at each step, 1-4? (see f g.)
Solution:
With simple matrix multiplications we have:
1
ψ1 = √ (|U i + |Di)
2
1
ψ2 = √ (eiϕU |U i + eiϕD |Di)
2
1
ψ3 = √ (eiϕU |Di + eiϕD |U i)
2
1
ψ4 = [eiϕU (|U i − |Di) + eiϕD (|U i + |Di)]
2
4
b) What is the measured contrast? [Hint: we def ne the contrast as the difference in the number of neutrons measured
at the detector U and the detector D that measure neutrons moving upward and downward respectively.]
Solution:
The contrast operator is def ned as C = |U ihU | − |DihD|. We can rewrite |ψi4 as
|ψi4 =
i
1 i(ϕU +ϕD )/2 h i(ϕU −ϕD )/2
(|U i − |Di) + e−i(ϕU −ϕD )/2 (|U i + |Di)
e
e
2
= ei(ϕU +ϕD )/2 [cos [(ϕU − ϕD )/2]|U i − i sin [(ϕU − ϕD )/2]|Di]
From this expression, it is easy to f nd the contrast as
2
2
hCi = cos [(ϕU − ϕD )/2] − sin [(ϕU − ϕD )/2] = cos(ϕU − ϕD ).
c) We now change the sample inside the interferometer so that the neutron will acquire a phase ϕ′U = ϕU − ϕD when
traveling upward and ϕ′D = 0 otherwise. How do your answers to questions a-b change?
Solution:
With this new sample, the state |ψi4 is
|ψi4 =
′
′
1 iϕ′U
1 ′
[e (|U i − |Di) + (|U i + |Di)] = eiϕU /2 [eiϕU /2 (|U i − |Di) + e−iϕU /2 (|U i + |Di)]
2
2
′
Notice that this state is exactly the same as written above to calculate the contrast. Since a global phase eiϕU /2 is
unimportant when calculating expectation values, we f nd the same contrast: hCi = cos (ϕ′U ) = cos(ϕU − ϕD ).
Problem 4:
Faulty neutron interferometer
30 points
Consider the same neutron interferometer as in Problem 3.c, but now assume that the mirrors are faulty: Instead of
ref ecting all the neutrons, they let pass some of them. We can describe the process as follow:
The mirrors are initially in their ground state, |ψimirror = |0i. When one neutron traveling upward impacts the mirror,
it is ref ected with probability p (leaving the mirrors in the ground state) or it continues upward with probability 1 − p,
leaving the mirrors in the state |1i. When one neutron traveling downward impacts the mirror, it is ref ected with
probability p (leaving the mirrors in the ground state) or it continues downward with probability 1 − p, leaving the
mirrors in the state |2i.
a) Describes formally this process giving the rules for the transitions |U i|0imirror → . . . and |Di|0imirror → . . . .
Solution:
The possible transitions described in this process are given by the propagator Unm acting on both neutron and mirror:
p
√
|U i|0imirror → Unm |U i|0imirror = p|Di|0imirror + 1 − p|U i|1imirror
p
√
|Di|0imirror → Unm |Di|0imirror = p|U i|0imirror + 1 − p|Di|2imirror
b) For which values of ϕ′U is the mirror+neutron system entangled (at the step 3)?
Solution:
Entanglement between the mirror and the neutron does not depend on ϕ′U (which def nes only a phase of the neutron
state), but it can depend on p.
′
We calculated |ψi2 = √12 (eiϕU |U i + |Di). Considering now the mirror system as well, we have:
′
1
|Ψi2 = √ (eiϕU |U i|0imirror + |Di|0imirror )
2
5
After the interaction with the mirror, we have:
p
p
′
√
√
1
|Ψi3 = √ [eiϕU ( p|Di|0imirror + 1 − p|U i|1imirror ) + |Di( p|U i|0imirror + 1 − p|Di|2imirror )]
2
p
′
′
1 √
= √ [ p(eiϕU |Di + |U i)|0imirror + 1 − p(eiϕU |U i|1imirror + |Di|2imirror )]
2
This state cannot be written as |ψineutron ⊗ |ϕimirror thus it is entangled. To conf rm this, we can take the partial trace
over the mirror:
1
ρ3 = Trmirror {|Ψ3 ihΨ3 |} = p|ψ3 ihψ3 | + (1 − p)11
2
′
(where |ψ3 i = (eiϕU |Di+ |U
i) is the state found in Problem 3 for perfect mirrors.) This reduced state is a mixed
state, as conf rmed by Tr ρ23 = [p2 + (1 − p)2 /2 + p] = 12 (p2 + 1) < 1.
Notice that in all this calculation, the value of ϕ′U is unimportant and the state is always entangled (unless p = 1 or
p = 0).
c) Write the Kraus operators that describe the faulty mirrors and the evolution |ψ2 i → ρ3 . What is ρ3 ? What type of
process is this Kraus sum describing for p → 0?
Solution:
The Kraus operators are Mk = hk|Unm |0i:
M0 =
and ρ3 =
P2
k=0
√
pσx ,
M1 =
p
1 − p|U ihU |,
M2 =
Mk ρ2 Mk , with ρ2 = |ψihψ|2 =. We obtain:
p
1 − p|DihD|
1
ρ3 = pσx ρ2 σx + (1 − p)(h0|ρ2 |0i |0ih0| + h1|ρ2 |1i |1ih1| = p|ψ3 ihψ3 | + (1 − p)11
2
Notice that the diagonal terms of the density operator are swapped but not reduced in intensity, while the off-diagonal
terms are reduced by an amount p. Thus, as the quality of the mirror decreases, p → 0 and the off-diagonal terms go
to zero; since the phase coherence of the state is lost, the process can be classif ed as a dephasing process. Notice that
it is different then what seen in class, since it is combined with the σx operator, inverting the populations.
d) What is the contrast obtained in this faulty interferometer?
Solution:
The contrast is now given by hCi = Tr {ρ4 C}, where ρ4 = UBS ρ3 UB† S . By linearity,
1
ρ4 = p|ψ4 ihψ4 | + (1 − p)11
2
Then hCi = Tr {Cp|ψ4 ihψ4 |} = p cos(ϕ′U ).
6
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22.51 Quantum Theory of Radiation Interactions
Fall 2012
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