Solutions to problem set 4 - Universitetet i oslo · FYS 4110 Modern Quantum Mechanics, Fall...

FYS 4110 Modern Quantum Mechanics, Fall Semester 2019 1

Solutions to problem set 4

4.1 Density operators

a) We have

〈σi〉 = Tr ρσi =1

2Tr(1 + rjσj)σi =

1

2rj Tr(1δij + iεjikσk) = ri (1)

b)

ρ = p1ρ1 + p2ρ2 =1

2(1 + r1 · σ) +

1

2(1 + r2 · σ) =

1

2[1 + (p1r1 + p2r2) · σ]

c) We know that the pure states are on the surfae of the Bloch sphere. Any mixed state will have aBloch vector of the form r = p1r1 + p2r2 with r1 and r2 unit vectors. This means that r is on theline between r1 and r2, at a position given by the probabilities, and therefore it is inside the Blochsphere.

d) We use the matrix representation:

ρ =1

2(1 + r · σ) =

1

2(1 + xσx + yσy + zσz) =

1

2

(1 + z x− iyx+ iy 1− z

)=

(ρ11 ρ12

ρ21 ρ22

)(2)

e) The density matrix is given by:

ρ =∑k

pk|ψk〉〈ψk| =1

3(| ↑x〉〈↑x |+ | ↑y〉〈↑y |+ | ↑z〉〈↑z |) (3)

where p1 = p2 = p3 = 1/3, is the (classical) probability of the state being up in x, y or zrespectively. One can do the computation of ρ in two ways:

• Matrix representation:

| ↑x〉 =1√2

(11

), 〈↑x | =

1√2

(1 1

)| ↑y〉 =

1√2

(1i

), 〈↑y | =

1√2

(1 −i

)| ↑z〉 =

(10

), 〈↑z | =

(1 0

)Then:

ρ =1

3

[1

2

(1 11 1

)+

1

2

(1 −ii 1

)+

(1 00 0

)]=

1

3

1

2

[(1 11 1

)+

(1 −ii 1

)+

(2 00 0

)]=

1

3

1

2

(4 1− i

1 + i 2

)=

1

2

(1 + 1

313 (1− i)

13 (1 + i) 1− 1

3

)Comparing with (2), we see that r =

(13 ,

13 ,

13

)


• Bra-ket formalism:We start of by expressing | ↑x〉 and | ↑y〉 in the |±〉 basis by solving the eigenvector equations:

σx| ↑x〉 = | ↑x〉 ⇒ (σ+ + σ−) (a|+〉+ b|−〉) = a|+〉+ b|−〉

We see that the only terms surviving σ±|∓〉, our equation becomes:

b|+〉+ a|−〉 = a|+〉+ b|−〉

Then choosing a = b = 1√2

gives

| ↑x〉 =1√2

(|+〉+ |−〉)

Next up:σy| ↑y〉 = | ↑y〉 ⇒ i (σ− − σ+) (a|+〉+ b|−〉) = a|+〉+ b|−〉

By the same argument as earlier, we arrive at:

ai|−〉+ bi|+〉 = a|+〉+ b|−〉

This gives: ai = b⇒ ba = i⇒ a = 1√

2, b = 1√

2i and

| ↑y〉 =1√2

(|+〉+ i|−〉)

Inserting this into (3):

ρ =1

3

1

2(|+〉+ |−〉) (〈+|+ 〈−|) +

1

3

1

2(|+〉+ i|−〉) (〈+| − i〈−|) +

1

3

1

22|+〉〈+|

=1

2

1

3((3 + 1) |+〉〈+|+ (1− i) |+〉〈−|+ (1 + i) |−〉〈+|+ (3− 1) |−〉〈−|)

=1

2

((1 +

1

3

)|+〉〈+|+ 1

3(1− i) |+〉〈−|+ 1

3(1 + i) |−〉〈+|+

(1− 1

3

)|−〉〈−|

)This can be read of as:

ρ =1

2

(1 +

1

3(1, 1, 1) · σ

)where r =

(13 ,

13 ,

13

)The Von Neuman entropy is given by equation 2.24 in the lecture notes, which reads:

S = −1

2log[(1 + r)1+r (1− r)1−r

]+ log 2

where r = |r| = 1√3.

S = −1

2log

[(1 +

1√3

)1+ 1√3(

1− 1√3

)1− 1√3

]+ log 2

≈ 0.744

when choosing log = log2.


f) The unit vector in the direction of r is n = r|r| = 1√

3(1, 1, 1). According to Eq (2.22) of the lecture

notes we have that

ρ =1

2(1 + r)|Ψn〉〈Ψn|+

1

2(1− r)|Ψ−n〉〈Ψ−n|

where |Ψ±n〉 are the eigenstates of σn = n ·σ. If we want the explicit form of these states we canrecall that (Eq (1.136)in the lecture notes)

|Ψn〉 =

(cos θ/2eiφ sin θ/2

)|Ψ−n〉 =

(−e−iφ sin θ/2

cos θ/2

)where (θ, φ) are the polar coordinates for n. With n 1√

3(1, 1, 1) we have cos θ = 1/

√3 and

φ = π/4. Using

cos θ/2 =1√2

√1 + cos θ =

√√3 + 1

2√

3

sin θ/2 =1√2

√1− cos θ =

√√3− 1

2√

3

we get

|Ψn〉 =1√2√

3

( √√3 + 1

eiπ/4√√

3− 1

)|Ψ−n〉 =

(−e−iπ/4

√√3− 1√√

3 + 1

).

4.2 Entropy of a thermal state

a)

ρ =1

Ze−βH , Z = Tr

(e−βH

),

dZ

dβ= −Tr

(e−βHH

)The Von Neumann entropy is given in the lecture notes as:

S = −Tr (ρ ln ρ) = − 1

ZTr[e−βH(−βH − lnZ)

]=β

ZTr[e−βHH

]+

1

ZlnZ) Tr

[e−βH

]= − β

Z

dZ

dβ+ lnZ = −β d

dβlnZ + lnZ. (4)

b)

H = ~ω∞∑n=0

(n+

1

2

)|n〉〈n|, En = ~ω

(n+

1

2

)

N(β)−1 =

∞∑n=0

e−βEn

∞∑n=0

e−β~ω/2e−β~ωn = e−β~ω/2∞∑n=0

e−β~ωn


We recognize∑∞

n=0 e−β~ωn as a konvergent sum of a geometric series.

∞∑n=0

e−β~ωn =1

1− e−β~ω

Thus:

N(β) =1− e−β~ω

e−β~ω/2

= eβ~ω/2 − e−β~ω+β~ω/2

= eβ~ω/2 − e−β~ω/2

= 2 sinhβ~ω

2

Inserting this into (4) gives:

S(β) = βd

dβln

(2 sinh

β~ω2

)− ln

(2 sinh

β~ω2

)

= β

ddβ

(2 sinh β~ω

2

)2 sinh β~ω

2

− ln

(2 sinh

β~ω2

)=β~ω

2

2 cosh β~ω2

2 sinh β~ω2

− ln

(2 sinh

β~ω2

)=

β~ω2

cothβ~ω

2− ln

(2 sinh

β~ω2

)

c) We now write the entropy as a function of x = 2kBT~ω = 2

β~ω so we get:

S(x) =1

xcoth

1

x− ln

(2 sinh

1

x

)where log e→ ln e = 1. The plot for x ∈ [0, 5] is


Evaluating the limit T → 0⇒ x→ 0

limx→0+

S(x) = limx→0+

(1

xcoth

1

x− ln

(2 sinh

1

x

))= lim

x→0+

1

x

(coth

1

x−

ln(2 sinh 1

x

)1x

)

= limx→0+

(1

x

)(limx→0+

coth1

x− limx→0+

ln(2 sinh 1

x

)1x

)

= limx→0+

(1

x

)(limx→0+

coth1

x− limx→0+

ln(2 sinh 1

x

)1x

)

Looking at the last limit, we see that it approaches “∞/∞” and we need to apply L’Hopital’s rule.

limx→0+

ln(2 sinh 1

x

)1x

L′H= lim

x→0+

−2 cosh 1x

2 sinh 1x

1x2

− 1x2

= limx→0+

coth1

x

Inserting back into the limit of S(x):

limx→0+

S(x) = limx→0+

1

x

(coth

1

x− coth

1

x

)= 0

Onto the limit T →∞⇒ x→∞

limx→∞

S(x) = limx→∞

(1

xcoth

1

x− ln

(2 sinh

1

x

))= lim

x→∞

1

xcoth

1

x− limx→∞

ln

(2 sinh

1

x

)Looking at the first limit:

limx→∞

1

xcoth

1

x= lim

y→0+y coth y = lim

y→0+

y cosh y

sinh y

This is a limit that approaches 0/0, and we may use L’Hopital’s rule:

limy→0+

y cosh y

sinh y

L′H= lim

y→0+

cosh y + y sinh y

cosh y= lim

y→0+(1 + y tanh y) = 1

Then,

limx→∞

S(x) = 1 +− limx→∞

ln

(2 sinh

1

x

)︸︷︷︸

=−∞

=∞

So the entropy approaches 0 for T → 0, which makes sense since there is only one possible statethe system can be in the limit, and the system is therefore ordered. In the other limit, there is aninfinite number of states available, so the system has an infinite number of states to ”choose” from,meaning that the entropy approaches infinity.

4.3 Matrix representation of tensor products


a =

(a1

a2

), b =

(b1b2

)|c〉 = |a〉 ⊗ |b〉 ⇒ |c〉 =

∑ij

aibj |ij〉, |ij〉 = |i〉A ⊗ |j〉B

a) We have

c =

c1

c2

c3

c4

=

a1b1a1b2a2b1a2b2

=

(a1ba2b

)(5)

For the basis vectors, we can assume

|1〉A =

(10

), |2〉A =

(01

)And similarily in the B space. We can use the result (5), to have:

|ij〉 = |i〉A ⊗ |j〉B =

(i1ji2j

)Then:

|11〉 =

1000

, |12〉 =

0100

, |21〉 =

0010

, |22〉 =

0001

b) We have

C =

A11B11 A11B12 A12B11 A12B12

A11B21 A11B22 A12B21 A12B22

A21B11 A21B12 A22B11 A22B12

A21B21 A21B22 A22B21 A22B22

=

(A11B A12BA21B A22B

)

c)

σ1 =

(0 11 0

), σ2 =

(0 −ii 0

), σ3 =

(1 00 −1

).

We show three examples

σ1 ⊗ σ2 =

0 0 0 −i0 0 i 00 −i 0 0i 0 0 0

=

(0 σ2

σ2 0

)

σ1 ⊗ σ3 =

0 0 1 00 0 0 −11 0 0 00 −1 0 0

=

(0 σ3

σ3 0

)


σ2 ⊗ σ3 =

0 0 −i 00 0 0 ii 0 0 00 −i 0 0

=

(0 −iσ3

iσ3 0

)

d) We have

Cc =

(A11B A12BA21B A22B

)(a1ba2b

)=

(A11a1Bb +A12a2BbA21a1Bb +A22a2Bb

)We use that A⊗ B|a〉 ⊗ |b〉 = A|a〉 ⊗ B|b〉 and that the matrix representing A|a〉 is

Aa =

(A11a1 +A12a2

A21a1 +A22a2

)Then the matrix representing A⊗ B|a〉 ⊗ |b〉 is(

(A11a1 +A12a2)Bb(A21a1 +A22a2)Bb

)which is the same as Cc

4.4 Entanglement in the Jaynes-Cummings model

a) We have|ψ(t)〉 = c−n (t)|−, n+ 1〉+ c+

n (t)|+, n〉

which gives the density matrix

ρ = |ψ(t)〉〈ψ(t)|= |c−n (t)|2|−, n+ 1〉〈−, n+ 1|+ c−n (t)c+

n (t)∗|−, n+ 1〉〈+, n|+ c−n (t)∗c+

n (t)|+, n〉〈−, n+ 1|+ |c+n (t)|2|+, n〉〈+, n|

Tracing over the photon mode we find

ρTLS =∑m

〈m|ρ|m〉 = |c−n (t)|2|−〉〈−|+ |c+n (t)|2|+〉〈+|.

This is diagonal, and we have the probabilities for the two states

p+ = |c+n (t)|2 = sin2 Ωnt

2sin2 θn

p− = |c−n (t)|2 = 1− sin2 Ωnt

2sin2 θn


The entanglement entropy is

S = −p+ ln p+ − p− ln p−

This is maximal when p+ and p− are as equal as possible. If sin2 θn > 1/2, which means thatθn > π/4, we can get p+ = p− = 1

2 with

Smax = −1

2ln

1

2− 1

2ln

1

2= ln 2.

This will happen when

sin2 Ωnt

2sin2 θn =

1

2

which means

t =2

Ωnarcsin

[1√

2 sin θn

]=

2

Ωnarcsin

[Ωn√2gn

]If sin2 θn < 1/2 we have p+ < 1

2 and maximal when Ωnt2 = π

2 + mπ, with m an integer.p+max = sin2 θn and

Smax = − sin2 θn ln sin2 θn − cos2 θn ln cos2 θn.

b) For the Rabi model (in the rotating frame) we have

|ψ(t)〉 = c0(t)|0〉+ c1(t)|1〉

withc0(t) = cos

Ωt

2+ i sin

Ωt

2cos θ, c1(t) = i sin

Ωt

2sin θ.

This is a pure state and the Bloch vector has components

mRx = 2 Re(c∗0c1) = sin 2θ sin2 Ωt

2

mRy = 2 Im(c∗0c1) = sin θ sin Ωt

mRz = |c0|2 − |c1|2 = 1− 2 sin2 θ sin2 Ωt

2.

For the Jaynes-Cummings model we use

ρTLS =1

2(1 + mJC · σ)

We know thatρTLS = p−|−〉〈−|+ p+|+〉〈+| = 1

2(1 + (p− − p+)σz)

from which we read out

mJCx = mJC

y = 0

mJCz = p− − p+ = 1− 2 sin2 θn sin2 Ωnt

2.


In the Rabi model, the state is always pure, and the Bloch vector presesses in a circle on the surfaceof the Bloch sphere. In the Jaynes-Cummings model, the qubit is entangled with the photon mode,and the reduced density matrix describes a mixed state. The Bloch vector oscillates along the axisof the Bloch sphere with mJC

z = mRz This is shown in the following figure with the Rabi model in

Blue and the Jaynes-Cummings model in red.

c) In the limit n→∞ we have that gn = 2λ√n+ 1 grows. This means that

Ωn =√

∆2 + g2n → gn

and sin θn = gnΩn→ 1. So the amplitude and frequency of the oscillations grow, but the Bloch

vector is always on the axis of the Bloch sphereand entanglement is not reduced.

An idea for a better classical limit is to assume that the photon mode starts in a coherent stateinstead of an energy eigenstate. We know that coherent states are the link to classical mechanicsfor the harmonic oscillator, and we can hope that it will extend to the Jaynes-Cummings model aswell. In the figure the result is shown in green


As we can see, it works to some extent, but it becomes a spiral instead of a circle. Here I usedan average photon number of 9 in the initial state. Maybe it should be bigger for the limit, butnumerics gets slow. More work is needed......

Solutions to problem set 4 - Universitetet i oslo · FYS 4110 Modern Quantum Mechanics, Fall...

Documents

Transcript of Solutions to problem set 4 - Universitetet i oslo · FYS 4110 Modern Quantum Mechanics, Fall...