15.5.1 POVM as a projective measurement in a direct sum
Multiplying the normalized two component vectors of (15.31) byp2/3, we form the 3 × 2 matrix M =˜
p2/3 −p1/6 −p1/6
0 p1/2 −p1/2
Let us call|v1i and |v2i the three-dimensional vectors whose components are the first rows of the matrix M . They obey˜ ||v1||2=||v2||2= 1 andhv1|v2i = 0. We complete the matrix ˜M to a 3× 3 matrix M by adding a third row made of the components of a vector|v3i
|v3i = (p1/3,p1/3, p1/3)
which is normalized and orthogonal to |v1i and |v2i. By construction, M is and orthogonal matrix, so that the vectors |uαi whose components are given by the columns of M are normalized and mutually orthogonal. Now, let us consider a projective measurement inH(3)
P1=|u1ihu1| P2=|u2ihu2| P3=|u3ihu3|
Assume that the state operator ρ has non vanishing matrix elements only in H(2). Then the probabilty of result α is
p(α) = Tr (ρ|uαihuα| = huα|ρ|uαi = h˜α|ρ|˜αi
15.5.3 A POVM with two arbitrary qubit states
1. The two projectors are Pa⊥ =
sin2α − sin α cos α
− sin α cos α cos2α
Pb⊥ =
cos2α − sin α cos α
− sin α cos α sin2α
To build a POVM with a third vector|ci we must have
A +|λ|2B −2 sin α cos α + Bλµ∗
−2 sin α cos α + Bλ∗µ A +|µ|2B
= I
in order that (15.23) be satisfied, whence|λ| = |µ| = 1/√
2. Choosing λ = µ = 1/√
2 we obtain
A = 1
1 + sin 2α = 1 1 + S B = 2 sin 2α
1 + sin 2α = 2S 1 + S
95
The POVM is
Qa⊥ = 1
1 + S|a⊥iha⊥| Qb⊥ = 1
1 + S|b⊥ihb⊥|
Qc = S
1 + S|cihc|
2. The state operator of the qubits sent by Alice is ρ = 1
2|aiha| +1 2|bihb|
If Bob measures the result a⊥, he knows with certainty that the spin was in the state |bi, because he would have got zero had the spin been in state|ai. The probability for finding a⊥ is
p(a⊥) = Tr (Qa⊥ρ) = 1
1 + STr (|a⊥iha⊥|ρ) =1
2(1− S)
If sin α = 1/2, then p(a⊥) = p(b⊥) = 1/4 and in 50% of the cases Bob will make the right guess. In quantum cryptography, Eve uses α = π/8, so that S = 1/√
2, (1− S)/2 ≃ 0.145. Thus, if she uses a POVM, Eve can be sure of the state sent by Alice in 58% of the cases. In the remaining 42%, she decides randomly, with a 50% probality of success. Thus she will get the correct result in (58+21)%=79% of the cases.
15.5.7 Superposition of coherent states
1. The term [H0, ρ] does not contribute to the evolution of ρnnbecause H0is diagonal in the{|ni} basis.
Furthermore
hn|a†aρ + ρa†a|ni = 2nρnn
hn|aρa†|ni = (n + 1)ρn+1,n+1 whence the time evolution of ρnn
dρnn
dt =−nΓρnn+ (n + 1)Γρn+1,n+1
If we choose n = 0, we find dρ/dt = Γρ11, which means that the population of the ground state increases at a rate proportional to that of the first excited state times Γ. The corresponding physical process is the spontaneous emission of a photon, so that Γ is the rate for spontaneous emission. The evolution equation for ρn+1,n is obtained from
hn + 1|[H0, ρ]|ni = ~ω0ρn+1,n
hn + 1|aρa†|ni = p(n + 1)(n + 2) ρn+2,n+1
hn + 1|{a†a, ρ}|ni = (2n + 1)ρn+1,n+1 so that
dρn+1,n
dt =−iω0ρn+1,n+ Γp(n + 1)(n + 2) ρn+2,n+1−1
2Γ(2n + 1)ρn+1,n
2. Using (2.54) we obtain
eλ∗aa†e−λ∗a= a†+ λ∗[a, a†] = a†+ λ∗ that is
a†e−λ∗a = e−λ∗a
a†+ λ∗
97
where we have used the invariance of the trace under circular permutations to derive the last line. This equation can be written schematically as
∂ which can be rewritten as
3. Let us examine the different terms in the RHS of (15.79). From the results of the preceding question a†aρ → ∂ Assembling all these results, we finally get the partial differential equation
∂ To implement the method of characteristics we write
dt
or solving for λ0, λ∗0
λ0= λ exp[−(Γ/2 − iω0)t] λ0= λ∗exp[−(Γ/2 + iω0)t]
λ exp[−(Γ/2 − iω0)t] and λ exp[−(Γ/2 + iω0)t] are constants along the characteristics. The partial differ-ential equation for C(λ, λ∗; t) tells us that this function is constant along the characteristics
C(λ, λ∗; t) = C0(λ, λ∗; t = 0) = C0(λ, λ∗)
4. The state operator at time t = 0 is C0(λ, λ∗) = Tr
|zihz| eλa†e−λ∗a
=hz|eλa†e−λ∗a|zi = exp(λz∗− λ∗z) We then have at time t
C(λ, λ∗; t) = exph
z∗λe−(Γ/2−iω0)t, λ∗ze−(Γ/2+iω0)ti which can be written as
C(λ, λ∗; t) = exp [λz∗(t)− λ∗z(t)]
with
z(t) = ze−(Γ/2+iω0)t Thus C(λ, λ∗; t) corresponds to the coherent state
|z(t)i = |z e−iω0te−Γt/2i
5. When|Φi is a superposition of coherent states
|Φi = c1|z1i + c2|z2i then
C(λ, λ∗; t = 0) =|c1|2e(λz1∗−λ∗z1)+|c2|2e(λz∗2−λ∗z2)+ c1c∗2hz2|z1ie(λz2∗−λ∗z1)+ c∗1c2hz1|z2ie(λz∗1−λ∗z2) The last two terms originate in the fact that|Φi is a coherent superposition, while these two terms would be absent in an incoherent superposition of the two coherent states. At time t we have
C(λ, λ∗; t) = |c1|2e[λz∗1(t)−λ∗z1(t)]+|c2|2e[λz∗2(t)−λ∗z2(t)]
+ c1c∗2hz2|z1ie[λz2∗(t)−λ∗z1(t)]+ c∗1c2hz1|z2ie[λz∗1(t)−λ∗z2(t)]
Note that the scalar products in the last two terms are hz2|z1i and hz1|z2i, and not hz2(t)|z1(t)i and hz1(t)|z2(t)i. In order to recover the same form as at t = 0 form, we must write, for example
hz2|z1i = hz2|z1i
hz2(t)|z1(t)ihz2(t)|z1(t)i = η(t)hz2(t)|z1(t)i
We can then describe the final state as a linear superposition of two coherent states, but the coherence is reduced by a factor
|η(t)| = exp
−1
2|z1− z2|2 1− e−Γt
≃ exp
−Γ
2|z1− z2|2
The coherence is then damped with a rate which is |z1− z2|2 larger that the damping rate Γ of the individual coherent states. The decoherence time is then
τdec= 2 Γ|z1− z2|2
99 5. At time t = 0, we have for the oscillator the superposition
|Φ(t = 0)i = c1|0i + c2|zi The global state vector at t = 0 is
|Ψ(t = 0)i = c1|0 ⊗ |0Fi + c2|z ⊗ 0Fi
where|0Fi is the state vector (vacuum state) of the radiation field, since at T = 0 there are no available photons (or phonons). The first component of |Ψi stays unchanged under the time evolution, because spontaneous emission cannot exist. By contrast, a photon will be emitted on average after a time∼ Γ|z|2 due the second component of |Ψi. Indeed, the time evolution of ρnn in question 1 tells us that the decay amplitude of an excited state|ni is nΓ, and the average number hni is equal to |z|2,|z|2=hni in the coherent state |zi. As soon as one photon is emitted, the two components of |Φi become entangled to orthogonal states of the environment, and the reduced state matrix of the oscillator loses all phase coherence. The decoherence time is thus the average time for the emission of a single photon, and τdec ≃ 1/|z|2Γ.
15.5.11 The Fokker-Planck-Kramers equation for a Brownian particle
1. Let us show the equivalence of the two formulae for w(x, p; t) w(x, p; t) = 1 We use the completeness relation and (9.22) to write
|x −y
and an analogous formula forhx + y/2|. Plugging the two formulae in (W1) and integrating over y leads to
2. Let us consider a Gaussian wave packet (Exercise 9.7.3)
ϕ(x) =
A straightforward calculation shows that w(x, p) = 1
Let us now consider the superposition (15.143)
ϕ(x)≃ 1
The first two terms would correspond to an incoherent superposition of two Gaussian wave packets, but the last one reflects the coherence of the two wave packets.
3. We limit ourselves to the second term [X,{P, ρ}] in the RHS of (15.142), as the other two terms can be dealt with using exactly the same techniques. Using the (W1) form of w we obtain at once
w2 = 1
2π~
Z +∞
−∞
e−ipy/~hx +y
2|{P, ρ}|x −y 2i ydy
= i
2π
∂
∂p Z +∞
−∞
e−ipy/~hx +y
2|{P, ρ}|x − y 2idy We then use the W2 form
w2 = 1
2π~
∂
∂p Z +∞
−∞
e−ipy/~hx +y
2|{P, ρ}|p −y 2idy
= i
2π
∂
∂pp Z +∞
−∞
e−ixy/~hp + y
2|{P, ρ}|x − y 2idy
= i~ ∂
∂p[pw(x, p; t)]
4. One has only to observe that w(x, p; t) must vanish when x→ ±∞.