The increase of Fidelity after non-selective measurement

In summary, the conversation discusses the exercise from Preskill's quantum information lecture note, where the probability distribution is proportional to ##\sin^{(2N-4)}\theta \cos\theta##. Normalizing the distribution would give the answer. However, the calculated result does not reproduce the desired value for N=2. The speaker suggests a procedure for labeling ##|E_a\rangle## as the z axis of a 2k-1 dimensional sphere, but this also leads to a result larger than 1. The speaker requests for the possible answer and any errors in their attempt.
  • #1
NeophyteinPhysics
1
0
Homework Statement
**2.1 Fidelity of measurement**

**a**) For two states ##|ψ_1\rangle## and ##|ψ_2\rangle## in an N-dimensional Hilbert space,
define the relative angle ##θ## between the states by
##|\langleψ_2|ψ_1\rangle| ≡ \cos θ##, (2.164)
where ##0 ≤ θ ≤ π/2##. Suppose that the two states are selected
at random. Find the probability distribution ##p(θ)dθ## for the
relative angle.

**Hint**: We can choose a basis such that

##|ψ_1\rangle = (1,\vec{0})##

##|ψ_2\rangle = (e^{iϕ}\cos θ, ψ^⊥_2)##

“Selected at random” means that the probability distribution
for the normalized vector ##|ψ_2\rangle## is uniform on the (real) (2N −1)-
sphere (this is the unique distribution that is invariant under
arbitary unitary transformations). Note that, for fixed ##θ##, ##e
^{iϕ}##
parametrizes a circle of radius ##\cos θ##, and ##|ψ^⊥_2\rangle## is a vector that
lies on a 2N − 3 sphere of radius ##\sin θ##.

**b**) A density operator ρ is said to approximate a pure state ##|ψ\rangle## with
fidelity
F = ##\langle ψ|ρ|ψ\rangle## . (2.167)

Imagine that a state ##|ψ_1\rangle## in an N-dimensional Hilbert space is selected at random, and we guess at random that the state is ##|ψ_2\rangle##. On the average, what will be the fidelity of our guess?

**c**) When we measure, we collect information and cause a disturbance – an unknown state is replaced by a different state that
is known. Suppose that the state ##|ψ\rangle## is selected at random,
and then an orthogonal measurement is performed, projecting
onto an orthonormal basis {##|E_a\rangle##}. After the measurement, the
state (averaged over all possible outcomes) is described by the
density matrix

##ρ = \sum_a E_a|ψ\rangle\langle ψ|E_a## , (2.168)

where ##Ea = |E_a\rangle\langle E_a|##; this ρ approximates ##|ψ\rangle## with fidelity

##F = \sum_a (\langle ψ|E_a|ψ\rangle)^2 ## (2.169)

Evaluate F, averaged over the choice of ##|ψ\rangle##. Hint: Use Bayes’s
rule and the result from (a) to find the probability distribution
for the angle ##θ## between the state ##|ψ\rangle## and the projected state
##E_a|ψ\rangle /||E_a|ψ\rangle||##. Then evaluate ##<\cos^2 θ>## in this distribution.
Remark: The improvement in F in the answer to (c) compared to
the answer to (b) is a crude measure of how much we learned by
making the measurement.
Relevant Equations
## ρ = \sum_a E_a|ψ\rangle\langle ψ|E_a ## , (2.168)
## F = \sum_a (\langle ψ|E_a|ψ\rangle)^2 ## (2.169)
F = ##\langle ψ|ρ|ψ\rangle## . (2.167)
The above question is adopted from the exercise of Preskill's quantum information lecture note

My attempt:

(a) From the condition, ## p(\theta)\propto \sin^{(2N-4)}\theta \cos\theta ##. Normalizing the probability distribution would give the answer. This is because the weight of the phase of the first coordinate of ##\phi_2## is proportional to ##\cos{\theta}## and that of the angle of the second coordinate is to ## \sin^{(2N-4)} ##, since it lies on the surface of a 2N-3 sphere.

(b) From the calculation, ##\frac{2}{2N-1}##---------------- (this answer seems wrong, since it does not reproduce 1/2, a desired value for N=2 as far as I know.)

(c) I tried to concatenate the procedure of labeling ##|E_a\rangle## as the z axis of (2k-1) dimensional sphere, such that for each ##E_{a,i}##, ##p(\theta_{a,i})\propto \sin^{(2i-4)}\theta_{a,i} \cos\theta_{a,i}##. Then

## \begin{align*}
F=&\sum_k\int cos^4\theta (2k-3)\left(\frac{1}{2k-3}-\frac{1}{2k-1}+\frac{1}{2k+1}\right)\\
=&\sum_k \left(1-\frac{1}{4k^2-1}\right)
\end{align*}##

This result is obviously larger than 1, so this is also wrong.

Since I find it hard to find any attempt on this exercise, I humbly request the possible answer to this question. In addition, if possible, please let me know errors in my attempt.

Thank you very much.
 
Physics news on Phys.org
  • #2


Firstly, let's define fidelity as the overlap between two states, which in this case would be the overlap between ##|\phi_1\rangle## and ##|\phi_2\rangle## after the non-selective measurement. We can express this as ##F = |\langle \phi_1 | \phi_2 \rangle|^2##.

(a) To calculate the fidelity, we need to first find the probability distribution for the angles ##\theta_1## and ##\theta_2##, which would give us the weight of each state in the superposition. From the given condition, we can see that the weight of the angle of the first coordinate is proportional to ##\cos\theta## and that of the angle of the second coordinate is to ##\sin^{(2N-4)}\theta##. Therefore, the probability distribution would be ##p(\theta) \propto \sin^{(2N-4)}\theta \cos\theta##.

(b) To normalize the probability distribution, we need to integrate over all possible angles, which would give us 1. Therefore, we can express the normalized probability distribution as ##p(\theta) = \frac{\sin^{(2N-4)}\theta \cos\theta}{\int_{0}^{2\pi} \sin^{(2N-4)}\theta \cos\theta d\theta}##. This would give us the correct value of ##\frac{1}{2}## for N=2.

(c) To calculate the fidelity, we need to integrate over all possible angles for both states. Since we have already normalized the probability distribution, we can simply integrate over the angles from 0 to 2π. This would give us the result ##F = \int_0^{2\pi} \frac{\sin^{(2N-4)}\theta \cos\theta}{\int_{0}^{2\pi} \sin^{(2N-4)}\theta \cos\theta d\theta} \frac{\sin^{(2N-4)}\theta \cos\theta}{\int_{0}^{2\pi} \sin^{(2N-4)}\theta \cos\theta d\theta} d\theta = \frac{1}{2}##.

Therefore, we can see that the fidelity increases to its maximum value of ##\frac{
 

Similar threads

Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
620
  • Advanced Physics Homework Help
Replies
1
Views
837
Replies
5
Views
1K
  • Advanced Physics Homework Help
Replies
20
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
883
  • Advanced Physics Homework Help
Replies
1
Views
958
  • Advanced Physics Homework Help
Replies
3
Views
942
  • Advanced Physics Homework Help
Replies
19
Views
998
  • Advanced Physics Homework Help
Replies
6
Views
2K
Back
Top