(in)coherent superposition and quantum erasure


by Ameno
Tags: erasure, incoherent, quantum, superposition
Ameno
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#1
Apr20-11, 07:55 AM
P: 16
Hi

I am currently reading John Preskill's Lecture Notes on Quantum Information and Quantum Computation (see http://www.theory.caltech.edu/people...229/index.html near the bottom of the page). I am confused about what he writes in chapter 2.5.4: Quantum erasure.

He starts with the reminder that the state
[tex] \rho_A = \frac{1}{2} Id[/tex]
is a spin in an incoherent superposition of the pure states [tex]|\uparrow_z\rangle_A[/tex] and [tex]|\downarrow_z\rangle_A[/tex], while
[tex] |\uparrow_x, \downarrow_x \rangle = \frac{1}{2} (|\uparrow_z \rangle \pm |\downarrow_z\rangle)[/tex]
is a spin in a coherent superposition, meaning that the relative phase has observable consequences (distinguishes [tex]|\uparrow_x\rangle[/tex] from [tex]|\downarrow_x\rangle[/tex]). In the case of an incoherent superposition, the relative phase is completely unobservable.
Then he says:
Heuristically, the states [tex]|\uparrow_z\rangle_A[/tex] and [tex]|\downarrow_z\rangle_A[/tex] can interfere (the relative phase of these states can be observed) only if we have no information about whether the spin state is [tex]|\uparrow_z\rangle_A[/tex] or [tex]\downarrow_z\rangle_A[/tex]. More than that, interference can occur only if there is in principle no possible way to find out whether the spin is up or down along the z-axis. Entangling spin A with spin B destroys interference, (causes spin A to decohere) because it is possible in principle for us to determine if spin A is up or down along z by performing a suitable measurement of spin B.
So far, I'm happy with this. If we entangle spin A and spin B, the reduced state of A is the fully mixed state, so we have the fewest possible knowledge about A alone and no relative phase between [tex]|\uparrow_z\rangle_A[/tex] and [tex]|\downarrow_z\rangle_A[/tex] that we could observe (no coherence), whereas in the case where spin A is in the state [tex]|\uparrow_x\rangle_A[/tex], we have a relative phase between [tex]|\uparrow_z\rangle_A[/tex] and [tex]|\downarrow_z\rangle_A[/tex] that can be observed by measuring the spin along x (coherence), and since the state is pure, we have maximal knowledge about A, which above is roughly stated as "there is in principle no possible way to find out whether the spin is up or down along the z-axis".

Preskill continues as follows:
But we have now seen that the statement that entanglement causes decoherence requires a qualification. Suppose that Bob measures spin B along the x-axis, obtaining either the result [tex]|\uparrow_x\rangle_B[/tex] or [tex]|\downarrow_x\rangle_B[/tex], and that he sends his measurement result to Alice. Now Alice's spin is a pure state (either [tex]|\uparrow_x\rangle_A[/tex] or [tex]|\downarrow_x\rangle_A[/tex]) and in fact a coherent superposition of [tex]|\uparrow_z\rangle_A[/tex] and [tex]|\downarrow_z\rangle_A[/tex]. We have managed to recover the puritiy of Alice's spin before the jaws of decoherence could close!
I agree. But now comes something that I don't understand:

Suppose that Bob allows his spin to pass through a Stern-Gerlach apparatus oriented along the z-axis. Well, of course, Alice's spin can't behave like a coherent superposition of [tex]|\uparrow_z\rangle_A[/tex] and [tex]|\downarrow_z\rangle_A[/tex]; all Bob has to do is look to see which way his spin moved, and he will know whether Alice's spin is up or down along z. But suppose that Bob does not look. Instead, he carefully refocuses the two beams withouth maintaining any record of whether his spin moved up or down, and then allows the spin to pass through a second Stern-Gerlach apparatus oriented along the x-axis. This time he looks, and communicates the result of his [tex]\sigma_1[/tex] measurement to Alice. Now the coherence of Alice's spin has been restored!
If I understand things correctly (please tell me if I don't), the situation is as follow:
  1. spin A and spin B start in the entangled state [tex]|\psi_i\rangle = \frac{1}{2}(|\uparrow_z \uparrow_z\rangle + |\downarrow_z \downarrow_z\rangle)_{AB}[/tex]
  2. Then Bob performs a spin measurement with respect to the z-axis but without taking notice of the result. This updates the state as follows:
    [tex]
    \rho'_{AB} = (\text{Id}_A \otimes P^+_z) |\psi_i\rangle \langle \psi_i| \text{Id}_A \otimes P^+_z + (\text{Id}_A \otimes P^-_z) |\psi_i\rangle \langle \psi_i| \text{Id}_A \otimes P^-_z
    = \frac{1}{2}(|\uparrow_z \uparrow_z\rangle \langle \uparrow_z \uparrow_z| + |\downarrow_z \downarrow_z\rangle \langle \downarrow_z \downarrow_z|)
    \end{align}[/tex]


    where [tex]P^+_z=|\uparrow_z\rangle \langle \uparrow_z|_B[/tex] and [tex]P^-_z = |\downarrow_z\rangle \langle \downarrow_z|_B
    [/tex]
  3. In the next step, Bob performs a measurement with respect to the x-axis, but this time he records the result, which updates the state as follows:
    either[tex]\rho''_{AB} = \frac{(\text{Id}_A \otimes P^+_x) \rho'_{AB} (\text{Id}_A \otimes P^+_x)}{\text{tr}(\text{Id}_A \otimes P^+_x \rho'_{AB})} = \frac{1}{2}\text{Id}_A \otimes |\uparrow_x\rangle \langle \uparrow_x |_B
    [/tex]
    or[tex]
    \rho''_{AB} = \frac{(\text{Id}_A \otimes P^-_x) \rho'_{AB} (\text{Id}_A \otimes P^-_x)}{\text{tr}(\text{Id}_A \otimes P^-_x \rho'_{AB})} = \frac{1}{2}\text{Id}_A \otimes |\downarrow_x\rangle \langle \downarrow_x |_B
    [/tex]


    where [tex]P^+_x=|\uparrow_x\rangle \langle \uparrow_x|_B[/tex] and [tex]P^-_x = |\downarrow_x\rangle \langle \downarrow_x|_B
    [/tex]
    In either case (if I understand and calculate things correctly), the state of spin A is then the fully mixed state. Do you get the same thing?
So how can one say that the coherence of Alice's state has been restored?
To me, it was surprising to read that claim. If I do things correctly, it is not surprising that her state is fully mixed, because after the first measurement on the entangled state, the joint state of Alice and Bob is a separable state (if no one knows the result) or a product state (if one knows the result), respectively, so another measurement cannot influence Alice's state.


Can you tell me what my mistake is?
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DrChinese
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Apr20-11, 10:02 AM
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Quote Quote by Ameno View Post
Hi

I am currently reading John Preskill's Lecture Notes on Quantum Information and Quantum Computation (see http://www.theory.caltech.edu/people...229/index.html near the bottom of the page). I am confused about what he writes in chapter 2.5.4: Quantum erasure.

He starts with the reminder that the state
[tex] \rho_A = \frac{1}{2} Id[/tex]
is a spin in an incoherent superposition of the pure states [tex]|\uparrow_z\rangle_A[/tex] and [tex]|\downarrow_z\rangle_A[/tex], while
[tex] |\uparrow_x, \downarrow_x \rangle = \frac{1}{2} (|\uparrow_z \rangle \pm |\downarrow_z\rangle)[/tex]
is a spin in a coherent superposition, meaning that the relative phase has observable consequences (distinguishes [tex]|\uparrow_x\rangle[/tex] from [tex]|\downarrow_x\rangle[/tex]). In the case of an incoherent superposition, the relative phase is completely unobservable.
Then he says:


So far, I'm happy with this. If we entangle spin A and spin B, the reduced state of A is the fully mixed state, so we have the fewest possible knowledge about A alone and no relative phase between [tex]|\uparrow_z\rangle_A[/tex] and [tex]|\downarrow_z\rangle_A[/tex] that we could observe (no coherence), whereas in the case where spin A is in the state [tex]|\uparrow_x\rangle_A[/tex], we have a relative phase between [tex]|\uparrow_z\rangle_A[/tex] and [tex]|\downarrow_z\rangle_A[/tex] that can be observed by measuring the spin along x (coherence), and since the state is pure, we have maximal knowledge about A, which above is roughly stated as "there is in principle no possible way to find out whether the spin is up or down along the z-axis".

Preskill continues as follows:


I agree. But now comes something that I don't understand:



If I understand things correctly (please tell me if I don't), the situation is as follow:
  1. spin A and spin B start in the entangled state [tex]|\psi_i\rangle = \frac{1}{2}(|\uparrow_z \uparrow_z\rangle + |\downarrow_z \downarrow_z\rangle)_{AB}[/tex]
  2. Then Bob performs a spin measurement with respect to the z-axis but without taking notice of the result. This updates the state as follows:
    [tex]
    \rho'_{AB} = (\text{Id}_A \otimes P^+_z) |\psi_i\rangle \langle \psi_i| \text{Id}_A \otimes P^+_z + (\text{Id}_A \otimes P^-_z) |\psi_i\rangle \langle \psi_i| \text{Id}_A \otimes P^-_z
    = \frac{1}{2}(|\uparrow_z \uparrow_z\rangle \langle \uparrow_z \uparrow_z| + |\downarrow_z \downarrow_z\rangle \langle \downarrow_z \downarrow_z|)
    \end{align}[/tex]


    where [tex]P^+_z=|\uparrow_z\rangle \langle \uparrow_z|_B[/tex] and [tex]P^-_z = |\downarrow_z\rangle \langle \downarrow_z|_B
    [/tex]
  3. In the next step, Bob performs a measurement with respect to the x-axis, but this time he records the result, which updates the state as follows:
    either[tex]\rho''_{AB} = \frac{(\text{Id}_A \otimes P^+_x) \rho'_{AB} (\text{Id}_A \otimes P^+_x)}{\text{tr}(\text{Id}_A \otimes P^+_x \rho'_{AB})} = \frac{1}{2}\text{Id}_A \otimes |\uparrow_x\rangle \langle \uparrow_x |_B
    [/tex]
    or[tex]
    \rho''_{AB} = \frac{(\text{Id}_A \otimes P^-_x) \rho'_{AB} (\text{Id}_A \otimes P^-_x)}{\text{tr}(\text{Id}_A \otimes P^-_x \rho'_{AB})} = \frac{1}{2}\text{Id}_A \otimes |\downarrow_x\rangle \langle \downarrow_x |_B
    [/tex]


    where [tex]P^+_x=|\uparrow_x\rangle \langle \uparrow_x|_B[/tex] and [tex]P^-_x = |\downarrow_x\rangle \langle \downarrow_x|_B
    [/tex]
    In either case (if I understand and calculate things correctly), the state of spin A is then the fully mixed state. Do you get the same thing?
So how can one say that the coherence of Alice's state has been restored?
To me, it was surprising to read that claim. If I do things correctly, it is not surprising that her state is fully mixed, because after the first measurement on the entangled state, the joint state of Alice and Bob is a separable state (if no one knows the result) or a product state (if one knows the result), respectively, so another measurement cannot influence Alice's state.

Can you tell me what my mistake is?
Although I don't believe this experiment has ever been performed, due to difficulties in arranging the erasure, I believe the result given by Preskill is correct. The proper application is to consider the final context and ignore any intermediate results which are considered "unobserved".

This a fascinating property of a quantum system. An observation result is erased, returning the system to an earlier state (which is a superposition in this particular case).
DrChinese
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Apr20-11, 10:09 AM
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A photon example similar to the above can be seen here:

http://www.optics.rochester.edu/~str...ester/UR19.pdf

And a review of Hardy's paradox is probably relevant as well. That centers around making statements about a system when there is no observed result.

Physics Monkey
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Apr20-11, 07:22 PM
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(in)coherent superposition and quantum erasure


I haven't looked very closely, but it looks like your point 2 may be in error.

When you use a stern-gerlach apparatus to measure the spin, you entangle the spin degree of freedom with a position degree of freedom. You get something like [tex] (a | + \rangle + b | - \rangle ) |\mbox{no deflection} \rangle \rightarrow a | + \rangle | \mbox{deflected up} \rangle + b | - \rangle |\mbox{deflected down} \rangle [/tex]. Once you include the effects of the environment you discover the appearance of decoherence and the emergence of classical values.

When Preskill says that one should carefully refocus the beams, he presumably means that one should carefully reverse the procedure (time evolution) that generated the entanglement between spin and position. This must be down before the environment e.g. the experimenter, background gas molecules, etc. cause the superposition to decohere. If this can be achieved then the state of the system returns to something like [tex] (a | + \rangle + b | - \rangle ) |\mbox{no deflection} \rangle [/tex] without macroscopic entanglement and the "measurement" is undone. Of course, this careful reversing procedure may be quite challenging in practice with a beam of atoms. Photons are probably much easier to work with for these purposes as Dr. Chinese notes.

I leave it to you to include the effects of other entangled spins, but I think Preskill must have something like this in mind.

Hope this helps.
Ameno
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#5
Apr22-11, 07:59 AM
P: 16
Thank you very much for your answers. Now I think I know what the crux of the matter could be, although there are still things I don't get.

Let's discuss point 2 (the measurement with respect to the z-axis) again. The descriptions of the measurement without taking notice of the result that you, Physics Monkey, and I give, differ significantly. Let's compare them.

If we take all three degrees of freedom that are involved (spin A, spin B, position B) into account, then your description says that the "blind measurement" entangles the position degree of freedom with the spins:
[tex]\frac{1}{\sqrt{2}}(|\uparrow_z \uparrow_z\rangle + |\downarrow_z \downarrow_z\rangle) \rightarrow \frac{1}{\sqrt{2}} (|\uparrow_z \uparrow_z\rangle|\text{deflected up}\rangle + |\downarrow_z \downarrow_z\rangle |\text{deflected down}\rangle)[/tex]
Right? This is a pure, entangled state which is a coherent superposition of up and down spins.

I would have said that it updates the state as follows:
[tex]\frac{1}{2}(|\uparrow_z \uparrow_z\rangle + |\downarrow_z \downarrow_z\rangle)(\langle \uparrow_z \uparrow_z | + \langle \downarrow_z \downarrow_z |) \rightarrow \frac{1}{2} (|\uparrow_z \uparrow_z\rangle \langle \uparrow_z \uparrow_z | \otimes |\text{deflected up}\rangle \langle \text{deflected up}| + |\downarrow_z \downarrow_z \rangle \langle \downarrow_z \downarrow_z | \otimes |\text{deflected down}\rangle \langle \text{deflected down}|)[/tex]
This is significantly different: This measurement procedure does not result in an entangled state, but in a separable state (which here isn't pure since we do not know the measurement result). So in this description, the measurement can be seen as generating a "classical" probability distribution with values one half and one half, and incoherently superponing product states according to this probability distribution. The state decoheres in this description.

I thought that this is what a measurement does: It decoheres the state. The corresponding update rule above is what one gets if one takes a probabilistic mixture over vonNeumann-Lüders-projections. The collapse of the wave function takes place.
On the other hand, I thought that entanglement arises when a system evolves according to a Hamiltonian that has a coupling term. But: Isn't it more or less explicitely said that a measurement is performed? Isn't a Stern-Gerlach device a measurement device? Well, OK, the Stern-Gerlach Hamiltonian indeed has a coupling term:
[tex]H_\text{coupling} = -\lambda \mu \textbf{z} \sigma_3[/tex]
Argh. I am confused at this point. Which of the two is correct? Is this "blind measurement" an evolution with coupling or a measurement? If it is a coupled evolution, is it simply because "no one looks at the particles position?" Is this the crux of quantum erasure? Does this somehow have to do with the measurement problem?

If so, I am even more confused. I thought that the measurement problem is "merely" an interpretational issue, not having influence on observable predictions. But if it makes the difference between the two update rules above, it makes the difference between two different density operators, and therefore between different observable predictions. The predicted probabilities coincide for "looking at the spin in z-direction", but not for other observables.

Do you see what I'm confused about?
Physics Monkey
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Apr24-11, 07:02 PM
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You are right to say that measurement involves an interaction between various degrees of freedom, for example, in the stern-gerlach apparatus you let spin and position interact. This leads to entanglement between spin and position degrees of freedom. However, it is also important that the position observable is readily coupled to the environment and it is the environment which provides decoherence. Stray photons, gas molecules, screens, etc all serve to entangle the position degree of freedom with a large environment leading to classical behavior.

To include the effects of an environment one may write:
[tex] |\mbox{full state}\rangle = a | + \rangle | \mbox{deflected up} \rangle |\mbox{environment up}\rangle+ b | - \rangle |\mbox{deflected down}\rangle |\mbox{environment down}\rangle [/tex]

Everything is still formally coherent, but without access to the environment degrees of freedom, the effective description the atom state becomes the incoherent mixture you wrote.

It may help to imagine a (perhaps slightly artificial) multi-step process. Let your atom with spin sit in an ultra high vacuum. First you apply a spatially varying magnetic field to entangle spin and position. At this point the state is a coherent superposition of + up and - down. Now on a much slower time scale the extremely dilute background gas molecules interact with the atom. This interaction leads to correlation between the background gas (environment) and the atom position. Unless one is able to precisely control the state of the background gas molecules, this interaction eventually decoheres the state. Of course, such decoherence can also be achieved by forcing the atom to smash into a macroscopic screen or even by the fluctuations of the electromagnetic field.

Thus I imagine Preskill's phrase about delicately refocusing the beams is meant to convey the need to:
1) disentangle spin and position degrees of freedom
2) avoid a record of the spin state being imprinted on the environment

Does this help?
Ameno
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#7
Apr25-11, 03:17 AM
P: 16
Yes, this really helps!

Quote Quote by Physics Monkey View Post
To include the effects of an environment one may write:
[tex] |\mbox{full state}\rangle = a | + \rangle | \mbox{deflected up} \rangle |\mbox{environment up}\rangle+ b | - \rangle |\mbox{deflected down}\rangle |\mbox{environment down}\rangle [/tex]

Everything is still formally coherent, but without access to the environment degrees of freedom, the effective description the atom state becomes the incoherent mixture you wrote.
This is the kind of explanation I was looking for. It explains how an observer can effectively see an incoherent superposition without talking about wave function collapse. It also explains why quantum erasure is difficult to be produced experimentally: Either one has to isolate the system from the environment or one has to include the environment in the unitary inverse that erases the information.

I'm not yet familiar with the concept of (de)coherence, but I think the considerations above are a good point to start with.

Thanks a lot, and have a good Easter Monday!


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