(in)coherent superposition and quantum erasure

Ameno

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?

DrChinese

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

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

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

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

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

Yes, this really helps!

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!