Weak measurement and the measurement problem

  • #1
302
30
Summary:
Why doesn't weak measurement solve the measurement problem experimentally? Formulating the question on the example of a Bell-type experiment with weak measurement.
When I think about it my biggest issue with the interpretation of QT can be boiled down to this question. There are the axioms of measurement in QT, which have their limiting nature and claim to be the only way to extract information from a quantum system - but that is just a claim with enough reasons to challenge it (it's science after all). So I guess what I don't understand is how the theory dodges that bullet. I try to formulate my issue via an example:

Let's take Bell type experiment where two entangled photons are generated, one moves left one right. For simplicity let's say that we know the exact state of the generated photons and let's assume the left one is up, its partner is in a down state along a preset axis. The right part of the experiment shall be just the default detector setup ##S_R## set along a configurable axis (R stands for Robert, because i don't like Bob).

Now for the left photon we try to do a something akin to a weak measurement using a two-photon interference. For that purpose let's call the left photon the target and introduce another photon with same wavelength which we call the probe (the ancilla in terms of weak measurement). Let's assume it is possible to synch them up perfectly to enable an interference between both to occur. Furthermore the probe photon state can be configured to have any polarization. Now the idea is to go over all probe polarization angles and record the angular distribution of coincidences of detecting both photons (target and probe) in the same detector (i.e. measuring the interference).

The way the Hond Ou Mandel effect is described, i would now expect the coincidences to peak whenever the probing polarization is parallel to the polarization the target was originally produced with. So the interference can be seen as a measure of how indistinguishable two photon states are. As far as I understand the effect, even if the targets originated from an ensemble composed of two or more different states, the interference would of course weaken with the number of states but show two or more peaks at angles parallel to those bases for each of the states which renders it pure allowing us to reconstruct the ensembles composition.

Now let's now look at the classical QT treatment of measurement for the right part of the experiment. To cut to the chase, in the formalism of measurement, the detection of the right photon would change the global state such that the target photon would be changed to be in a eigenstate along R's axis and also turns into an ensemble of two. And here is a problem: while a probability distribution of a classical measurement of the target left photon would be insensitive to that, particle interaction generally is and so is interference since both interact with the actual state, not its projection to probabilities. Doing the two photon interference calculation with the updated state now gives a peak around a different angle. Expressed via operators it would seem the projection operator to the double photon interference state does not commute with the operator for ##S_R##. The thing is that because we knew everything about the probe photon there was two know, measuring it only gives us additional information about it's short affair it had with the target which we wouldn't get otherwise - to put it in words true to the origin meaning of ancilla.

So on the one hand it would seem this would allow us to put the axioms of measurement to an experimental test and be done with the measurement problem. Measure measurement. On the other hand I know what that would imply specifically as this is a Bell-type experiment (which is ironically the easiest setup i could find to express my thoughts). So how does the QT formalism dodge that bullet of not having a "spooky action at a distance"?
 

Answers and Replies

  • #2
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
19,518
10,278
I'm not sure that your example is very illuminating, because it's obviously a non-entangled two-photon state,
$$|\Psi \rangle=\hat{a}^{\dagger}(\vec{k},\lambda=1) \hat{a}^{\dagger}(-\vec{k},\lambda=2)|\Omega \rangle.$$
Since here is no entanglement the polarization states of the two photons are predetermined, i.e., the photon going in direction ##\vec{k}## is horizontally (labelled with ##\lambda=2##) and that going in direction ##-\vec{k}## is vertically (\labelled with ##\lambda=-1##) polarized. Here ##|\Omega \rangle## is the vacuum state and ##\hat{a}^{\dagger}(\vec{k},\lambda)## a creation operator of a photon with momentum ##\vec{k}## and two linear-polarization states labelled with ##\lambda \in \{1,2 \}##.

For a Bell experiment you should have entangled photons, e.g., one in the polarization-singlet state, i.e.,
$$|\Psi' \rangle=\frac{1}{\sqrt{2}} \left [\hat{a}^{\dagger}(\vec{k},\lambda=1) \hat{a}^{\dagger}(-\vec{k},\lambda=2) - \hat{a}^{\dagger}(\vec{k},\lambda=2) \hat{a}^{\dagger}(-\vec{k},\lambda=1) \right]|\Omega \rangle.$$
 
  • Like
Likes Demystifier and Killtech
  • #3
302
30
For a Bell experiment you should have entangled photons, e.g., one in the polarization-singlet state, i.e.,
$$|\Psi' \rangle=\frac{1}{\sqrt{2}} \left [\hat{a}^{\dagger}(\vec{k},\lambda=1) \hat{a}^{\dagger}(-\vec{k},\lambda=2) - \hat{a}^{\dagger}(\vec{k},\lambda=2) \hat{a}^{\dagger}(-\vec{k},\lambda=1) \right]|\Omega \rangle.$$
you are of course absolutely right. I was a little bit too occupied with setting up the HOM interference that I forgot to properly think about the initial target photons state. Even so, it seems the entangled state still is suited to produce a HOM interference for as long as the initial polarization isn't completely arbitrary.
 
  • #4
Demystifier
Science Advisor
Insights Author
Gold Member
12,437
4,782
In the HOM experiment, it's not that one photon interferes with another photon. Instead, it's that one 2-photon state interferes with another 2-photon state.

More importantly, it's not clear to me what exactly is the problem (which you call the "measurement problem") that you try to solve.
 
  • #5
302
30
For a Bell experiment you should have entangled photons
More importantly, it's not clear to me what exactly is the problem (which you call the "measurement problem") that you try to solve.
I think I figured out how to correct my setup to make sense in terms of my question. Initially I mistook a superposition for an ensemble which made me look for a different initial state.

Because if the initial state is entangled in the way vanhees wrote, the HOM interference will just yield an coincidence distribution uniform in the polarization angle - so no interference at all. However, as it is a superposition state, then in theory the probing photon can be produced to be in the identical state, too (i.e. it needs to be similarly entangled). Now indeed the HOM interference should reappear, albeit I won't have useful parameters to play with for the probing photon. Therefore the setup now becomes only a check for the persistence of entanglement.

The general idea is to try to isolate the aspect/information that is affected by measurement in theory and also take advantage of the theory describing that aspect/information to takes part in interactions, therefore using an interaction as an sensor for that change. My thought was that in probability theory and the theory of stochastic processes you can classify information whether it interacts by its the time evolution of its probability being non linear. That means it cannot be though as independent of the described system and is therefore an irreducible aspect of the real state of the system, which in principle by that nature can be measured at least indirectly.
 

Related Threads on Weak measurement and the measurement problem

  • Last Post
Replies
0
Views
667
P
  • Last Post
Replies
6
Views
873
  • Last Post
Replies
20
Views
5K
  • Last Post
Replies
10
Views
7K
Replies
5
Views
1K
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
3
Views
2K
Replies
17
Views
7K
Replies
3
Views
286
Top