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I am trying to understand the concept of basis change for a pair of entangled particles (in mathematical sense as well as what it means in case of detection on detectors in a Mach-Zehender Interferometer) but I am struggling. Here is what I understand:

Let Alice create a pair of entangled particles be in Horizontal and Vertical polarization ($H$ and $V$) as follows:

$$(1/\sqrt2) (HH+VV).$$

Now, she sends one of this particles to Bob far away and Bob selects a new basis (Left and Right polarization) as follows:

  • $L=(1/\sqrt4)(H-iV)$,
  • $R=(\sqrt3/\sqrt4)(H+iV)$

Now, the entangled particle equation can be rewritten as (please correct if wrong): $1/\sqrt6(R(H-iV)) + 1/\sqrt2(L(H+iV))$

Queries: After choosing this new basis Bob decides to measure each particle in $L/R$ basis, and later Alice decides to send each corresponding particle through classical Mach-Zehnder Interferometer (with equal arms length, two $H/V$ polarizers, two detectors $Det_c$ and $Det_d$ for constructive and destructive interference respectively):

  1. For each particle what is the probability of Bob detecting the particle in $L$ vs $R$ polarization respectively.
  2. For all particles with Bob detected with $L$ polarization, how does the corresponding entangled particle detected in the interferometer with Alice.
  3. Similarly, For all particles with Bob detected with $L$ polarization, how does the corresponding entangled particle detected in the interferometer with Alice.

I am trying to understand what this equation means and what will show up in practical terms in an experiment. But I am lost. My background is not in physics thus I am struggling without being able to make much sense out of it. Can someone please explain (assume a layman)?

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2 Answers 2

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  1. You've made this harder than it needs to be by choosing $R$ and $L$ with different lengths. You can always replace a state vector with a parallel vector of the same length without affecting any probabilities.

  2. Therefore it's better to just put $L=H-iV$ and $R=H+iV$. Now the original state is just $LR+RL$ (or actually $1/\sqrt2$ times this, but multiplying by a constant won't change any probabilities.) If you prefer working with vectors of length 1, you can put coefficients of $1/\sqrt2$ in front of the new $L$ and $R$.

  3. The answer to your question 1 is that if your experiment returns either $LR$ or $RL$, then it must do so with equal probabilities, because $LR$ and $RL$ both occur with coefficients of the same magnitude (in this case $1$) in the expression from point 2). (In general the probabilities are proportional to the squares of the magnitudes of the coefficients.)

3a) If you insist on sticking with your non-equal-length vectors, the calculation is a little subtler but gives the same result.

  1. After Bob's measurement, the pair is in either state $LR$ or $RL$. Therefore Alice measures either $R$ or $L$. As far as Alice is concerned, the two states are equally likely, so she is equally likely to measure either $R$ or $L$.
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  • $\begingroup$ Thank you. I deliberately chose them to be different lengths to make it more clearer. But I am confused with the answer or maybe i am missing something, here is how: Bob chooses L/R but Alice on her side sends the particle through H/V and not L/R (as I mentioned). Thus what do we see in terms of Alice (H/V Mach-Zehnder interferometer) on the detectors is the query $\endgroup$ Jul 20, 2022 at 12:08
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    $\begingroup$ Ah. Well, after Bob's measurement, the particles are either in state LR = L(H+iV) = $LH + i LV$or $RL=R(H-iV)=RH-iRV$. Either way, there is a 50/50 chance she will measure $H$ and a 50/50 chance she will measure $\pm iV$ (note that $+iV$ and $-iV$ are both alternate names for $V$). [She actually measures one of $LH,LV,RH$ or $RV$, but $LH$ and $RH$ both look the same to her, as do $RH$ and $RV$.] $\endgroup$
    – WillO
    Jul 20, 2022 at 12:36
  • $\begingroup$ The last bit of my comment above should refer to $LV$ and $RV$, not $RH$ and $RV$. $\endgroup$
    – WillO
    Jul 20, 2022 at 13:05
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There are some details here that are a little odd.

For example you cannot change the basis for one where the eigen states have different lenghts, all of them must have a lenght of 1. Let's think about it.

This is one of the states you propose:

𝐿=$\frac{1}{\sqrt{4}}(𝐻-i𝑉)$

Imagine you want to measure the probability of a particle in state L of being in the state L.

You will find that:

$ \langle L | L \rangle = 1/2$

That cannot be because we are saying the particle is in that specific state, so the probability must be 1. Thus, the problem is just your normalization constant.

I think that maybe you were trying to do something interesting that leaded you to a new state with new probabilities. Which is not bad, just be careful with your calculations.

By the way, those are actually not Left and Right polarizations. Those states represents circular polarization

for right and left polarizations the state must be

𝐿=$\frac{1}{\sqrt{2}}(V-H)$

R=$\frac{1}{\sqrt{2}}(V+H)$

you see using the i in the state is like having the following state

$\frac{1}{\sqrt{2}}(V+\exp^{i\pi/2}H)$

which is a phase of $\pi/2$ between both states that create the circular polarization. I am sure there should be a video or gift of this that make it easier to understand.

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