Understanding Entanglement basis change in a practical sense 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):

*

*For each particle what is the probability of Bob detecting the particle in $L$ vs $R$ polarization respectively.

*For all particles with Bob detected with $L$ polarization, how does the corresponding entangled particle detected in the interferometer with Alice.

*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)?
 A: *

*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.


*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 $R$.


*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.


*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 $L$.

