1
$\begingroup$

In Alain Aspect's talk "Bell’s Theorem: The Naive View of an Experimentalist", Aspect explains the strong correlations in an EPR experiment where both polarisers are in the same orientation.

I have difficulties with the actual mathematical formulation of the projection onto the partially measured "eigenspace". This derivation can be found in section 2.3 of the aforementioned paper.

Prerequisites

Consider two photons, entangled in the following Bell state:

$$ \left|\psi\right> = \frac{1}{\sqrt{2}}\left( \left|x\ x\right> + \left|y\ y\right> \right) $$

Here, $\left|x\ x\right>$ denotes the Kronecker product $\left|x\right>_1\otimes \left|x\right>_2$.

Each photon $i$ passes a polariser that is rotated by the angle $\theta_i$. After passing the polariser, the photons are detected in either of the two states $\left|\pm\right>_i$:

$$ \left|\pm\right>_i = \cos\left(\theta\right)\left|x\right>_i +\sin\left(\theta\right)\left|y\right>_i $$

Partial measurement of the Bell state

Let's consider only photon 1 and measure it. Let's say the result is $\left|+\right>_1$. This outcome has probability $\frac{1}{2}$. Alain Aspect now continues that the new state $\left|\psi'\right>$ after the measurement is obtained by

projection of the initial state vector $\left|\psi\right>$ onto the eigenspace associated to the result $+$.

However, this step is missing a detailed derivation and I failed to derive the proper projection onto the new basis.

My approach is as follows:

Let $\hat{P}$ be the projection operator with \begin{align} \hat{P} &= \left|+\right>_1 \left<+\right|_1 \otimes \mathbb{1}_2 \\ &=\left(\cos^2\theta_1 \left|x\right>_1\left<x\right|_1+ \cos\theta_1\sin\theta_1 \left|x\right>_1\left<y\right|_1+ \sin\theta_1\cos\theta_1 \left|y\right>_1\left<x\right|_1+ \sin^2\theta_1 \left|y\right>_1 \left<y\right|_1\right) \otimes \mathbb{1}_2. \end{align} or in matrix notation where $\left|x\right> = (1\ 0)^T, \left|y\right> = (0\ 1)^T$ \begin{align} \hat{P} &= \left( \begin{matrix} \cos^2\theta_1 & \cos\theta_1\sin\theta_1 \\ \cos\theta_1\sin\theta_1 & \sin^2\theta_1 \end{matrix} \right) \otimes \mathbb{1}\\ &= \left( \begin{matrix} \cos^2\theta_1 & 0 & \cos\theta_1\sin\theta_1 & 0 \\ 0 & \cos^2\theta_1 & 0 & \cos\theta_1\sin\theta_1 \\ \cos\theta_1\sin\theta_1 & 0 & \sin^2\theta_1 & 0 \\ 0 & \cos\theta_1\sin\theta_1 & 0 & \sin^2\theta_1 \end{matrix} \right). \end{align} In matrix notation, the Bell state is given by $$ \left|\psi\right> = \frac{1}{\sqrt{2}}\left[ \left( \begin{matrix} 1 \\ 0 \\ 0 \\ 0 \end{matrix} \right) + \left( \begin{matrix} 0 \\ 0 \\ 0 \\ 1 \end{matrix} \right) \right]. $$

Now the new state $\left|\psi'\right>$ is given by

\begin{align} \left|\psi'\right> &= \hat{P}\left|\psi\right> \\ &= \frac{1}{\sqrt{2}} \left( \cos^2\theta_1\left|x\ x\right> \cos\theta_1\sin\theta_1\left|x\ y\right> \cos\theta_1\sin\theta_1\left|y\ x\right> \sin^2\theta_1\left|y\ y\right> \right) \\ &\equiv \frac{1}{\sqrt{2}} \left|+\ +\right> \equiv \frac{1}{\sqrt{2}} \left|+\right> \otimes \left|+\right> \end{align}

This result appears to be close to that of Alain Aspect. However, my approach brings up the factor $\frac{1}{\sqrt{2}}$ in the final state. This would also mean that the final state is not normalised ($\left|\left<\psi'|\psi'\right>\right|^2 = \frac{1}{2}$). This so happens to be exactly the probability of measuring $\left|+\ +\right>$ in the first place: $\left|\left<+\ +|\psi\right>\right|^2 = \frac{1}{2}$.

My questions are

  1. Is my formulation of the projection operator $\hat{P}$ correct?
  2. Why is the final state not normalised?
$\endgroup$
2
  • 2
    $\begingroup$ When you project a state onto some subspace, you will always get something with a different norm unless that state is already in the subspace, so something like a factor of $1/\sqrt{2}$ is hardly surprising. If the projection actually corresponds to something physical like a measurement, then the post-measurement state is just the normalized version of the projection (as per the usual projection measurement postulate of QM!), which you either effect by hand or state (as part of the postulate!) that the post measurement state is $\hat{P}|\psi\rangle/\sqrt{\langle\psi|\hat{P}|\psi\rangle}$. $\endgroup$
    – march
    Oct 30, 2023 at 15:21
  • 2
    $\begingroup$ @march That is actually the answer to the question; so consider to put that comment into an answer. $\endgroup$ Oct 30, 2023 at 19:07

1 Answer 1

1
$\begingroup$

When you project a state onto some subspace, you will always get something with a different norm unless that state is already in the subspace, so something like a factor of $1/\sqrt{2}$ is hardly surprising. If the projection actually corresponds to something physical like a measurement, then the post-measurement state is just the normalized version of the projection (as per the usual projection measurement postulate of QM!), which you either effect by hand or state (as part of the postulate) that the post measurement state is $$ |\psi_{\textrm{post-measurement}}\rangle=\frac{\hat{P}|\psi\rangle}{\sqrt{\langle\psi|\hat{P}|\psi\rangle}}\,. $$ Following through with this calculation will result in exactly the state $\lvert+\,+\rangle$. Otherwise, it appears to me that OP's math is correct!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.