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Suppose you have two polarization-entangled photons A and B in the following state:

\begin{equation} \Phi=\frac{1}{\sqrt{2}}\bigl(\left|H_{A},H_{B}\right\rangle + \left| V_{A},V_{B}\right\rangle\bigr) \end{equation}

Suppose the photon A passes through a linear polarizer oriented at the -45 angle.

Q1. Does that affect the polarizations of both photons at the same time

\begin{equation} \Phi=\left|-45_{A},-45_{B}\right\rangle \end{equation}

or just the polarization of the photon A?

\begin{equation} \Phi=\frac{1}{\sqrt{2}}\bigl(\left|-45_{A},H_{B}\right\rangle + \left| -45_{A},V_{B}\right\rangle\bigr) \end{equation}

Q2. And are the photons A and B still entangled (until they hit detectors) or was the entanglement broken by the photon A passing through the polarizer?

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You can describe the act of passing through a linear polarizer as a projection of the polarization degree of freedom of the first photon along the associated angle.

If for example the polarizer only lets through photons with the $\newcommand{\ket}[1]{\lvert #1\rangle}\ket +\equiv(\ket H+\ket V)/\sqrt2$ polarization, then the associated projector operator is $$\Pi_+\otimes I\equiv\ket+\!\langle+\rvert\otimes I,$$ where you have a projection of the polarization of the first photon, describing passing through the polarizer, and an identity on the state of the second photon, as this photon is unaffected by the polarizer.

To see the effect of this operation on your state, you just need to rewrite it in the appropriate basis. Straightforward calculations will lead you to write $$\ket\Phi=\frac{1}{\sqrt2}(\ket{+_A}\otimes\ket{+_B} + \ket{-_A}\otimes\ket{-_B}).$$ In other words, the state is also maximally entangled when measured in the $\ket{\pm}\equiv(\ket{H}\pm\ket V)/\sqrt2$ basis (this is no coincidence: a maximally entangled state can be showed to be maximally entangled in any basis).

Passing through the polarizer you get (with some probability) the final state $\ket{+_A}\otimes\ket{+_B}$ (if you want to use as reference the polarization state at a $-45^\circ$ angle, just swap all $\ket{+}$s with $\ket{-}$s in the answer).

In conclusion, yes, the polarizer does "affect" the polarization of both photons because of their shared entanglement, and no the photons are not entangled anymore after the polarizer. To be clear, this does not mean that actions performed on one of the photons somehow instantaneously affect the other photon. Rather, it means that the outcome of the measurements on the two photons are correlated, so that the outcome of measuring the first photon changes our description of the second one, to reflect the newfound knowledge of the outcome of future measurements on it. So in this sense, the polarizer only affects the photon that passes through it, and the other photon is not affected by it.

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    $\begingroup$ This is better, but it is still misleading. Given the knowledge of the result of the projective measurement on the $A$ side, then yes, the state on $B$ changes. But if one is performing an experiment on $B$ without any knowledge of what's happening on $A$ and without the ability to post-select the runs where $A$ got a particular outcome, then $B$ only sees a maximally-mixed state that is unaffected by what's happening on $A$. $\endgroup$ – Emilio Pisanty Mar 3 at 21:31
  • $\begingroup$ @EmilioPisanty I mean, sure. I didn't realize that was being put in question here. I added a clarification on that note $\endgroup$ – glS Mar 4 at 10:18
  • $\begingroup$ Thank you for a very clear explanation. However, is the answer to my question #2 in your last paragraph entirely correct? Does entanglement end when one of the photons passes through a polarizer, and not when it is detected after passing through a polarizer? For example, if one of the photons passes through a Mach-Zehnder interferometer that uses polarizing beam splitters instead of ordinary beam splitters, wouldn't entanglement still exist after the 1st PBS? It'd be great if you could expand on your answer to the question #2 a little bit. $\endgroup$ – triclope Mar 6 at 23:02
  • $\begingroup$ @triclope well, I would say it's mostly a matter of semantics at that point. Obviously, the entanglement only disappears when a measurement is made, so the question is what exactly is meant by "the photon passes through the polarizer". If you mean that you observed that the photon passed through it, then a measurement was made and entanglement is no more. If you instead refer to the state of the photon after evolution through the "polarizer", and you are describing the polarizer as a unitary evolution (e.g. as a PBS), then entanglement is still there until you measure $\endgroup$ – glS Mar 7 at 0:28

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