I'm wondering if anyone can help me understand how a polarizer changes the quantum state of two polarization-entangled photons. I haven't found a clear description in the literature.

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 polarizer oriented at the +45 angle.

Does that convert $ H_{A} $ in the equation above to $ \frac{1}{\sqrt{2}}\bigl(\left|+45\right\rangle\bigr) $, which is $ \frac{1}{2}\bigl(\left|H_{A} + V_{A}\right\rangle\bigr) $, and $ V_{A} $ also to $ \frac{1}{2}\bigl(\left|H_{A} + V_{A}\right\rangle\bigr) $? I used $ \frac{1}{\sqrt{2}} $ because only a half of the inputlight passes through the polarizer.

Is, therefore, the resulting state the following, or did I misunderstand it completely?

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

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


Your final result is correct up to a normalization. Another way to see this is that a polarizer at a $45^\circ$ angle would simply be a projector onto the $|+_A\rangle \equiv \frac1{\sqrt2}(|H_A\rangle+|V_A\rangle)$ state for the $A$ photon. i.e. the state of the photon after going through it is simply $|+_A \rangle \langle+_A| \times |\text{initial state} \rangle$ in the Hilbert space of photon $A$. So the action of putting only photon $A$ through the $45^\circ$ polarizer on the joint Hilbert space of both photons is simply described by the transformation: $$|+_A\rangle\langle+_A|\otimes \mathbb 1_B$$ Now simply act this transformation on your initial state to get the final state: $$|\psi\rangle = \Big(|+_A\rangle\langle+_A|\otimes \mathbb 1_B \Big)\frac1{\sqrt 2} \Big(|H_A H_B \rangle + |V_A V_B \rangle \Big)$$ $$=\frac1{\sqrt 2}|+_A \rangle \langle +_A|H_A \rangle \otimes 1|H_B \rangle+\frac1{\sqrt 2}|+_A \rangle \langle +_A|V_A \rangle \otimes 1|V_B \rangle$$ Using our definition of $|+_A \rangle$ gives: $$|\psi \rangle=\frac12|+_A \rangle \otimes |H_B \rangle + \frac12|+_A \rangle \otimes |V_B \rangle$$ $$= \frac1{2\sqrt 2}|H_A H_B \rangle + \frac1{2\sqrt 2}|V_A H_B \rangle + \frac1{2\sqrt 2}|H_A V_B \rangle+\frac1{2\sqrt 2}|V_A V_B \rangle$$ Which is identical to your result. However, note that the projector $|+_A \rangle \langle +_A|$ doesn't keep the norm of the initial state vector, it in fact multiplies its norm by $\langle +_A | \psi_0 \rangle$. So you do need to normalize the resulting state at the end: $$|\psi \rangle = \frac1{2}\Big[ |H_A H_B \rangle + |V_A H_B \rangle + |H_A V_B \rangle+|V_A V_B \rangle \Big]$$


As flippiefanus points out in his comment, the above state is nothing but: $$|\psi \rangle = |+_A \rangle \otimes |+_B \rangle$$ So the initial entanglement of the two photons puts the second photon in the $|+ \rangle = \frac1{\sqrt 2}(|H \rangle + |V \rangle)$ state as well, even though the polarizer only acted on the first photon.

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    $\begingroup$ Perhaps you can add that the B-system then also becomes a plus state $|+_B\rangle$, as one can see from your third last expression. This then shows that the entanglement induces the state on the other side to take on a specific form. $\endgroup$ – flippiefanus Feb 9 at 4:19
  • $\begingroup$ That's a very nice observation. Thanks! I'll add it to the answer. $\endgroup$ – Sahand Tabatabaei Feb 9 at 4:20
  • $\begingroup$ Thank you for a great explanation! To clarify: $\left|\phi\right\rangle$ describes measurement of coincidences: detected A photons that have passed through the polarizer and the B photons entangled with them, right? If we could measure the photons absorbed by the polarizer and their coincidences, would we just replace $\left|+_{A}\right\rangle$ with $\left|-_{A}\right\rangle$ in your equation? That would be the same as describing the state if the polarizer is rotated to -45, I think. Or would there also be some relative phase shift between $\left|+\right\rangle$ and $\left|-\right\rangle$? $\endgroup$ – triclope Feb 9 at 16:13
  • $\begingroup$ Yes I think so. Looking at an absorbed photon (instead of a transmitted one) should correspond to the projector $|- \rangle \langle - |$. So you'd simply replace $|+ \rangle$ to $|- \rangle$. $\endgroup$ – Sahand Tabatabaei Feb 9 at 20:59

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