# Description of quantum state of entangled photons after polarizer

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:

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

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?

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

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

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.
• 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. – flippiefanus Feb 9 at 4:19
• 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 $\left|-\right\rangle$? – triclope Feb 9 at 16:13
• 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 $|- \rangle$. – Sahand Tabatabaei Feb 9 at 20:59