Quantum state of entangled photons

[boxfullofvacuumtubes]

Homework Statement

Suppose two polarization-entangled photons A and B in the following Bell 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}

1. What is the state if the photon A passes through a double-slit?

2. What is the state if the photon A passes through a double-slit and the photon B passes through a linear polarizer oriented at the +45 angle?

2. The attempt at a solution

My attempt to solve (1):

A double-slit in the photon A's path creates a phase shift $$e^{i\Delta\phi}$$ because of unequal paths from each slit to a particular place on a screen. As the photon A can now take a path through one or the other slit, and there is a phase shift between the two,

\begin{equation}
\left|H_{A},H_{B}\right\rangle \longrightarrow \frac{1}{\sqrt{2}}\bigl(\left|H_{A},H_{B}\right\rangle\bigr) + e^{i\Delta\phi}\frac{1}{\sqrt{2}}\bigl(\left|H_{A},H_{B}\right\rangle\bigr)
\end{equation}

Similarly:

\begin{equation}
\left|V_{A},V_{B}\right\rangle \longrightarrow \frac{1}{\sqrt{2}}\bigl(\left|V_{A},V_{B}\right\rangle\bigr) + e^{i\Delta\phi}\frac{1}{\sqrt{2}}\bigl(\left|V_{A},V_{B}\right\rangle\bigr)
\end{equation}

Therefore:

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

My attempt to solve (2):

The photon B passing through a 45-degree polarizer has a 50% probability of being absorbed and a 50% probability of converting:

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

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

Putting (1) and (2) together:

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

The rest is easy, just manual work, but I'm wondering if I made any mistakes up to this point. Anyone willing to help?

 

[mark726]

Your approach seems correct so far. However, there are a few things to consider:

1. The phase shift, which you denoted as $$e^{i\Delta\phi}$$, should depend on the position of the double-slit, not just on the fact that there are two paths. This means that the phase shift will change for different points on the screen.

2. When the photon B passes through the polarizer, it will not convert or be absorbed with equal probability. The probability of conversion will depend on the angle of the polarizer and the polarization of the photon B.

3. When writing the state after the polarizer, you need to take into account the state of photon A as well. In your equation, you only have the state of photon B after the polarizer.

4. Lastly, it might be helpful to write the state in terms of the basis states for both photons (i.e. $$\left|H_{A}\right\rangle, \left|V_{A}\right\rangle, \left|H_{B}\right\rangle, \left|V_{B}\right\rangle$$) rather than just the polarization states. This will make it easier to keep track of the different combinations.

I hope this helps and good luck with your calculations!