- #1
boxfullofvacuumtubes
- 20
- 2
Hi all,
I'm trying to understand how to describe the quantum state of entangled photons, including their phase, if one of them encounters a double-slit.
Here's a simple example:
Suppose you have 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}
Suppose the photon A passes through a double-slit.
Is my understanding correct that 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, is the following true?
\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}
\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, can the resulting state be described by this?
\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}
Or, is this a wrong way of including the phase shift in the quantum state of entangled photons?
I'm trying to understand how to describe the quantum state of entangled photons, including their phase, if one of them encounters a double-slit.
Here's a simple example:
Suppose you have 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}
Suppose the photon A passes through a double-slit.
Is my understanding correct that 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, is the following true?
\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}
\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, can the resulting state be described by this?
\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}
Or, is this a wrong way of including the phase shift in the quantum state of entangled photons?