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Let the pure states be in superposition of horizontal and vertically polarized basis states. They are arriving at the point detector one at a time. So, a pure state is $|\Psi\rangle = \alpha|V\rangle + \beta|H\rangle$. The formula for counting rate is as follows, $$G({\bf r},t) \propto \langle \Psi_F^{(0)}|\hat{E}^{(-)}({\bf r}, t) \hat{E}^{(+)}({\bf r}, t) | \Psi_F^{(0)} \rangle $$. Let us assume it is a single mode field. So, $$\hat{E}^{(+)} = iE_V \hat{a}e^{-i\omega t}$$ and $$\hat{E}^{(-)} = -iE_V \hat{a}^\dagger e^{i\omega t}$$. So,$$G({\bf r},t) \propto \langle \Psi_F^{(0)}|(-iE_V \hat{a}^\dagger e^{i\omega t}) (iE_V \hat{a}e^{-i\omega t}) | \Psi_F^{(0)} \rangle $$ $$ G({\bf r},t) \propto E_V^2 \langle \Psi_F^{(0)}| \hat{a}^\dagger \hat{a} | \Psi_F^{(0)} \rangle $$ $$ G({\bf r},t) \propto E_V^2 \langle \Psi_F^{(0)}| \hat{a}^\dagger \hat{a} | \Psi_F^{(0)} \rangle $$ $$ G({\bf r},t) \propto E_V^2 \langle \Psi_F^{(0)}| \hat{n} | \Psi_F^{(0)} \rangle $$ $$ G({\bf r},t) \propto E_V^2 \langle \Psi_F^{(0)}| \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix} | \Psi_F^{(0)} \rangle $$. Let's consider the other case now. If the incoming states are mixed states, let's assume the density matrix to be $$\rho = \frac{1}{2} \left( \begin{pmatrix} 1 \\ 0 \\ \end{pmatrix} . \begin{pmatrix} 1 & 0\\ \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \\ \end{pmatrix} . \begin{pmatrix} 0 & 1\\ \end{pmatrix} \right) = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \\ \end{pmatrix}$$. Here, $|H\rangle = \begin{pmatrix} 1 \\ 0 \\ \end{pmatrix}$ and $|V\rangle = \begin{pmatrix} 0 \\ 1 \\ \end{pmatrix}$. So, the formula for photon detection rate is as follows. $$G({\bf r},t) \propto Tr[\rho \hat{E}^{(-)} \hat{E}^{(+)}]$$ $$G({\bf r},t) \propto E_V^2 Tr[\rho \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix}]$$ $$G({\bf r},t) \propto E_V^2 Tr[\begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \\ \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix}]$$ $$G({\bf r},t) \propto E_V^2 Tr[\begin{pmatrix} \frac{1}{2} & 0 \\ 0 & 0 \\ \end{pmatrix} ]$$ $$G({\bf r},t) \propto \frac{1}{2} E_V^2 $$. The observable for both pure and mixed state look same. Does it mean that the physical design of the detection system will be same for both pure and mixed state?

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Hi Omar - is there something in particular you think might be wrong with your calculations? If so, it'd be much better if you edit the question to focus on that. (If you do, I'll be happy to reopen this.) Questions that just ask someone to check your work aren't really what this site is for. – David Z Mar 14 '13 at 17:30
@DavidZaslavsky, I added my confusion at the last sentence. – Omar Shehab Mar 14 '13 at 17:52
up vote 1 down vote accepted

Yes, the photon statistics for pure and mixed states in this example will be the same, as long as $\vert\alpha\vert^2 = \frac{1}{2}$.

Specifically, the states

$\vert\psi\rangle = \alpha\vert H\rangle + \beta\vert V\rangle$, and

$\vert\psi\rangle\langle\psi\vert = \vert\alpha\vert^2 \vert H\rangle\langle H\vert + \vert\beta\vert^2 \vert V\rangle\langle V\vert$

have the same expectation value in terms of the photon number operator.

To measure the difference between them requires a measurement which incorporates the coherence. A common technique to achieve this is optical homodyne tomography.

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