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If $| \alpha\rangle $ represents a coherent state (the normalized right eigenstate of the destruction operator $a$ in Quantum Mechanics; $\alpha$ is a complex number), then it is known that:

\begin{equation} \int | \alpha\rangle\!\langle \alpha| \frac{d^2\alpha}{\pi} = I \end{equation} where $I$ refers to the identity operator.

Can any operator acting on the appropriate Hilbert space be represented in the Glauber-Sudarshan P-representation, and if it can, how to prove that this is the case? (I am especially interested about the representation of density operators)

By the Glauber-Sudarshan representation, I mean the following: \begin{equation} \int P(\alpha,\alpha^*)|\alpha\rangle\!\langle\alpha|\frac{d^2 \alpha}{\pi} \end{equation}

Both the integrals are over the entire complex plane.

glS
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Harsha
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1 Answers1

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The answer is no, and the details are clearly spelled out in Glauber's Les Houches lectures (circa 1964). Glauber introduces a "T-representation" which can represent any operator in the Fock space of harmonic oscillator states, a less general "R-representation" which can represent any density operator, and the still less general "P-representation" which can "represent virtually all [states] studied in optics".

R Kelly
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