Linearizing Quantum Operators

Question

I was reading an article on harmonic generation and came across the following way of decomposing the photon field operator. $$ \hat{A}={\langle}\hat{A}{\rangle}I+ \Delta\hat{a}$$

The right hand side is a sum of the "mean" value and the fluctuations about the mean. While I understand that the physical picture is reasonable, is this mathematically correct? If so what are the constraints this imposes? In literature this is designated as a "linearization" process.

My understanding of a linear operator is that it is simply a homomorphism. I have never seen anything done like this and I'm having a hard time finding references which justify this process.

I would be grateful if somebody can point me in the right direction!

Cheers,

-Ant

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Comments

This question has been cross-posted on two sites simultaneously, see http://physics.stackexchange.com/q/16080/2451

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@Qmechanic: Argh! I wish people wouldn't do that. It makes it impossible to keep track of whether it has been answered or not.

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Yeah, I think SE sites generally doesn't allow it, cf. http://meta.stackoverflow.com/q/64068

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Sorry guys. I am new here and I was not sure which site would be appropriate. I saw tons of F=ma questions on the Physics site, so I was not sure if my question would be answered there. Both versions have given me different answers, so thank you.

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@Antillar: that's just because the Physics site has been around longer and deals with a broader range of material. If you look for F=ma questions there you will of course find plenty of them ;-)

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I don't see how this can be justified based on calculus reasoning alone? The first thing that comes to my mind is Cosets, but I am not sure how to take that anywhere.

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Imagine the operator $\hat A$ has eigenfunctions. Then in space of its eigenfunctions it is like a regular function, not operator. And even if $\hat A$ is always an operator. The definitions work as long as they are reasonable, reversible, etc. So it may be not so necessary to search for a group theoretic reason. You just introduce new variables and you work with them, that's it.

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Answer 1

The operation you mentioned $$\Delta \hat{a} = \hat{A}-\langle \hat{A}\rangle \hat{I}$$ is just shifting of the annihilation operator. Typically people even drop the identity and write $$\hat{a}_{new}=\hat{a}_{old}-\alpha_0,$$ where $\alpha_0$ is a complex number. In your question the shift is such that $\langle \hat{a}_{new} \rangle = 0$, which may be convenient for calculations. In particular when the pump beam has many photons, the quantum part (i.e. related to $\hat{a}_{new}$) may be neglected.

Additionally, $\hat{a}_{new}$ is an annihilation operator with the same anti-commutation relations as $\hat{a}_{old}$.

From mathematical point of view it is a perfectly legit operation. Moreover, both operators have the same domain, and spectrum only shifted by $\alpha_0$. To see that domain is the same take any $|\psi\rangle \in \text{dom}(\hat{a}_{old})$. Then denoting $|\phi\rangle := \hat{a}_{old} |\psi\rangle$, we check that $\hat{a}_{new} |\psi\rangle = |\phi\rangle-\alpha_0|\psi\rangle$ is a well-defined vector.

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