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  Linearizing Quantum Operators

+ 3 like - 0 dislike

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!



This post has been migrated from (A51.SE)
asked Oct 23, 2011 in Theoretical Physics by Antillar Maximus (45 points) [ no revision ]
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This question has been cross-posted on two sites simultaneously, see http://physics.stackexchange.com/q/16080/2451

This post has been migrated from (A51.SE)
@Qmechanic: Argh! I wish people wouldn't do that. It makes it impossible to keep track of whether it has been answered or not.

This post has been migrated from (A51.SE)
Yeah, I think SE sites generally doesn't allow it, cf. http://meta.stackoverflow.com/q/64068

This post has been migrated from (A51.SE)
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.

This post has been migrated from (A51.SE)
@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 ;-)

This post has been migrated from (A51.SE)
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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.

This post has been migrated from (A51.SE)
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.

This post has been migrated from (A51.SE)

1 Answer

+ 5 like - 0 dislike

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.

This post has been migrated from (A51.SE)
answered Oct 24, 2011 by Piotr Migdal (1,260 points) [ no revision ]

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