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Linearizing Quantum Operators

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!

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marked as duplicate by Qmechanic Jan 13 '13 at 13:49

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3  
Why not look at it as at a definition of $\Delta \hat a$? The it can always be done (providing that the mean value is finite). It does not implies even any linearization. If the fluctuation magnitude is small with respect to the mean value, it can be considered "linearization", like $f(x)\approx f(x_0)+ f'(x_0)\cdot (x-x_0)$, but as a definition it is valid for any $\Delta \hat a$. –  Vladimir Kalitvianski Oct 23 '11 at 20:24
    
Greetings Vladimir, as you have suggested, it makes sense from a pure calculus point of view (continuous and differentiable functions), but would it necessarily apply in the case of quantum operators? I am looking for a Group theoretic reason to justify this operation. The problem I am having is that the author in (pra.aps.org/abstract/PRA/v49/i3/p2157_1) decomposes $$\frac{d\hat{A}_1}{dz}=-\alpha \hat{A}_1^{\dagger }\hat{A}_2 e^{-{i\Delta kz}}$$ into separable differential equations, one involving only the average values and the other involving only fluctuations. –  Antillar Maximus Oct 23 '11 at 21:24
    
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. –  Antillar Maximus Oct 23 '11 at 21:25
    
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. –  Vladimir Kalitvianski Oct 23 '11 at 21:40
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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. –  Joe Fitzsimons Oct 24 '11 at 10:24

1 Answer 1

up vote 5 down vote accepted

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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