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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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I am aware of the standard calculus reasoning, i.e a Taylor series expansion about the mean and dropping higher order terms, but that does not necessarily mean that any functional expansion is separable. I mean, the author states that a diff eq:$$\frac{d\hat{A}_1}{dz}=-\alpha \hat{A}_1^{\dagger }\hat{A}_2 e^{-{i\Delta kz}}$$ can be solved by treating the average and fluctuations separately. I don't see how you can decouple them? –  Antillar Maximus Oct 23 '11 at 21:28
    
I don't see an arXiv version of this article but I edited the link into the question body. –  David Z Oct 23 '11 at 22:05
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This question has been cross-posted on two sites simultaneously, see theoreticalphysics.stackexchange.com/q/365/189 (= physics.stackexchange.com/q/27041/2451) –  Qmechanic Oct 24 '11 at 7:42

1 Answer 1

up vote 1 down vote accepted

This type of decomposition is done all the time, and it is weird looking in the operator formalism. It is most natural in the path integral, where it is known as the background field method.

The path integral is over classical values, so that you can always write the field formally as the sum of a classical background and a fluctuating quantum part. The integral over the quantum part reproduces the correct answer for the background, because the integral is translation invariant in field space--- you are allowed to shift the zero value. The background field method is usually used for quick one loop calculations in nonabelian gauge theories, but you can do the decomposition for photons too.

If you are adament that you want to do it in the operator formalism, you can just declare that you redefined the operators by subtracting a multiple of the identity. It isn't natural, but it's equivalent to background field.

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Thank you for the answer. Can you kindly point me to a book or an article where I can read about this formalism? I have no background in QFT beyond second quantization. –  Antillar Maximus Oct 24 '11 at 14:16
    
@Antillar: For the path integral, the classical reference is Feynman and Hibbs, but it's a little dated. Feynman's 1947 Review of Modern Physics is better. Polchinski's string theory book has a condensed path-integral appendix, and Mandelstam/Yourgrau "Variational Principles.." is back in print. These three places (and Wikipedia) are the only sources that do path integrals correctly. Other authors do not understand how the canonical commutation relation come from the path integral. The background field method is unpublished work of Feynman (I believe), and does not have a good original source. –  Ron Maimon Oct 24 '11 at 14:23
    
Ron, thank you. I will check out those references. How does the background field method justify decoupling fluctuations from the mean-field to solve them separately? This is the real problem I am having. It almost seems like a vector space formalism, only in this case you describe the space of $\hat{A}$ in terms of two basis, i.e $I$ and $\Delta\hat{a}$. Sorry about my obsession with Groups/Vector Spaces. :) –  Antillar Maximus Oct 24 '11 at 14:41
    
@Antillar: the photon obeys a linear equation, and the background is decoupled from the interacting fluctuations. Perhaps these sources will not help--- they are too general--- they are not talking about electromagnetism specifically. If you are comfortable with a path integral to the point where the EM field is not mysterious, you can always decompose the field as an exercise. I am sorry I do not know the best reference. –  Ron Maimon Oct 24 '11 at 18:44

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