# Structure of the Phonon Free Propagator

## Main Question or Discussion Point

Hey Everyone,

I've a quick a question regarding the make-up of bosonic Green's functions, taking the free propagator for phonons as example. According to Mahan, 3rd ed. it is given by:

D(q, $\lambda$, t-t')=-i$\langle$0|TA$_{q}$(t)A$_{-q}$(t')|0$\rangle$

with A$_{q}$=a$_{q}$+a$^{+}_{q}$

[ Eqs. 2.66 & 2.67]

The operators showing up in the propagator ( i.e. A ) are not the simple creation & annihilation operators ( i.e. a, as would be the case for fermions ) but linear combinations thereof. What's the reason for this? Does this imply that phonons are only produced in pairs? Is this the same for other bosonic propagators ( I'm only familiar with phonons ). Hints, Corrections & Solutions greatly appreciated!

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DrDu
I think this is done like it is because A_q is the observable amplitude of the phonon. In contrast to this the fermionic field operators aren't observables. The coupling to the phonons occurs either via the fields A_q or via the momenta $\partial A_q /partial t$. So it makes sense to work with the linear combinations of the creation/ anihilation operators. Note that these are not bilinear operators in anihilation/creation but linear combinations. Hence there is no pair production involved. Rather the operator can produce coherent superpositions of states with different quanta.

OK, that sounds like a physical reason, I can appreciate that. Thanks for your answer! Do you know if this is done analogously for other bosons?

DrDu
Most notably photons in QED.

DrDu
At least in relativistic field theories, the combination of creation and anihilation operators in the field operators for the simplest bosons is due to them not carrying charge, in contrast to e.g. electrons. Obviously there are also charge carrying bosons, but it turns out that they can be described in terms of two "chargeless" bosonic fields, so this does not change the mechanism. I am not quite sure how a non-relativistic argument in many body theory would look like.

DrDu