Structure of the Phonon Free Propagator

1. Sep 4, 2012

dejo-ro

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!

2. Sep 4, 2012

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.

3. Sep 5, 2012

dejo-ro

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?

4. Sep 5, 2012

DrDu

Most notably photons in QED.

5. Sep 5, 2012

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.

6. Sep 5, 2012

DrDu

I think the basic reason is the following:
The time dependent Dirac equation for the electron is a first order differential equation while the Klein Gordon equation is second order. Hence a solution of the Dirac equation requires only one field amplitude a while the KG equation requires two field amplitudes a and b.
For a chargeless field, a and b must be complex conjugate.
The time independent Schroedinger equation can be reduced to a first order differential equation in a similar way
http://www.itp.uni-bremen.de/~noack/PauliDirac.pdf [Broken]
(sorry, in German, but I think the formulas should be clear)
and the argument can be taken over.

Last edited by a moderator: May 6, 2017