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I'm currently reviewing some of my notes on Quantum Field Theory (the version of Greiner) and I was wondering if QFT always works in the Hartree-Fock approximation ? Or at least that's what it seems to me!

We have our field-operators $\hat{\psi}(\vec{r},t)$ and $\hat{\psi}^\dagger(\vec{r},t)$ which annihilate or create a particle in $(\vec{r},t)$. By using the appropriate commutation-relations we get fermions or bosons. But these are ONE-PARTICLE operators which obey the correct commutation-relations, or which give the right symmetry (using Fock-space structure).

Now intuïtively I can see this for the free-particle Hamiltonians that this will give an exact result since we'll be able to rewrite them as:

$\hat{H}_0=\sum\limits_nE_n\hat{a}^\dagger_n\hat{a}_n,$

which indeed yields a result in the sense of product functions (since every eigenfunction of $\hat{a}^\dagger_n\hat{a}_n$ is also one of $\hat{H}_0$.

Now the problem starts when we get two-particle (of many-particle) interactions since the Hamiltonian isn't diagonizable in any easy way. This forces us to use perturbation theory and hence the scattering matrix. Upon applying Wick's theorem we can chop up the n-th order term of the scatering matrix into operators of the form $\hat{a}^\dagger_n\hat{a}_n$ which we can calculate in terms of our product basis. Which can also be expressed in terms of a product basis set.

Now long question short: Do we always work in the Hartree-Approximation when we're doing QFT, or am I mistaken?

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A keyword for web-searching seems to be "Many-Body Quantum Field Theory". For instance, in this ref, there is a discussion (for a self-energy diagram), Chapter $12.6.1$, page $209$, equation $(12.29)$ –  Trimok Dec 18 '13 at 13:09
    
There are some QED calculations based on perturbation from non-relativistic result, e.g. arXiv:1212.3196. The non-relativistic part is done by explicitly correlated calculation, not Hartree-Fock. Nevertheless that's somehow mixed QM+QFT calculation, not pure QFT. –  user26143 Dec 23 '13 at 15:14
    
Yeah, I guess I'm mixing up the free-field situation versus the interacting field situation. But I'm not so very sure how to see trough that one ... –  Nick Dec 23 '13 at 21:50
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No, the Hartree approximation is only the very simplest of the approximations used. Moreover, it only works for bosoonic fields, for QED or QCD, which contain fermions, one needs at least the Hartree-Fock approximation.

The Hartree approximation and Hartree-Fock approximation are called mean field approximations as the influence of all other particles to a single particle is accounted for only in an averaged way. Mean field approximations are often reasonable first approximations but don't show important features of realistic QFTs, such as anomalous dimensions.

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Isn't it also easier to say (due to my reasoning above) that these product states give a good interpretation in the FREE PARTICLE case, where the definition of "a particle" makes sence and also these Hartree-Fock considerations apply. While in the interacting case these states can mix up and you can get sums of different states which aren't Hartree-Fock ? –  Nick Dec 27 '13 at 21:56
    
No. Nonproduct states are valid states even in the free case. However, in the free case, they will stay product states under the dynamics, while in the nonfree case they won't. In particular, in the nonfree case, a product state cannot be an exact product state. –  Arnold Neumaier Jan 1 at 13:43
    
How would you create a non-product state with the creation-operators? Would you define a 2-particle creations-operator? –  Nick Jan 3 at 1:29
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@Nick: One creates product states and sums them up. –  Arnold Neumaier Jan 6 at 15:01
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