How to normalize wave functions in QFT? such as \lambda \phi 4 theory?

by mings6
Tags: functions, lambda, normalize, theory, wave
mings6 is offline
Mar10-12, 06:05 AM
P: 11
In quantum mechanics, most wave functions are normalized with \int |\phi|^2 dx^3 =1. But I did not see any field in the quantum field theory is normalized. I understand they maybe just plain waves and does not need to be normalized. But in some cases, if we do not expand the field as plain wave, how to normalize them? For instance, in the \lambda \phi 4 theory, the field \phi has the dimension of GeV. Should we use \int |\phi|^4 dx^4 =1 instead of \int |\phi|^2 dx^3 =1 or \int |\phi|^4 dx^3 =1?
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martinbn is offline
Mar10-12, 06:28 AM
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The wave function in quantum mechanics represents the state of the system. The field in quantum field theory is not the state, so why do you expect that it be normalized? On the other hand the states in QFT are assumed normalized.
mings6 is offline
Mar11-12, 09:24 AM
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If we want to solve the \lambda \phi 4 equation in the same way as solve the Dirac equation, I think we should normalize the result. So we can say that the trivial expresion of <initial | final> = <\phi|\phi>=1 means, when no perturbation, the amplitude from \phi to its own is 1. For instance, the free particle solution of the Dirac equation is normalized to 1 by a factor \sqrt{m/E}.

Demystifier is offline
Mar12-12, 03:19 AM
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How to normalize wave functions in QFT? such as \lambda \phi 4 theory?

Normalization of a field is needed only when that field ITSELF is interpreted as a probability density amplitude. The field \phi in \phi^4 is usually not interpreted in that way.
vanhees71 is offline
Mar12-12, 07:07 AM
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In QFT the normalization of the field operators are defined via the equal-time commutation relations,

[itex][\phi(t,\vec{x}),\Pi(t,\vec{y})]=\mathrm{i} \delta^{(3)}(\vec{x}-\vec{y}).[/itex]

Here [itex]\Pi[/itex] is the canonical field momentum for [itex]\phi[/itex]. In [itex]\phi^4[/itex] theory, it's

[itex]\Pi(x)=\frac{\partial \mathcal{L}}{\partial \dot{\phi}(x)}=\dot{\phi}(x).[/itex]

For the asymptotically free states, symbolized by external legs in Feynman diagrams, the states are to be normalized in the usual way to [itex]\delta[/itex] distributions (supposed you have taken account of wave-function renormalization in the external legs, i.e., left out all self-energy insertions in them, see Weinberg, QT of Fields, vol. 1).

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