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Commutation relations in QFT [duplicate]

So I have just started learning QFT. So you take a classical field and turn the degrees of freedom into operators. All fine, just like normal quantum. However I am confused about the commutation ...
4
votes
2answers
408 views

Relation between spinors and anticommutation relation of fermions

I read that the state of a pair of particles is the tensor product of the single states of both, and you will get a wavefunction with the parameters of both, if you swap the parameters you will get a ...
2
votes
2answers
103 views

Can the necessity of using anti-commutators for Dirac fields and commutator for Klein-Gorden be deduced from the field equations?

We all learned to use the commutator for quantizing the KG field and the anti-commutator for the Dirac field. We are told (which is correct) so that KG-excitations are bosons and Dirac-excitations ...
3
votes
1answer
433 views

Wrong sign anticommutation relation for the Dirac field?

Consider the Dirac Lagrangian $$\mathcal{L}=\psi ^{\dagger }\gamma ^{0}\left( \mathrm{i}\gamma ^{\rho }\partial _{\rho }-m\right) \psi .$$ The conjugate momenta to $\psi ^{a}$ are defined, as usual, ...
12
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1answer
2k views

Making sense of the canonical anti-commutation relations for Dirac spinors

When doing scalar QFT one typically imposes the famous 'canonical commutation relations' on the field and canonical momentum: $$[\phi(\vec x),\pi(\vec y)]=i\delta^3 (\vec x-\vec y)$$ at equal times ($...
1
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1answer
272 views

Anticommutation relations and bispinor field

In a case of free Dirac field we have $$ \hat {H} = \int \epsilon_{\mathbf p}\left( \hat {a}^{+}_{s}(\mathbf p )\hat {a}_{s}(\mathbf p ) - \hat {b}_{s}(\mathbf p )\hat {b}_{s}^{+}(\mathbf p ) \right)...