# 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 relations.

For the Klein-gordon/spin 0 scalar field we say $$[\psi_\alpha(x),\psi_\beta^\dagger(y)]=\delta_{\alpha\beta}\delta(y-x)$$ however for the dirac/spin half field we say $$\{\psi_\alpha(x),\psi_\beta^\dagger(y)\}=\delta_{\alpha\beta}\delta(y-x).$$

In Tong's lecture notes he seems to justify this by appealing to the fact that it works. However I don't find this very satisfying. Is there a mathematical reason for seeing what the commutation relations are a priori or do you just have to see what works.

## marked as duplicate by Qmechanic♦ quantum-field-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 19 at 9:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Look up the spin statistics theorem – Slereah Jul 19 at 8:54
• The way I see it is that Pauli's exclusion principle is satisfied when there is anticommutation (Dirac field). Since we know that Pauli's exclusion principle holds from empirical observations or other theories (e.g. solid state physics) it would be nice if the fields would also satisfy that. Mathematically I don't think there's a much deeper reason since ultimately QFT models physics. Perhaps you could say: well we could either commute or anticommute and see that it leads to the correct description and that's it. I may be wrong though. – Bonsay Jul 19 at 9:00
• Possible duplicates: physics.stackexchange.com/q/17893/2451 , physics.stackexchange.com/q/65819/2451 and links therein. – Qmechanic Jul 19 at 9:16
• You end up with negative energy states.See Peskin pages 52-58. – saad Jul 19 at 9:21