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"Don't panic!"

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First of all, sorry if this is in the wrong forum, wasn't quite sure which one to post it in given the question.

My question is, given two space-time points ##x^{\mu}## and ##y^{\mu}##, if the events that occur at these points are simultaneous, i.e. ##x^{0}=y^{0}##, are the two events necessarily

If one has two fields ##\phi## and ##\phi^{\dagger}##, say, then the commutation relation between them is given by [tex][\phi (t,\mathbf{x}),\phi^{\dagger} (t,\mathbf{y})]=i\delta^{(3)}(\mathbf{x}-\mathbf{y})[/tex]

Now is the reason for this being equal to a ##\delta##-function because of locality? i.e. given that the two fields are evaluated at the same time, then as locality demands that they can only "communicate" if they are separated by a time-like separation, they must necessarily be evaluated at the

My question is, given two space-time points ##x^{\mu}## and ##y^{\mu}##, if the events that occur at these points are simultaneous, i.e. ##x^{0}=y^{0}##, are the two events necessarily

*space-like*separated? The reason I ask is that I'm trying to understand the notion of*equal-time commutation relations*in QFT (in which the commutator is non-zero in the case where ##\mathbf{x}=\mathbf{y}##).If one has two fields ##\phi## and ##\phi^{\dagger}##, say, then the commutation relation between them is given by [tex][\phi (t,\mathbf{x}),\phi^{\dagger} (t,\mathbf{y})]=i\delta^{(3)}(\mathbf{x}-\mathbf{y})[/tex]

Now is the reason for this being equal to a ##\delta##-function because of locality? i.e. given that the two fields are evaluated at the same time, then as locality demands that they can only "communicate" if they are separated by a time-like separation, they must necessarily be evaluated at the

*same spatial point*, as if ##\mathbf{x}\neq\mathbf{y}## then there would be a*space-like*separation between the two fields (as ##\Delta s^{2}=(x^{0}-y^{0})^{2}-(\mathbf{x}-\mathbf{y})^{2}=-(\mathbf{x}-\mathbf{y})^{2}<0##), and they would therefore commute (in order to obey locality)?
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