What's the significance of the phase in a coherent state?

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vancouver_water
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In a coherent state defined by [tex]|\alpha\rangle = \exp{\left(-\frac{|\alpha|^2}{2}\right)}\exp{\left(\alpha \hat{a}^\dagger\right)} |0\rangle[/tex] there is a definite phase associated with the state by [tex]\alpha = |\alpha| \exp{\left(i\theta\right)}[/tex] where the number operator and phase operator are conjugates, [tex]-i\partial_{\theta} = \hat{n}.[/tex] The meaning of the number operator is obvious but what is the significance of the phase in this state? What would be a consequence of picking a new phase for this state?
 
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Take the expectation value of x and p and you will see.
 
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So i get [itex]\langle x\rangle \propto \Re{(\alpha)}[/itex] and [itex]\langle p \rangle \propto \Im{(\alpha)}[/itex]. Which gives the relation [itex]\langle p \rangle = m\omega\langle x \rangle \tan\theta[/itex]. So it gives the relation between expectations of x and p.
 
vancouver_water said:
So i get [itex]\langle x\rangle \propto \Re{(\alpha)}[/itex] and [itex]\langle p \rangle \propto \Im{(\alpha)}[/itex]. Which gives the relation [itex]\langle p \rangle = m\omega\langle x \rangle \tan\theta[/itex]. So it gives the relation between expectations of x and p.
Yes. Are you familiar with the phase space from classical mechanics, or action-angle variables?
 
DrDu said:
Yes. Are you familiar with the phase space from classical mechanics, or action-angle variables?
With phase space yes, but not with action angle variables. I'll read about them though, Thanks!
 
Also take in mind that alpha is time dependent as coherent states aren't energy eigenstates.