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

  • #1
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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  • #2
DrDu
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Take the expectation value of x and p and you will see.
 
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  • #3
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.
 
  • #4
DrDu
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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.
Yes. Are you familiar with the phase space from classical mechanics, or action-angle variables?
 
  • #5
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!
 
  • #6
DrDu
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Also take in mind that alpha is time dependent as coherent states aren't energy eigenstates.
 

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