- #1
Orbb
- 82
- 0
Hi everyone,
I have the following question about coherent states: It is known that the creation operator has no eigenket. However, the action of a creation operator [tex]a^{\dagger}[/tex] on a coherent ket [tex]|\alpha\rangle[/tex] can be written as
[tex]a^{\dagger}|\alpha\rangle = \left( \frac{\partial}{\partial \alpha} + \frac{\alpha^*}{2}\right)|\alpha\rangle.[/tex]
My question now concerns
[tex]e^{\lambda a^{\dagger}} |\alpha\rangle} = e^{\lambda a^*/2}e^{\lambda\partial_{\alpha}}|\alpha\rangle,[/tex]
which follows from the equation above. I wish to find an explicit expression for that one. It came to my mind that there may be an analogy with the displacement operator acting on position eigenstates
[tex]e^{\lambda \partial_x}|x\rangle = |x+\lambda\rangle.[/tex]
But does it hold? Or is there another way? Writing down the explicit form of a coherent state doesn't help me much because this way I can't get rid of the creation operator in the exponential.
Thank you very much for any thoughts!
I have the following question about coherent states: It is known that the creation operator has no eigenket. However, the action of a creation operator [tex]a^{\dagger}[/tex] on a coherent ket [tex]|\alpha\rangle[/tex] can be written as
[tex]a^{\dagger}|\alpha\rangle = \left( \frac{\partial}{\partial \alpha} + \frac{\alpha^*}{2}\right)|\alpha\rangle.[/tex]
My question now concerns
[tex]e^{\lambda a^{\dagger}} |\alpha\rangle} = e^{\lambda a^*/2}e^{\lambda\partial_{\alpha}}|\alpha\rangle,[/tex]
which follows from the equation above. I wish to find an explicit expression for that one. It came to my mind that there may be an analogy with the displacement operator acting on position eigenstates
[tex]e^{\lambda \partial_x}|x\rangle = |x+\lambda\rangle.[/tex]
But does it hold? Or is there another way? Writing down the explicit form of a coherent state doesn't help me much because this way I can't get rid of the creation operator in the exponential.
Thank you very much for any thoughts!