# Coherent states

vanhees71
Gold Member
2019 Award
Of course, the ground state is an eigenstate of ##\hat{b}## with eigenvalue 0. Just calculate the derivative to see it.

Cthugha
Of course, the ground state is an eigenstate of ##\hat{b}## with eigenvalue 0. Just calculate the derivative to see it.
Now I am somewhat puzzled. Are you sure about that? Maybe I just missed something, but isn't ##\hat{b}## the equivalent of the creation operator? I always thought that the annihilation operator had the plus sign.

vanhees71
Gold Member
2019 Award
You are right. I've misread the sign.

The annihilation operator is
$$\hat{a}=\sqrt{\frac{m \omega}{2}} \hat{x}+\mathrm{i} \frac{1}{\sqrt{2m \omega}} \hat{p}.$$
In position representation that maps to
$$\hat{a}=\sqrt{\frac{m \omega}{2}} x +\frac{1}{\sqrt{2 m \omega}} \partial_x = \frac{1}{\sqrt{2}} (\xi + \partial_{\xi}).$$
The creation operator is its adjoint,
$$\hat{a}^{\dagger}=\frac{1}{\sqrt{2}}(\xi-\partial_{\xi})=\hat{b},$$
which of course has no eigenstate.

The annihilation operator has many eigenstates, the coherent states. The special one with eigenvalue 0 is the ground state. This gives the eigenvector equation
$$\hat{a} u_0(\xi)=0 \; \Rightarrow \; \partial_{\xi} u_0=-\xi u_0 \xi \; \Rightarrow \; u_0(\xi)=N \exp(-\xi^2/2)$$
with ##N## a normalization factor.