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- Thread starter LagrangeEuler
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Cthugha

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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.

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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.

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