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New answers tagged linear-algebra

1

Using the expansion into Fock states: $$|\alpha\rangle=e^{-|\alpha|^2/2}\sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n!}}|n\rangle$$ and the identity $$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$ It is straightforward to work out that: $$\langle \beta |\alpha \rangle=e^{-|\alpha|^2/2}e^{-|\beta|^2/2}e^{\alpha\beta^*}$$ As you can see, if $\beta=\alpha$ this ...

0

The coherent state is $$|\alpha\rangle~=~e^{-|\alpha|^2/2}e^{\alpha a^\dagger}|0\rangle$$ by extension then we can compute $\langle-\alpha|\alpha\rangle$ as $$\langle-\alpha|\alpha\rangle~=~\langle 0|e^{-|\alpha|^2/2}e^{-\alpha a} e^{-|\alpha|^2/2}e^{\alpha a^\dagger}|0\rangle$$ $$=~e^{-|\alpha|^2}\langle 0|e^{-\alpha a} e^{\alpha a^\dagger}|0\rangle$$ ...

0

They are different states. The state $|\alpha\rangle$ is 180 degrees out of phase with respect to the state $|\alpha\rangle$ in the coherent sense (not the quantum mechanical sense).

2

Here is what I understand: if you have a particle at state $|x \rangle$, active translating it by $a$ means moving the particle to state $| x + a \rangle$. Passive transformation means you keep the particle in the same place, and change the coordinate by new variable $x = x' + a$ (note that the coordinate system is translated backwards $-a$). I am not very ...

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