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If $|0 \rangle$ is the vacuum state in quantum mechanics and $\alpha$ is any complex number, what is $\langle 0 | \alpha | 0 \rangle$? I need to have that $\langle 0 | \alpha | 0 \rangle = \alpha$, but this seems to imply that $\langle 0 | 0 \rangle = 1$. Is this true? If so, why?

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1 Answer 1

The usual convention is that $\langle 0 \vert 0 \rangle$ is indeed $1$. This is true because we are free to pick any overall normalization for $\lvert 0 \rangle$ given its definition of "the eigenvector of $\hat{a}$ with eigenvalue $0$." Scaling an eigenvector keeps it an eigenvector with the same eigenvalue. Thus we declare it to have a normalization that is easy to work with. In general ladder operator eigenvectors have this property: $\langle n \vert n \rangle = 1$. If there is another convention in use, it would hopefully be clearly stated.

If you are asking why $1$ rather than $0$, the answer is that only the $0$-vector should have such a property, and $\lvert 0 \rangle$ is not the $0$-vector of the Hilbert space.

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I would add that this fixes the normalization of all other state vectors, e.g. if they are constructed using creation operators from the vacuum state. –  alexarvanitakis Apr 29 '13 at 12:19

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