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I am confused about what the rules are for simplifying an inner product in Dirac notation.

Let

$ \langle\lambda|\alpha e^{i\theta}|\chi\rangle$

where $\alpha$ is a complex numbers, with $\theta$ real.

Is it valid to say that this is equivalent to:

$ \alpha e^{i\theta}\langle\lambda|\chi\rangle$

And if instead of $\alpha e^{i\theta}$ I had an operator, say

$ \langle\lambda|U|\chi\rangle$

I assume that in that later case, I cannot take the operator out, because $U$ acts on $|\chi \rangle$, and an operator acting on a ket is not the same as acting on a scalar product? Or rather an operator doesn't act on a scalar, but on a ket?

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    $\begingroup$ Yes, numbers come out of the product. Indeed, operators normally do not come out of the products, and act either on the kets, or the bras by adjoint action. When their eigenvalues on either are simple, by abuse of notation they may be thought to be pulled out of such products, but one must be careful. Is that what confuses you? Do you know how to think of operators as matrices acting on vectors? $\endgroup$ Commented Apr 9, 2020 at 16:29

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One of the properties of a vector is that it can be scaled by a scalar. In this case, the vector $|\chi\rangle$ is scaled by the complex number $\alpha e^{i\theta}$. This vector still “points” along the same direction as $|\chi\rangle$ except that it is scaled by a factor of $\alpha e^{i\theta}$. What this means is that the inner product of $|\chi\rangle$ with any other vector will also be scaled by the same amount. This is why we can take it out of the inner product.

When an operator $U$ acts on the vector $|\chi\rangle$ however, in general the resulting vector will “point” in some other direction. This means that the inner product will not be preserved. So we can’t take the operator out. Except of course in the case when $|\chi\rangle$ is an eigenvector of $U$.

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