1. ## Hilbert Space

Suppose we want to compute $\langle \alpha| \beta \rangle$. Let $|\alpha \rangle = \sum_{n} a_{n}| \phi_{n} \rangle$ and $|\beta \rangle = \sum_{n} b_{n}| \phi_{n} \rangle$.

So $\langle \alpha| \beta \rangle = \alpha^{*} \beta$. In expanding $|\beta \rangle$, how do we get $\langle \alpha| \beta \rangle = \sum_{n} b_{n} \langle \alpha| \phi_{n} \rangle$? Because by linearity this is equaled to $\sum_{n} \alpha \langle b_{n}| \phi_{n} \rangle$. But shouldn't is be $\alpha^{*}$? Because then we eventually get:

$\langle \alpha| \beta \rangle = \sum_{n} b_{n} \left(\sum_{m} a_{m}^{*} \langle \phi_{m}| \phi_{n} \rangle \right) = \sum_{n} a_{n}^{*} b_{n}$.

2. Originally Posted by manjohn12
Suppose we want to compute $\langle \alpha| \beta \rangle$. Let $|\alpha \rangle = \sum_{n} a_{n}| \phi_{n} \rangle$ and $|\beta \rangle = \sum_{n} b_{n}| \phi_{n} \rangle$.

So $\langle \alpha| \beta \rangle = \alpha^{*} \beta$. In expanding $|\beta \rangle$, how do we get $\langle \alpha| \beta \rangle = \sum_{n} b_{n} \langle \alpha| \phi_{n} \rangle$? Because by linearity this is equaled to $\sum_{n} \alpha \langle b_{n}| \phi_{n} \rangle$. But shouldn't is be $\alpha^{*}$? Because then we eventually get:

$\langle \alpha| \beta \rangle = \sum_{n} b_{n} \left(\sum_{m} a_{m}^{*} \langle \phi_{m}| \phi_{n} \rangle \right) = \sum_{n} a_{n}^{*} b_{n}$.
I'll have to make a few assumptions in order to answer this question.

1. I assume that the vectors $| \phi_{n} \rangle$ are supposed to form an orthonormal basis for the Hilbert space.

2. I assume that you are using the physicists' convention that the inner product is linear in the second variable and conjugate-linear in the first variable (mathematicians always use the opposite convention: linear on the left and conjugate-linear on the right).

Under those assumptions,
\begin{aligned}\langle \alpha| \beta \rangle &= \sum_{n} b_{n} \langle \alpha| \phi_{n} \rangle\quad(\text{linearity on the right}) \\
&= \sum_{n} b_{n} \Bigl(\sum_{m} \langle a_{m}\phi_{m}| \phi_{n} \rangle \Bigr)\quad(\text{de}\text{finition of }\alpha) \\
&= \sum_{n} a_{n}^{*} b_{n}\quad(\text{conjugate-linearity on the left})\end{aligned}