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Thread: Hilbert Space

  1. #1
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    Hilbert Space

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

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

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

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

    $\displaystyle \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 $\displaystyle | \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,
    $\displaystyle \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} $
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