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I am studying Quantum Mechanics, and the book by Griffths introduces some concepts that I have never come across in my Math courses. I will try to summarize my questions, and hopefully someone will be able to give me some directions.

First of all, the inner product $\langle .,.\rangle$ is defined as the operation with the following properties:

$\langle \alpha,\beta \rangle = \langle \beta,\alpha \rangle^{*}$

$\langle \alpha,\alpha \rangle \geq 0$ and $\langle \alpha,\alpha \rangle=0 \iff | \alpha \rangle = |0\rangle$

$\langle\alpha|(b|\beta\rangle+c|\gamma\rangle)=b\langle \alpha,\beta\rangle + c\langle \alpha,\gamma\rangle$

Then, for an orthonormal basis $\langle \alpha,\beta\rangle = a_{1}^{*}b_{1}+a_{2}^{*}b_{2}+...+a_{n}^{*}b_{n} $. So is this just a convention? Could $\langle \alpha,\beta\rangle $ also be $b_{1}^{*}a_{1}+...+b_{n}^{*}a_{n}$?

Then a Hermitian Transformation is defined as a transformation such that: $$ \langle \hat{T}^{\dagger}\alpha|\beta\rangle = \langle\alpha|\hat{T}\beta\rangle $$ Then the author goes on to say that, in particular: $ \langle \alpha|c\beta\rangle = c\langle\alpha|\beta\rangle$ but $ \langle c\alpha|\beta\rangle = c^{*}\langle\alpha|\beta\rangle$ for any scalar $c$. My other question is: how can these last two equalities be derived from the definition of scalar product and Hermitian Transformation?

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  • $\begingroup$ The first equation’s is written in page 92.You can check it yourself(after understanding former section). $\endgroup$ – Harry May 27 at 14:06
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Then, for an orthonormal basis $⟨α,β⟩=a_{∗1}b_1+a_{∗2}b_2+...+a_{∗n}b_n$. So is this just a convention? Could $⟨α,β⟩$ also be $b_∗1a1+...+b_{∗n}a_n$?

Yes, a Hermitian form could be conjugate-linear in the second argument instead. The latter is probably more usual for mathematicians, the former for physicists.

My other question is: how can these last two equalities be derived from the definition of scalar product and Hermitian Transformation?

For the first, use the third axiom of the inner product with $|\gamma\rangle=0$. Then we get the second by $\langle c\alpha|\beta\rangle=(\langle \beta|c\alpha\rangle)^*=c^*\langle \beta|\alpha\rangle^*=c^*\langle\alpha|\beta\rangle$. These identities don't depend on the notion of Hermitian transformation at all.

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    $\begingroup$ It's "convenient" to associate the complex conjugation with the linear operator (the object in the dual space) instead of the vector. For physicists, then, since the kets are the states, it makes more sense to complex conjugate the coefficients from the bras. (Which is not to detract from @Kevin Carlson's answer, just to elaborate on convenience.) $\endgroup$ – Eric Towers Mar 10 '14 at 3:06

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