I was listening to a lecture, and it referred to states being orthogonal. It would say H|a>=n|a> and in this |a> would be orthogonal. What does this mean?
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7$\begingroup$ You've asked several basic linear algebra questions while trying to learn QM. I suggest you go back and learn linear algebra separately, because otherwise QM is going to be a long, long slog. This is like trying to learn mechanics without algebra. $\endgroup$– knzhouCommented Aug 6, 2016 at 20:16
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$\begingroup$ Watch the James Binney (Oxford University) lectures on youtube and there is a free ebook that is based on the lectures. Like knzhou says, it's a long slog, best of luck with it. $\endgroup$– user108787Commented Aug 6, 2016 at 20:28
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$\begingroup$ It's not that I don't know linear algebra, its just some stuff fails to come into memory. I know most of it, allowing me to move through quite simply, but if I were to try and find these terms in my notes it would take quite some time. $\endgroup$– PhiCommented Aug 6, 2016 at 22:15
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2 Answers
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One state is not orthogonal. Orthogonality is a relation between two states $| \phi \rangle$ and $| \varphi \rangle$. Two such states are orthogonal iff $$ \langle \phi | \varphi \rangle = 0$$ A set of states $\left\{ |\varphi \rangle_i | i \in I \right\}$ for an index set $I$ can be mutually orthogonal if the above holds for any pair.
Eigenvectors to different eigenstates of an operator can also be shown to be orthogonal. This has been discussed on Physics SE many times.
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$\begingroup$ Thanks for answering, and I recalled this definition after posting my question and looking through my year old notes. Sorry for the bad wording, and again thanks for answering. $\endgroup$– PhiCommented Aug 6, 2016 at 22:17
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This is because in linear algebra it can be shown that if your $H$ is normal, then there exists orthonormal basis in which it is diagonal.
answered Aug 6, 2016 at 20:14