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In Shankar QM it is stated that since the $\boldsymbol K$ operator is Hermitian, vectors, which are expanded in the $\boldsymbol X$ basis with components $f(x) = \langle x | f \rangle$, must have an expansion in the $\boldsymbol K$ basis (i.e. the eigenbasis of $\boldsymbol K$).

Why is this?

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    $\begingroup$ This has nothing to do with being Hermitian. You can express vectors and operators in whatever basis you want. The definition of a basis is that you can do this. I think you aren't quite asking what you mean to ask. $\endgroup$ – DanielSank Jan 24 '15 at 4:19
  • $\begingroup$ That's what I thought, but he seemed to go out of the way to make the distinction. It was in a comment which set up the Fourier Transform as the Transformation from $\boldsymbol X$ to $\boldsymbol K$ $\endgroup$ – kypalmer Jan 24 '15 at 4:26
  • $\begingroup$ @DanielSank well the basis does have to be complete to do that. I'm guessing the $\boldsymbol K$ basis is supposed to be the eigenbasis of the operator, so the question boils down to whether the eigenbasis of a Hermitian operator is complete. kypalmer can you confirm? (and edit your question accordingly) $\endgroup$ – David Z Jan 24 '15 at 4:27
  • $\begingroup$ @DavidZ: What's a "basis" if it's not complete? To me "basis" means "complete set of vectors". $\endgroup$ – DanielSank Jan 24 '15 at 4:29
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    $\begingroup$ Again, you're asking why a vector can be expanded in two different bases. The answer is because a basis is defined as a set of vectors in which you can expand any other vector. $\endgroup$ – DanielSank Jan 24 '15 at 4:42
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Theorem. If $\hat{K}$ is Hermitian then the eigenvectors of $\hat{K}$ can be chosen so as to form an orthonormal basis $|\xi_i\rangle$.

(The "can be chosen" wording completely handles the cases of degenerate eigenvalues, and the fact that an eigenvector times a constant is still an eigenvector)

(the converse is not true. In fact replace the word "Hermitian" with "Normal" and the theorem still holds. See: the wikipedia page on the spectral theorem. Also, I'm mainly considering the finite dimensional case)

Theorem. If $\{|\xi_i\rangle\}$ forms an orthonormal basis, then $I=\sum |\xi_i\rangle \langle \xi_i|$, where $I$ is the identity operator.

So, if your vector is $|\varphi\rangle$, you can also write it as $|\varphi\rangle=I|\varphi\rangle=\sum |\xi_i\rangle \langle \xi_i|\varphi\rangle$

if you're in a basis given by $\{ |n\rangle \}$, then $\langle n|\varphi\rangle$ are your components and $I=\sum |\xi_i\rangle \langle \xi_i|=\sum |n\rangle \langle n|$, so:

$|\varphi\rangle=II|\varphi\rangle=\sum_{i,n}|\xi_i\rangle \langle \xi_i|n\rangle \langle n|\varphi\rangle$

which is a change of basis represented by matrix multiplication, with $\langle \xi_i|n\rangle$ being the matrix elements.

My point in all this is that those two theorems when combined with Dirac notation are very nice!

Bonus:

Consider two particularly interesting bases:

  1. The basis of delta functions $\{|x\rangle\}$ where $\langle x | y\rangle =\delta(x - y)$. The $|x\rangle$ vectors are vectors with specific position in space.
  2. The basis of exponentials $\{|\nu \rangle\}$ where $\langle x | \nu \rangle = e^{i 2 \pi \nu x}$. These have specific spatial frequency. Note that $\langle \nu | \mu \rangle = \delta(\nu - \mu)$.

Now consider the equation we had at the end of the previous section:

\begin{align} |f\rangle &= \int dx d\nu | \nu \rangle \langle \nu | x \rangle \langle x | f \rangle \\ \langle \nu' | f \rangle &= \int dx d\nu \langle \nu'| \nu \rangle e^{-i 2 \pi \nu x} f(x)\\ &= \int dx d\nu \delta(\nu - \nu') e^{-i 2 \pi \nu px} f(x) \\ \tilde{f}(\nu') &= \int dx \, e^{-i 2 \pi \nu' x} f(x) \, . \end{align} So yeah, the Fourier transform is a change of basis.

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    $\begingroup$ By the way, using this nice notation you can explain Green's functions and essentially every use of linear algebra in physics. $\endgroup$ – DanielSank Jan 24 '15 at 4:52
  • $\begingroup$ So, basically, as long as both $ | \xi_i \rangle$ and $ |n\rangle$ are complete bases, then they are equivalent in their ability to describe any vector. Is this right? Does any of this change for an infinite dimensional function space? $\endgroup$ – kypalmer Jan 24 '15 at 4:59
  • $\begingroup$ @kypalmer: Glad you asked. Check out my edit to the answer. The "reason" the exponentials $\exp{i 2\pi \nu x}$ form a complete basis is that they are the eigenvectors (eigenfunctions) of the Hermitian operator $-iD$ where $D$ is the derivative with respect to $x$! I have written a very extensive and pedagogical document on this topic. If you would like me to email you a copy you can find my email address on my profile page. Don't be shy! I wrote it to help people like you. $\endgroup$ – DanielSank Jan 24 '15 at 5:07
  • $\begingroup$ There it is! Good extension to infinite dim. $\endgroup$ – kypalmer Jan 24 '15 at 5:13

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