I asked a question on math exchange Are properties of linear operators reflected in matrix representations with different output and input basis?.

In that question I asked: if I had a Hermitian operator $H$, then when is its matrix representation Hermitian? With the help of comments I was able to work out that, in general, if I write a matrix representation $H$ in some orthonormal basis $B$ (same input/output basis), then the matrix representation $H_B$ is also hermitian. The same could not be said about a non-orthonormal basis.

But if I take the example of

$$H=|\psi\rangle \langle \psi|+|\phi\rangle \langle \phi|$$

where $|\psi\rangle$ and $|\phi\rangle$ are not orthogonal but normalized, and if I write a matrix representation of $H$ in the basis $\{|\psi\rangle ,|\phi\rangle \}$ , $$\begin{pmatrix} 1 & \langle \psi|\phi \rangle \\ \langle \phi|\psi \rangle & 1 \end{pmatrix} $$ it is also Hermitian.

I was able to see that this was a special case where in non-orthogonal basis also matrix representation is hermitian (if I am not wrong). It is so because we are able to write the hermitian operator as $\sum_i |i\rangle \langle i|$ where $\{|i\rangle \}$ is some non-orthogonal basis. When can we write a hermitian operator in such an outer product notation Also, is it correct that only in orthonormal basis matrix representation is also hermitian?

It is a variation of the question I asked on math exchange, I am posting this question here because it involves bra ket notation, and users on math exchange have previously pointed it out that it was not suitable for that site.

  • 1
    $\begingroup$ I do not know why you are so interested in "matrix representations" of a linear operator; however a few comments: 1) a linear operator on a finite dimensional space is always a matrix (at least you may say it is isomorphic to a matrix) so it is hermitian if, roughly speaking, it is hermitian as a matrix; 2) in infinite dimension spaces, the finite rank operators are the only ones that could make sense in a matrix notation (and anyways not every finite rank operator, only the ones that have also finite dimensional $D(A)\setminus Ker(A)$); $\endgroup$ – yuggib Apr 11 '15 at 8:52
  • $\begingroup$ 3) even if you can write a finite rank operator with finite domain of "non-triviality" effectively as a matrix, you are losing information, i.e. you are just writing the restriction of your operator as an operator from the finite dimensional space $D(A)\setminus Ker(A)$ (with dimension $N$) to the finite dimensional space $Ran(A)$ (with dimension $M$). This is isomorphic to a $N\times M$ matrix, so it has a chance of being hermitian as a matrix only if $N=M$. However it seems to me a very non-standard and ineffective way of analyzing an operator. $\endgroup$ – yuggib Apr 11 '15 at 9:00
  • $\begingroup$ @yuggib I know that I dwelling too much into a trivial thing,sorry about that. But can't help it. I am just a beginner in quantum computation. I just started reading neilson chuang and these doubts arise when I am solving the exercises. As far as this question is concerned if a linear operator is hermitian its not necessary it's matrix representation is also hermitian , it depends on the basis (non-orthonormal/orthonormal) if considering only same output input basis. I really appreciate your efforts in answering my stupid doubts :) $\endgroup$ – sashas Apr 11 '15 at 9:01
  • $\begingroup$ Whilst the information and detail given by @yuggib are fine, don't let that (or he/she) make you think that questioning or doubting or trying to look at things in different ways is a bad or stupid thing! You have nothing to apologise for! $\endgroup$ – danimal Apr 11 '15 at 10:16
  • 1
    $\begingroup$ Related post by OP: physics.stackexchange.com/q/175287/2451 , physics.stackexchange.com/q/174885/2451 $\endgroup$ – Qmechanic Apr 11 '15 at 10:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.