Recently,my professor told me that for a qudit system we can consider the generalized $\hat{\sigma}_{z}$ observable as


where $\omega=e^{\frac{2i\pi}{d}}$. This makes the eigenvalues complex number, however as I know that for Hermitian observable eigenvalues are always real number. Then how come we have complex eigenvalues for a Hermitian matrix?

  • 4
    $\begingroup$ its not, in general, hermitian. You caan see this by putting it in matrix form: $$ \text{diag}(1,e^{2i\pi/d},e^{4i\pi/d},...)^\dagger=\text{diag}(1,e^{-2i\pi/d},e^{-4i\pi/d},...)\neq\text{diag}(1,e^{2i\pi/d},e^{4i\pi/d},...) $$ $\endgroup$ Sep 11, 2020 at 16:06
  • $\begingroup$ Hint: it is a unitary operator. Now, can you see its logarithm is antihermitean? $\endgroup$ Sep 11, 2020 at 17:42
  • $\begingroup$ I don't think they really meant "observable", more likely "generalized sigma_z operator". In quantum information, one rarely measures expectation values of observables (why would one?). $\endgroup$ Sep 12, 2020 at 18:49

1 Answer 1


Assuming the $\mid k \rangle $ are orthogonal, you have written down a unitary operator. A matrices is an examples of normal operators and normal operators have a spectral decomposition.

It is not so standard, but one can consider unitary or even normal operators as observables, but they are complex-values observables.

More standard is to use two real observables that are compatible to give the same information. Essentially the real and imaginary parts of the complex observable. In your case, one would be the sum of $\cos(k) \mid k \rangle\langle k \mid $ and the other the sum of $\sin(k) \mid k \rangle\langle k \mid $.

This issue comes up all the time with position on a finite system with periodic boundary conditions. Perhaps your professor had in mind one of the approaches above.


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