Can any physically realisable state be expressed as a linear combination of eigenstates of any Hermitian operator?

For example, as a linear combination of the eigenstates of the Hamiltonian or as a linear combination of the eigenstates of the spin operator $S^2$?

This is because if I have a physical state, any measurement of a Hermitian operator will give me a certain result and then the measurement operation will collapse the wavefunction into the eigenfunction relative to the eigenvalue obtained by the measurement, right?

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    $\begingroup$ With reference to your title and first sentence: A state is not a combination of operators. Do you mean combinations of eigenstates of operators? Do you mean an eingenstate if a combination of operators? $\endgroup$ Jan 27, 2022 at 13:39
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    $\begingroup$ As @ZeroTheHero suggests, you should tidy this question a bit. Also, you can in general write a state as superposition of a complete basis. If the operator's eigenstates are a complete basis then yes, you can do the decomposition. If the basis is not complete then it will depend on the circumstances. $\endgroup$
    – Dan
    Jan 27, 2022 at 13:50
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    $\begingroup$ In QM we can describe every state $\psi$ by its density matrix $|\psi\rangle\langle\psi|$ and that's a Hermitian operator. $\endgroup$
    – Kurt G.
    Jan 27, 2022 at 13:55
  • $\begingroup$ I've edited the question. $\endgroup$
    – Salmon
    Jan 27, 2022 at 13:57
  • $\begingroup$ @Dan I mean: since a measurement of a Hermitian operator on any state gives a result, doesn't this mean that the state before the measurement was in a superposition of eigenstates of the operator? even if they do not form a complete basis? $\endgroup$
    – Salmon
    Jan 27, 2022 at 14:09

1 Answer 1


Actually, in a finite dimensional Hilbert space, any pure state is the eigenstate of some hermitian observable. You don’t need to go to linear combos.

The difficulty is finding and interpreting this observable, but constructing it is not so hard. Take your state expressed in any basis, and declare $\vert 1\rangle$ to be this state. Find $(n-1)$ states so that $\langle i\vert j\rangle=\delta_{ij}$ (in $n$-dimensional space). Your hermitian operator is then $$ \hat A=\sum_{i=1}^n a_i \vert i\rangle\langle i\vert $$ with $a_i$ the (real) eigenvalue. By construction $\vert 1\rangle$ is an eigenstate with eigenvalue $a_i$.

  • $\begingroup$ Thank you for your answer. anyway, did my question makes sense? Since I can make any measure over a state of an hermitian operator and the measure will give a value, namely an eigenvalue of the operator, I can always express a state as a linear combo of some hermitian operator. In fact all the eigenstates of an hermitian operator form a complete set, right? $\endgroup$
    – Salmon
    Jan 29, 2022 at 22:17
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    $\begingroup$ yes that’s the key they form a complete set although you might need other operators to differentiate repeated eigenvalues or account for degrees of freedom outside the Hilbert space of states of your original operator… the obvious example is spin, which needs to be added to whatever to account for the spin of the system. $\endgroup$ Jan 29, 2022 at 22:19
  • $\begingroup$ Everything clear, thanks. $\endgroup$
    – Salmon
    Jan 29, 2022 at 22:21

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