# Quantum state as a linear combination of eigenstates of any operators

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?

• 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? Jan 27, 2022 at 13:39
• 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.
– Dan
Jan 27, 2022 at 13:50
• In QM we can describe every state $\psi$ by its density matrix $|\psi\rangle\langle\psi|$ and that's a Hermitian operator. Jan 27, 2022 at 13:55
• I've edited the question. Jan 27, 2022 at 13:57
• @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? Jan 27, 2022 at 14:09

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 $$a_i$$.