I am confused as to what hermiticity of an operator means when given a basis set.
My course notes say that hermitian operators in Hilbert space stay unchanged under it's complex conjugate:
$$<n|A|m>^* = <m|A|n>.$$ And that mathematically hermiticity imply that the eigenvalues are always real, and there exists an orthonormal set of eigenvectors.
The 'completeness relation' says that: $$\hat A\Sigma_n|n><n|=\Sigma_n\lambda_n|n><n|, $$ With $\hat A$ the operator, $|n><n|$ the orthonormal set and $\lambda$ the eigenvalues.

My question is:

Given an orthonormal basis $|\psi_n>$ in Hilbert space of an operator so that:
$$\hat A|\psi_n>=a_i|\psi_n>,$$
or a linear combination (eg. $\hat A|\psi_n>=a_i|\psi_n>+\space b_i|\psi_n>+...$),
Does this also mean that operator $\hat A$ is always hermitian, or is it only hermitian if the coëfficients $a_i$ are real? Are in this case $a_i$ eigenvalues or are they coëfficients determining the chance a certain observation $|a_i|^2/N^2<\psi_n|\psi_n>$, ($N$ is normalisation), takes place? If these are not the eigenvalues, how are they determined from this basis set?

  • $\begingroup$ Possible duplicate: Difficulties with bra-ket notation $\endgroup$
    – Qmechanic
    Sep 4, 2022 at 8:13
  • $\begingroup$ It looks like your mixing up eigenvalues and expansion coefficients. Also basis vectors and eigenvectors. Any set of vectors that span the space of interest can be used as basis set. The basis set does not have to be connected to any operator. We usually use the set of eigenvectors of a hermitian operator as basis since they have convenient properties like orthogonality but we don't have to. $\endgroup$
    – Hans Wurst
    Sep 4, 2022 at 9:24

2 Answers 2


Hermitian operator

Your definition is ok and equivalent to

$\langle a| A b \rangle = \langle A a| b \rangle$

Eigenspace of a Hermitian operator

It's possible to prove that a Hermitian operator has eigenvectors that form an orthogonal vector base, associated with real eigenvalues

$A | n \rangle = \lambda_n | n \rangle$.

Than you can normalize eigenvectors to get a orthonormal basis.

Generic wave function

Given a generic wave function $| \psi \rangle$ you can write it using the basis $\{ |n\rangle\}_n$, as a superposition of states

$| \psi \rangle = a_n |n\rangle $

so that you can write the action of the operator $A$ as

$ A | \psi \rangle = a_n A |n\rangle = a_n \lambda_n |n\rangle $.

Normalization condition and interpretation as a probability

With the normalization condition for the wave function $\langle \psi|\psi \rangle = 1$ (holding also for eigenvectors), you readily get

$1 =\langle \psi|\psi \rangle = \langle a_n n | a_m m \rangle = \sum_n a_n^2$.

If a system is ina state descirbe by wavefunction $|\psi \rangle$, the probability of measuring state $| m \rangle$ is

$|\langle \psi | m \rangle|^2 = |\langle a_n n | m \rangle|^2 = a_m^2$.

Eigenvalues and result of measurement process

On the other hand, eigenvalues of an operator are connected to the values you get from measurement process for the physical quantity associated with that oeprator. See Born's rule, https://en.m.wikipedia.org/wiki/Born_rule


The coefficients $a_i$ are the eigenvalues when $\hat{A}|\psi_n\rangle = a_n|\psi_n\rangle$. And, yes, an operator can have an orthonormal basis and not be hermitian. The eigenvalues must be real. Consider, for example, the operator \begin{align} M &= \left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right]. \end{align} It is visually not hermitian. In fact, it is anti-hermitian $M^\dagger = -M$. You can verify that this operator has an orthonormal basis: $v_1 = (|e_1\rangle - i|e_2\rangle)/\sqrt{2}$, and $v_2 = (|e_1\rangle + i|e_2\rangle)/\sqrt{2}$. The corresponding eigenvalues are $-i$ and $i$, respectively. You can prove the pattern that all anti-hermitian operators have imaginary eigenvalues with orthonormalizable eigenvectors (i.e. eigenvectors with the same eigenvalue are not, necessarily, normal).

The "linear combination" idea, $\hat{A}|\psi_n\rangle = \sum_n a_n|\psi_n\rangle$ doesn't work, though. Consider the matrix \begin{align} M &= \left[\begin{array}{cc} 2 & 1 \\ 0 & 2 \end{array}\right]. \end{align} It has one eigenvalue, 2, and one eigenvector $|e_1\rangle$, and that's it. You can write the action of $M$ on $|e_2\rangle$ as $3|e_1\rangle + 2|e_2\rangle$, which satisfies the "linear combination" condition stated, but this operator is not diagonalizable. This is an example from the theory of square matrices of something called a Jordon block (a square block with zeros everywhere except possibly the main diagonal where all of the values are identical and ones in the row above the main diagonal). It turns out that any square matrix in finite dimensions can be "diagonalized" into a sequence of Jordon blocks along the diagonal, given the right basis. (Anti-)Hermitian matrices are just a special case where all of the Jordon blocks have dimensionality 1.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.