2
$\begingroup$

I came across this simple proof:

We show that Hermitian operators have real eigenvalues. The definition of a Hermitian operator is

\begin{equation} \langle \phi_i | \hat A | \phi \rangle = \langle \phi_i | \hat A | \phi \rangle^* \tag{1} \end{equation}

Then if $|\psi\rangle$ is an eigenvector of $\hat A$, we have

$$ \hat A|\psi\rangle = \lambda|\psi\rangle \tag{2}$$

and therefore

$$ \langle\psi|\hat A|\psi\rangle = \lambda . \tag{3}$$

If $\hat A$ is hermitian, we my apply (1) so that

$$\langle\psi|\hat A|\psi\rangle = \langle\psi|\hat A|\psi\rangle^*$$ $$\lambda = \lambda^*.$$

What I am not getting, is the step from (2) to (3). Seems to me that would be true only if $\psi$ is normed ($\langle\psi|\psi\rangle = \int \psi^*\psi \text d \tau = 1$).

Is it true in general that eigenvectors/eigenfunctions of operators are normed and orthogonal?

$\endgroup$
  • $\begingroup$ See also Applications of the Spectral Theorem to Quantum Mechanics $\endgroup$ – John Rennie Jun 22 '18 at 14:59
  • 1
    $\begingroup$ @JohnRennie The duplicate question is asking about the general consequences of observables corresponding to Hermitian operators. This question seems to be asking a minor technical clarification about eigenvector normalisation. $\endgroup$ – gj255 Jun 22 '18 at 15:12
  • 2
    $\begingroup$ @SeanBone Eigenvectors are defined merely as non-zero vectors $|\psi\rangle$ such that $A |\psi\rangle = \lambda |\psi\rangle$. There is no constraint on normalisation, and you can easily check that if $|\psi\rangle$ is an eigenvector, then so is $c|\psi\rangle$ for any non-zero $c$. Hence in going from (2) to (3), there is an implicit assumption as you suspect. Commonly we take states in QM to be of unit norm, but not always. $\endgroup$ – gj255 Jun 22 '18 at 15:15
  • $\begingroup$ If the wave function is not normalised then predictions based on it will be wrong. There is no theoretical basis for normalisation. $\endgroup$ – my2cts Jun 22 '18 at 18:23
2
$\begingroup$

The property of orthogonality can always be imposed, but it is not required at all in the excerpt you've cited.

The normalization of the eigenvectors can always be assured (independently of whether the operator is hermitian or not), by virtue of the fact that if $Av=\lambda v$, then any multiple $w=\alpha v$ of that vector will obey $$ Aw = A\alpha v = \alpha A v = \alpha \lambda v = \lambda w. $$ Thus, given any eigenvector of any operator, you can always assume (for free) that it's been normalized to unity.

However, this is also not necessary for the manipulations you've cited: if you remove that normalization, then your equation $(3)$ becomes $$ \langle\psi|\hat A|\psi\rangle = \lambda \langle\psi|\psi\rangle, \tag{3'}$$ in which $\lambda \langle\psi|\psi\rangle $ is (by the properties of the inner product) a real and positive number. The rest of the manipulations are unaffected: you get to $$ \lambda \langle\psi|\psi\rangle = \lambda^* \langle\psi|\psi\rangle $$ and all you need to do is divide by $\langle\psi|\psi\rangle $.

$\endgroup$
0
$\begingroup$

One can always work in an orthonormal basis. By default, we do this when working with quantum mechanics for convenience.

Note that if $\hat{A}|\psi\rangle=\lambda|\psi\rangle,\,\hat{A}^\dagger|\phi\rangle=\mu|\phi\rangle$ then $\langle \phi|\hat{A}|\psi\rangle$ is equal to both $\lambda\langle\phi|\psi\rangle$ and $\mu^\ast\langle\phi|\psi\rangle$, so either the vectors are orthogonal or $\lambda=\mu^\ast$. We can even ensure orthogonality in this special case with a basis change called the Gram-Schmidt process. Finally, we can rescale eigenvectors to have unit norm. This allows such convenient results as $\operatorname{id}=\sum_i |i\rangle\langle i|$ so that $|\Psi\rangle=\sum_i \langle i|\Psi\rangle|i\rangle$.

$\endgroup$

Your Answer

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

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