UCSD's online QM notes, as usual, starts stating that QM operators are Hermitian and says that operator $O$ elements can be computed by

$$O_{ij} = \langle u_j|O|u_i\rangle$$

The $u_i$ are eigenvectors, which are orthogonal to each other, thanks to being Hermitian. So, I suppose that $O$ is a Hermitian operator. But where are the eigenvalues?

I expect that when you apply the operator to its eigenvector, a $\lambda$ must appear. Yet, I see it nowhere. Neither, text says that $O$ is a diagonal, as I may expect if I apply $\langle u_i|$ to $O|u_i\rangle = \lambda_i |u_i\rangle$, I should get $\lambda_i$ on the main diagonal and 0 everyelse. Are $u_i$ eigenvectors of the $O$ or some another operator?

  • $\begingroup$ Put it in this way. Let $\hat{O}$ be an operator. Insert completeness relation $$\sum_i |u_i \rangle \langle u_i | = \hat{1}$$, we have $$\hat{O}=\sum_{ij} |u_i \rangle \underline{\langle u_i |\hat{O} |u_j \rangle} \langle u_j | $$. The underline part is the representation of operator $\hat{O}$ in $\{u_i\}$ basis. If $u_i$ is the eigenfunction of operator $O$, namely $\hat{O} |u_i \rangle = \lambda_i |u_i \rangle$ we have $\hat{O}=\sum_i \lambda_i | u_i \rangle \langle u_i| $, which sometimes called spactral expansion of operator. The matrix element is $$O_{ij}=\lambda_i \delta_{ij} $$ $\endgroup$
    – user26143
    Commented Dec 10, 2013 at 16:41
  • $\begingroup$ The eigenvalue $\lambda_j$ depends on the operator $\hat{O}$ and the eigenfunction $|u_j\rangle$. For example, the quantum harmonic oscillator $\hat{O}\equiv\hat{H}= \hat{p}^2/2m+m\omega^2\hat{x}^2/2$ with $\lambda_j\equiv E_n=(n+\frac12)\hbar\omega$. $\endgroup$
    – Kyle Kanos
    Commented Dec 10, 2013 at 16:47
  • $\begingroup$ @user26143 I do not understand a couple of things. What is the completeness relation? $\endgroup$
    – Val
    Commented Dec 10, 2013 at 20:15
  • $\begingroup$ It is a standard result in many textbooks of quantum mechanics. E.g. Sakurai's modern quantum mechanics, section 1.3, p19. $\endgroup$
    – user26143
    Commented Dec 10, 2013 at 21:19
  • $\begingroup$ @user26143 Do you mean that this is not a simple consequence of basis orthonormality? $\endgroup$
    – Val
    Commented Dec 10, 2013 at 22:01

1 Answer 1


In those notes you linked to, the $|u_i\rangle$ form an orthonormal basis. They are not necessarily the eigenkets of the operator $O$. By choosing different bases $\{u_i\}$, the "representation" of the operator $O$ (e.g.., how you write it out in a matrix) will change.

You are correct that if one chooses the set of kets that are eigenkets of the operator, you'll get a diagonal matrix. But in those notes they are trying to keep it general.


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