Ok, so I'm beginning to study quantum mechanics. For reference, the book I'm using is "Konishi-Paffuti/Quantum Mechanics-A New introduction".

Now, I get that the quantum state of something (say, a particle) is described by its wave function, which we'll get, I assume, solving a PDE. Once I know the wavefunction, then it's just a matter of some integrals to know the expected value. From my understanding, say I have a function $f(q)$, its expected value is

$$<f(q)>=\int\psi^{*}f(q)\psi \space dq$$

And then I have the mean value of what I expect in my measurement. Now, in this picture where does the "eigenvalues of an operator" come in? Why do I need this assumption if I have an expected value already which I can compare with experimental data?

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    $\begingroup$ Did one of the answers below answer your question? If so, please accept one! If not, please leave a comment asking for more details. $\endgroup$
    – march
    Jul 5, 2015 at 5:05

2 Answers 2


Expanding on physicsphile's answer, there is an alternative way of computing the expectation value of $f(q)$, and that is to sum the possible values of $f(q)$ weighted by their respective probabilities as follows.

If $f(q)$ represents a physical dynamical variable, then it is a Hermitian (self-adjoint) operator and can therefore be diagonalized by some set of eigenfunctions $\phi_i(q)$ such that

$$f(q)\phi_i(q) = \lambda_i\phi_i(q),$$

where $\lambda_i$ is the eigenvalue. This set of eigenfunctions is an orthonormal (read: they are orthogonal and normalized) basis for the Hilbert space, and therefore any wave function can be expanded as a linear combination of these eigenfunctions,

$$\psi(q) = \sum_i a_i \phi_i(q).$$

Using your formula for the expected value, we have

$$\langle f(q)\rangle = \int dq~\psi(q)^*f(q)\psi(q) = \int \sum_i a_i^*\phi_i(q)^*f(q)\sum_j a_j\phi_j(q)=\sum_{i,j}a_i^*a_j\int \phi_i(q)^*f(q)\phi_j(q)$$

Now, $f(q)$ acting on $\phi_i(q)$ yields $\lambda_i\phi_i(q)$, and since this eigenvalue is just a number, it can be pulled out of the sum, yielding

$$\langle\phi_i(q)\rangle =\sum_{i,j}a_i^*a_j\lambda_i\int \phi_i(q)^*\phi_j(q).$$

Now, since the eigenfunctions are orthonormal, the integral evaluates to the Kronecker-delta

$$\delta_{ij} = \begin{cases} 1 & i=j\\ 0 & i\neq j \end{cases},$$

in which case the sum collapses to a single sum, yielding

$$\langle\phi_i(q)\rangle =\sum_{i}|a_i|^2\lambda_i.$$

When we recognize $|a_i|^2$ as exactly the expression in physicsphile's answer, you can see that you can interpret the integral you've written as a sum over the eigenvalues (i.e. the possible measured values of $f(q)$!) weighted by their probabilities $|a_i|^2$ in the state $\psi(q)$.


They give you more detail. Using the the eigenfunctions you can work out how probable different measurement results are. The probability of eigenvalue $i$ being measured is:

$$ \left|\int \psi \phi_i^*dq\right|^2 $$

where $\phi_i$ is the corresponding eigenvalue.


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