In quantum mechanics, my understanding of operators related to observables are as follows: By a postulate of QM, every observable can be represented as an Hermitian operator. The eigenfunctions of these Hermitian operators form a complete basis for the Hilbert space and the eigenvalues represent the values which can be measured with probability given by the squared absolute value of the expansion coefficients of the state vector in that basis. So for example, consider the eigenvalue equation for the position operator: $$\hat{X}|x\rangle = x|x\rangle,$$ this implies that if the state was $|\psi \rangle = |x \rangle$, then we are guaranteed to get $x$ after measurement, and if the state is a superposition of states then the probability of measuring the particle at a position $x$ is given by the squared absolute value of the expansion coefficient, so $|\langle x| \psi \rangle|^2$ (I understand that it is actually more like $|\langle x| \psi \rangle|^2dx$ since it is a continuous spectrum).

I want to know, if we consider a state given by a wave function $\psi(x)$, then we have that the position operator is defined by $$\hat{X}\psi(x) = x \psi(x).$$ Given this we can find the eigenfunctions. We can also compute the expectation values. My question is, what is the interpretation of $\hat{X}\psi(x) = x \psi(x)$? As I understand it is incorrect to interpret the operator acting on the state as the act of measuring the position so that the operator changes the state from $\psi(x)$ to $x \psi(x)$. So what is the meaning or interpretation of this equation? Does it have any utility independent from the expectation value and commutation relations?

Thanks for any assistance.

  • 1
    $\begingroup$ I'm not sure I understand the question. You seem to be aware that that equation is simply the definition of how the position operator acts on states represented by wavefunctions (in position space). So what exactly is your question about that definition? $\endgroup$
    – ACuriousMind
    Commented Oct 27, 2016 at 16:49
  • $\begingroup$ @ACuriousMind My question is, independently does it have any meaning? Or does it have to be used in some other context to have any use? On face value you would take it as representing the position measurement changing the waves function, which as I have learnt is not the case, so does it have any significant meaning on it's own? $\endgroup$
    – user100411
    Commented Oct 27, 2016 at 16:54
  • $\begingroup$ @ACuriousMind The OP's X operator is defined to act on kets not scalar functions. So it doesn't really make sense to let is "act on" a scalar function "psi" without context. The only context where it makes sense is where the psi is a coefficient of expansion in the x kets... $\endgroup$
    – hft
    Commented Oct 27, 2016 at 18:43
  • $\begingroup$ @hft I'm using $\hat{X} \psi(x) := \langle x| \hat{X}|\psi \rangle = x \psi(x)$. $\endgroup$
    – user100411
    Commented Oct 27, 2016 at 18:48
  • $\begingroup$ So, like I just said: "X\psi(x)" doesn't make any sense without context. But if you define it as "<x|X||psi>" then, sure, it equals x\psi(x). $\endgroup$
    – hft
    Commented Oct 27, 2016 at 22:35

2 Answers 2


Your psi function $\psi(x)$ in this case is the coefficient. Because the X eigenvalues are continuous, the coefficients you mentioned in your post are also continuous. For example, given a state $|\Psi\rangle$, this can be represented in the continuous $|x\rangle$ basis as:

$$ |\Psi\rangle = \int ~\mathrm dx~ \psi(x)|x\rangle $$

such that, for example, $$ \hat X|\Psi\rangle = \int~\mathrm dx~ \psi(x)\hat X|x\rangle = \int~\mathrm dx ~ x\psi(x)|x\rangle $$

And, just like you said in your post, the $$ |\psi(x)|^2 $$

are the probability amplitudes.

And the only way something like "$\hat X\psi(x) = x\psi(x)$" makes sense is in the context of an integration over the $|x\rangle$ kets as shown, for example, in the middle and right-hand side of the above equation.


If we consider the time independent Schrodinger equation as $$\hat{\mathcal{H}}\psi(x) = E\psi(x)$$ $\hat{\mathcal{H}}$ acts on $\psi(x)$ to give us energy eigenvalues $E$. So we have an operator (hamiltonian) acting on a wave function giving us a measurable quantity, being the eigenvalues.

$\hat{\mathcal{H}}\psi(x) = E\psi(x)$ on its own does not provide any useful information other than telling us that it is an eigenvalue problem, But it is when solving the differential equation that we are able to find the form of the wave function $\psi(x)$ and the eigen energies.

$\hat{X}\psi(x) = x\psi(x)$ is the same thing, but the position operator $\hat{X} = x$. The hamiltonian however is $\hat{\mathcal{H}} = \frac{-\hbar^2}{2m}\frac{d^2}{dx^2} + V(x)$, so we end up with differential equation:

$\frac{-\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} + V(x)\psi(x) = E\psi(x)$

All information about a particle is contained in its wave function $\Psi(x,t)$, but the only measurements of an operator which we can actually observe is one of its eigenvalues.


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