Interpretation of observable of quantum state equation

Question

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.

Answer 1

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.

Answer 2

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.