Why is the Schrödinger Equation valid for the component functions (wave function) of state vectors?

Question

I'm new to quantum mechanics and confused about the way the Schrödinger equation is used (more general eigenvalue equations of observables).

Let's take the time-independent Schrödinger equation (eigenvalue equation).

Suppose our system is n+1 dimensional and the eigenvectors of our hamiltonian (or any other observable) are $|x_i\rangle $ with eigenvalue $x_i$. A state vector takes the form: $|\psi\rangle = \sum_{i=0}^n \phi(x_i)|x_i\rangle$ with component/(wave) functionf $\phi(x_i) = \langle x_i | \psi \rangle$. The time-independent Schrödinger equation (or eigenvalue equation for any observable) reads: $$ \hat{H} |x_i \rangle = x_i |x_i \rangle $$ This is an eigenvalue equation for a $\mathbf{\text{state vector}}$.

My confusion is why that equation should also hold for the component (wave) function. Ie why is: $$ \hat{H} \phi(x_i) = x_i \phi(x_i) $$ true?

Thanks in advance!

Cheers,

Thomas

EDIT: I think it is only valid, if the Hamilton operator (or the observable) is itself expressed in the $|x_i\rangle$ basis (it then becomes a matrix (at least in the finite dimensional case). Even though I never saw that spelled out... So one should better write: $$ \hat{H}_{|x_i\rangle} \phi(x_i) = x_i \phi(x_i) $$

Answer 1

In standard bra-ket notation, one could write $$\hat H|\psi\rangle = E|\psi \rangle\implies \int \mathrm dx \int\mathrm dy\ |x\rangle\langle x|\hat H|y\rangle\langle y|\psi\rangle = E\int\mathrm dx \ |x\rangle\langle x|\psi\rangle$$ where we have inserted two copies of the identity operator $\mathbf 1= \int\mathrm dx \ |x\rangle\langle x|$ on the left and one copy on the right. For a generic Schrodinger Hamiltonian of the form $$\hat H = \frac{1}{2m} \hat P^2 + V(\hat X)$$ then we have $$\langle x|\hat H| y\rangle = \delta(x-y) \left(-\frac{\hbar^2}{2m} \nabla^2+V(x)\right)\equiv \delta(x-y) \hat h$$ where $\hat h$ is sometimes called the expression for the Hamiltonian in the position basis. Since $\langle x|\psi\rangle \equiv \psi(x)$, this becomes $$\int \mathrm dx \ |x\rangle \hat h\psi(x) = \int \mathrm dx \ |x\rangle E \psi(x) \iff \hat h\psi (x) = E \psi(x)$$

In an abuse of notation, sometimes the distinction between the abstract operator $\hat H$ and its representation in the position basis $\hat h$ is not made explicit, with the proper meaning of the symbols to be understood from context.

Answer 2

Algebraic relations are independent of the choice of basis, so the expression:

$$\hat{H}|x_i\rangle=x_i|x_i\rangle$$

doesn't require any choice of basis (only that $|x_i\rangle$ is an eigenvector of $\hat{H}$).

When you project this relation on $|x_i\rangle$, the left-hand side becomes $\langle x_i|\hat{H}|x_i\rangle$, which is by definition the $i$th matrix element of $\hat{H}$ on its diagonal. You can write it $\hat{H}_{|x_i\rangle}$ if you want, but it isn't a notation I've ever seen. $\langle x_i|\hat{H}|x_i\rangle$ is just fine.