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!



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: $$ \hat{H}_{|x_i\rangle} \phi(x_i) = x_i \phi(x_i) $$


2 Answers 2


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.


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


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.


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