So I followed this lecture:


which starts of with the statement:

If you have a Schrödinger equation for an energy eigenstate you have

$$-\frac{\hbar}{2m}\frac{d^2}{dx^2}\psi(x) + V(x)\psi(x) = E \psi(x)\tag{1}$$

Question 1: What does it mean to have a energy eigenstate in this context? All eigenstates I ever cared about were the eigenstates and eigenfunctions of Hamiltonians.

Question 2: Is equation (1) a general statement or specific to some conditions? Usually I assumeed that Schrödingers-equation is used for time-evolutions but this doesn't seem to be the case here.


This is the time independent Schrödinger equation. It is basically an eigenvalue problem

$$\hat{H}\psi=E\psi $$ where $$\hat{H}=-\frac{\hbar^2}{2m}\nabla^2+V(x)$$

is the Hamiltonian of the system. Since you yourself mentioned eigenstates of the Hamiltonian I'm going to guess you already know about why the Hamiltonian has this form. The solution of this equation are the eigenstates of the Hamiltonian operator, a set of eigenvectors and eigenvalues.

Probably, the Schrödinger equation you have in mind is the time dependent Schrödinger equation

$$i\hbar\frac{\partial}{\partial t}|\psi\rangle=\hat{H}|\psi\rangle $$

Why do we need two separate equations? Well, the true equation of motion of the state is the time dependent version, nevertheless, considering the appearance of the hamiltonian, it's useful to solve the time independent one and get the eigenstates. Why? Because once you have some states $|n\rangle$ such that $\hat{H}|n\rangle=E_n|n\rangle$ you can verify that

$$ |\psi(t)\rangle=\sum_n a_n\exp \left(-i\frac{E_n t}{\hbar} \right)|n\rangle$$

is a solution to the time dependent version for some coefficients $a_n$.

  • $\begingroup$ First: Thank you very much that was very helpful! Second: You said that the solution of the time independent version are ' energy eigenstates of the Hamiltonian operator, a set of eigenvectors and eigenvalues.' I always assumed that the vectors are the states ? At least that's the case in the bracket notation. Am I missing here something ? What is the difference between an eigenstate and an eigenvector? $\endgroup$ – CatoMaths Jan 11 '19 at 23:15
  • $\begingroup$ @CatoMaths no difference - sorry for using both terms, it can be confusing. "Eigenstate" is just a name we give to an eigenvector of a physical observable. It's just a name we give out of physical interpretation. The solution of the time independent equation is a set of eigenvectors $|n\rangle$ and eigenvalues $E_n$. The eigenvectors are usually called eigenstates, and the eigenvalues are interpreted as the energies of the corresponding eigenstates. $\endgroup$ – user2723984 Jan 11 '19 at 23:21
  • $\begingroup$ Also I should probably not have written "energy eigenstates of the Hamiltonian", usually one says "energy eigenstates" to refer to the eigenstates of the Hamiltonian, but writing both is redundant and confusing, I edited it. $\endgroup$ – user2723984 Jan 11 '19 at 23:24
  • $\begingroup$ Perfect, finally I am starting to understand! Thank you so much. If you don't mind I would have a minor question on this topic. We said that the solution of this equation is the eigenvector and corresponding eigenvalue. Let's call them $\mid \psi_n \rangle$ and $E_n$. I am having a hard time to make the connection between the normal notation $\psi_n$ and the bracket notation $\mid \psi_n \rangle$. This may sound stupid but I am used to be always in the backet notation so seeing $\psi_n$ now as non-vector is somewhat confusing to me. $\endgroup$ – CatoMaths Jan 11 '19 at 23:37
  • $\begingroup$ In bra-ket notation, vectors are denoted by $|\psi\rangle$. If you have an eigenbasis $|x\rangle$ of some observable, for example position, you can decompose $|\psi\rangle$ as $\int dx \langle x|\psi\rangle |x\rangle$ and we note by $\psi(x):=\langle x|\psi\rangle$, this is the wave function. It's not a vector in an Hilbert space, it's a function (or more generally a tempered distribution). Often physicists interchange the two in somewhat confusing ways, see this question of mine from when I was confused too :) $\endgroup$ – user2723984 Jan 12 '19 at 8:31

You have stated that the only eigenstates you care about are for the Hamiltonian. That equation IS the Hamiltonian for a non-relativistic particle.

The differential operator on the left hand side is H, the Hamiltonian operator, the E on the right hand side is the energy, a scalar number. Solving this equation for the allowed wavefunctions and energies provides you with a complete set of eigenfunctions and eigenvalues {psi_n, E_n}. By the way I think you are missing a square on your h_bar.

As for the time-dependent equation, its solutions can be built up from the eigenfunctions since the span Hilbert space. This is a common approach to solving time dependent PDE, and is used in acoustics, optics and all other wave mechanics.

As for it being a special case? The only things special about the equation you have posted are (1) it is non-relativistic, (2) it is 1-dim (in 3-dim the second derivative would be replaced with the Laplacian operator), and (3) no boundary conditions are explicitly mentioned, e.g. psi(x0) = 0 or psi(infinity) = 0, etc.


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