I just worked out the 1D free particle solution to the Schrödinger equation.

My wave function was \begin{equation} \psi(x,t) = Ae^{i(px-Et)/\hbar} \end{equation} So I plugged this into both sides of the time-dependent Schrödinger equation and started to verify. I did LHS and RHS separately.
I then ended up with \begin{equation} i\hbar\frac{\partial}{\partial t}\psi = \frac{1}{2}\frac{p^2}{m}\psi \end{equation}

which looks like the correct form for the free particle solution.

My Confusion

I don't understand where the operators went. Usually when I see the Hamiltonian defined in the time-dependent SE it reads

\begin{equation} i\hbar\frac{\partial}{\partial t}\psi = \hat{H}\psi = \frac{1}{2}\frac{\hat{p}^2}{m}\psi \end{equation}

But my answer seemingly is hatless. Above I defined $p = \hbar k$ which is the de Broglie relationship. But the article I got the original wave function from didn't say I needed to make the $p$ in $ \psi(x,t) = Ae^{i(px-Et)/\hbar}$ an operator. So I'm confused what should and should not be an operator.

My Question:

Can someone clarify what should and should not be an operator in my verification of the 1D solution to the SE for a free particle?


Your solution is right. What you get verifying it is that $\psi$ is also an eigenfunction of the momentum operator, which means

$$\hat p\psi=p\psi,$$

where $\hat p=-i\hbar\nabla$ is momentum operator, and $p$ is its eigenvalue.

Now, applying $\hat p$ twice and dividing by $2m$, you can get

$$\frac1{2m}\hat p^2\psi=\frac1{2m}p^2\psi,$$

which is just another form of time-independent Schrödinger's equation for free particle:

$$\frac1{2m}\hat p^2\psi=E\psi.$$

Here $\hat T=\frac1{2m}\hat p^2$ is the kinetic energy operator, and $E$ is its eigenvalue, i.e. energy of the particle in this eigenstate $\psi$.

  • $\begingroup$ Are all solutions to the Schrödinger equation eigenfunctions? I assume not. How do I tell if a given solution to the SE is an eigenfunction? $\endgroup$ Mar 20 '15 at 6:54
  • $\begingroup$ All solutions of time-independent Schrödinger's equation with appropriate boundary conditions are eigenfunctions of Hamiltonian — that's by construction. As for time-dependent SE, the solution can be chosen to be an eigenfunction of an operator of a conserved quantity in a given problem. In your case, this conserved quantity is momentum. If you happened to choose the solution as $\sin$ instead of $\exp$, this choice would correspond to conservation of parity (i.e. such a function has definite parity), while your original choice has definite momentum $p$. $\endgroup$
    – Ruslan
    Mar 20 '15 at 6:57

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