Probably missing something very obvious, sorry if this is a stupid question.

I have to show the function $\psi\left(x, y\right) = \frac{1}{\sqrt{2\pi\mathrm{i}t}} \int_{-\infty}^{\infty}\mathrm{e}^{\mathrm{i}\left(x - y\right)^{\, 2}/4t}\, \psi_{0}\left(y\right)\,\mathrm{d}y$ is a solution of the Schrodinger equation for a free particle, i.e. is a solution of $i\hbar \frac{\partial \psi(x,t)}{\partial t} = \frac{-\hbar^2}{2m} \frac{\partial^2 \psi(x,t)}{\partial x^2}$.

Working through I get that $i\hbar \frac{\partial \psi(x,t)}{\partial t} = -\hbar \frac{\partial^2 \psi(x,t)}{\partial x^2}$. That is, I can't get a $1/2m$ term no matter what I do, and my lecture notes don't contain any relations of $\hbar$ and $m$ so I'm not sure how to finish the proof.

I read in another question on free particles we can "let $\hbar^2/2m = 1$ without loss of generality but I can't find a source on this claim.

Any help or hints appreciated, thanks.

  • 2
    $\begingroup$ PS To write $\hbar$ write \hbar in between $$. $\endgroup$
    – jim
    Commented May 29, 2016 at 16:33
  • $\begingroup$ Thanks, wasn't sure of the notation. $\endgroup$
    – D. P
    Commented May 29, 2016 at 16:37

1 Answer 1


That function certainly can't solve that equation, since the parameters $\hbar$ and $m$ don't occur in it.

It's quite usual to set various constants or combinations thereof to $1$ in physics to simplify expressions. You can either do this simply by choosing units such that the constant expression takes the value $1$, or you can explicitly rescale time and space. See natural units.

Note that in the present case the choice $\frac{\hbar^2}{2m}=1$ that you mention would be unhelpful, as it would leave a factor of $\hbar$ on the left-hand side – what you need is $\frac\hbar{2m}=1$.

  • $\begingroup$ I thought as much, I forgot to mention that $\psi_0 : \mathbb{R} \to \mathbb{C}$ is a smooth function with compact support, however I don't see how this would change the solution either. Would the assumption $\overline{h}/2m = 1$ be reasonable in your opinion (I'm a mathematician, not a physicist so I'm not used to this kind of work)? If not I can't see how the professor expects us to solve this. $\endgroup$
    – D. P
    Commented May 29, 2016 at 16:29
  • 1
    $\begingroup$ @D.P Yes, quite reasonable; I think they simply set all (real-valued) constants to $1$ to simplify things. I wouldn't call it an "assumption", more like a choice of units or a simplification. It doesn't change the character of the solutions. $\endgroup$
    – joriki
    Commented May 29, 2016 at 16:40
  • $\begingroup$ I see, thanks for the clarification. $\endgroup$
    – D. P
    Commented May 29, 2016 at 16:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .