1
$\begingroup$

Im reading the following article https://arxiv.org/abs/1105.4014. There they derive the Schrödinger equation from Ehrenfest‘s Theorem. By starting with the following equations: \begin{equation} \left\langle\frac{\mathrm{d}}{\mathrm{d} t}\psi\right|\hat{p}|\psi\rangle+\langle\psi|\hat{p}\left|\frac{\mathrm{d}}{\mathrm{d} t}\psi\right\rangle=\left\langle-V'(\hat{x})\right\rangle \end{equation} \begin{equation} \left\langle\frac{\mathrm{d}}{\mathrm{d} t}\psi\right|\hat{x}|\psi\rangle+\langle\psi|\hat{x}\left|\frac{\mathrm{d}}{\mathrm{d} t}\psi\right\rangle=\frac{1}{m}\langle \hat{p}\rangle. \end{equation} Then they use the canonical commutator relation and remove the mean. My question is now why can the averaging dropped out? They justify this by saying that it must apply to all initial conditions, but for me it‘s not clear why than the averaging can be dropped out.

$\endgroup$

1 Answer 1

1
$\begingroup$

This step is a mathematical one; in general, if $$ \int f = \int g, $$ you can't conclude that $f=g$. But if $$ \int f \psi = \int g \psi $$ for a large class of functions $\psi,$ then you sometimes can conclude that $f=g$. For example, to argue that $f(x)=g(x)$ for some $x$, you can let $\psi$ be a sharply peaked function at $x$ (so nonzero near $x$, zero elsewhere, think of an approximation to the dirac delta).

Your situation is a little more difficult, here instead we want to conclude from $$ \int \psi^* \hat{A} \, \psi = \int \psi^* \hat{B} \, \psi $$ for some operators $\hat{A}$ and $\hat{B}$ and a large class of wavefunctions $\psi$, that $\hat{A} = \hat{B}$. Here the argument to convince yourself of this is more complicated, and depends on the operators $\hat{A}$ and $\hat{B}$, but I think this is roughly what is intended.

$\endgroup$
5
  • $\begingroup$ Thanks! Do you know a theorem when this is true? $\endgroup$
    – Silas
    Commented Apr 27, 2022 at 20:23
  • $\begingroup$ @Silas yes, it holds for example if $f$ is a continuous (real or complex valued) function on $\mathbb{R}^n$ and $\psi$ is allowed to be any so called test function on $\mathbb{R}^n$. See chapter 9.1 of Folland's book on Fourier analysis for this. $\endgroup$
    – Steven
    Commented Apr 28, 2022 at 6:04
  • $\begingroup$ @Silas this was for the simple case, I actually remember now that the general case should follow relatively easily using a result such as this: math.stackexchange.com/q/1307747/606584 $\endgroup$
    – Steven
    Commented Apr 28, 2022 at 6:52
  • $\begingroup$ Thank you very much that was exactly what I needed! The only problem I still have is that I think that we don‘t get from all initial conditions all wavefunctions. Or can we build the wavefunctions by superpositioning? $\endgroup$
    – Silas
    Commented Apr 28, 2022 at 7:25
  • $\begingroup$ @Silas I don't want to read the full paper, but this is probably related to the fact that the wave function at $t=0$ determines the wave function for all $t>0$. An intuitive argument for this follows directly from the SE. $\endgroup$
    – Steven
    Commented Apr 28, 2022 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.