6
$\begingroup$

The velocity operator is defined as $\mathbf{v}=i[H,\mathbf{r}]$ for the Hamiltonian $H$ satisfying $H\psi=\epsilon \psi$. This can be obtained from the Ehrenfest theorem. I'm wondering if $\mathbf{v}=i[H,\mathbf{r}]$ still holds for $H$ satisfying the generalized Schrodinger equation $H\psi=\epsilon S\psi$, where $S$ is the overlap operator?

$\endgroup$
  • $\begingroup$ +1. But where do you get that this only works if $H\psi = \epsilon \psi$ ? $\endgroup$ – Nike Dattani Sep 11 at 4:15
6
$\begingroup$

Consider a time-independent operator $\mathbf{r}$, its expectation value, and the time derivative of the expectation value (by applying the product rule for derivatives):

\begin{align} \tag{1} \langle \mathbf{r}\rangle &\equiv \int \psi^* \mathbf{r}\psi \,\,\, dr\\ \frac{\rm{d}}{\textrm{d}t}\langle \mathbf{r}\rangle &= \int \left(\frac{\rm{d}}{\textrm{d}t}\psi^*\right) \mathbf{r}\psi \,dr + \int \psi^* \mathbf{r}\left(\frac{\rm{d}}{\textrm{d}t}\psi\right)\,dr\tag{2}. \end{align}

Now let's plug in the Schrödinger equation (with $\hbar=1$ since that's what you had) and its conjugate:

\begin{align}\tag{3} \frac{\rm{d}}{\textrm{d}t}\psi &= -\textrm{i}H\psi \\ \frac{\rm{d}}{\textrm{d}t}\psi^* &= \textrm{i}H\psi^*. \tag{4}\\ \end{align}

The result is:

\begin{align} \frac{\rm{d}}{\textrm{d}t}\langle \mathbf{r}\rangle &= \int \left(\textrm{i}H\psi^* \right) \mathbf{r}\psi \,dr + \int \psi^* \mathbf{r}\left(-\textrm{i}H\psi \right)\,dr\tag{5} \\ \langle \mathbf{v}\rangle &\equiv\textrm{i} \left[H ,\langle \mathbf{r}\rangle \right].\tag{6} \end{align}

I never explicitly mentioned anyhting about eigenvalues and never used $H\psi = \epsilon\psi$ or $H\psi = \epsilon S \psi$: I just used:

  • the definition of an expectation value $\langle \cdot \rangle$,
  • the product rule for derivatives,
  • the Schrödinger equation (see the derivation of the Schrödinger equation in Section 2.2 of this PDF to see why this is the equation of motion for $\psi$ if we want its norm to be preserved as Born's rule relates wavefunction norms to probabilities: i.e. unitary evolution), so we have used:
    • Born's rule,
    • the law that probabilities must sum up to 1,
  • the definitions:
    • $\langle \mathbf{v}\rangle \equiv \frac{\rm{d}}{\textrm{d}t}\langle \mathbf{r}\rangle$,
    • $[A,B] \equiv AB-BA$,
    • $\textrm{i}^2=-1.$

On page 20 of the same aforementioned PDF you can see how the Heisenberg equation of motion is derived, if you want to do something similar without expectation values.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ By the definition of an expectation value, you are assuming orthogonal wave functions, right? How about nonorthogonal wave functions? $\langle \mathbf{r} \rangle = {\langle \psi|\mathbf{r}| \psi\rangle / \langle \psi|\mathbf{S}| \psi\rangle }$ $\endgroup$ – xmW Sep 11 at 4:29
  • $\begingroup$ For nonorthogonal wave functions, Eq. 3 turns out to be $id\psi / dt = S^{-1}H\psi$ ? $\endgroup$ – xmW Sep 11 at 4:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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