# Velocity operator for a Hamiltonian that satisfies the generalized Schrodinger equation

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?

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

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.

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