3
$\begingroup$

Hamiltonian, being a hermitian operator, is an observable for energy. I wonder if Schrodinger equation can be defined for any $K$, also a hermitian operator, but an observable for another quantity, i.e.

$$\dot{\Psi}=-\frac{i}{\hbar}K\Psi.$$

$\endgroup$
8
$\begingroup$

You can solve that equation if you want, but the solution won't necessarily mean anything. The Hamiltonian is directly tied to time evolution even in classical mechanics, by the equation $$\frac{df}{dt} = \{H, f\}$$ where the right-hand side is the Poisson bracket. Even in Lagrangian mechanics, the Hamiltonian pops out as the conserved quantity from time translation symmetry. And in special relativity, the Hamiltonian has to be linked to time if momentum is linked to space.

The point is, the link between Hamiltonians and time is really deep, so it's not clear what your alternate Schrodinger equation would mean. If you really had a wavefunction that satisfied your equation, then $K$ would just be equal to the Hamiltonian, and it wouldn't make sense to call it anything else.


However, if we didn't want to focus on time evolution, we could instead replace $t$ with a formal parameter $\alpha$ and define $$\frac{d}{d\alpha} \psi = - \frac{i}{\hbar} K \psi.$$ This equation makes sense for general $K$, and solving it can tell you a bit about the physical meaning of $K$. For example, if you chose $K$ to be the momentum, you would get $$\psi(x, \alpha) = \psi(x - \alpha).$$ That is, the transformation generated by the momentum is just spatial translation, as you know from classical mechanics. If you switch to Heisenberg picture and do the same thing, the result is directly analogous to how generating functions (here, $K$) correspond to one-parameter families of canonical transformations in classical mechanics.

$\endgroup$

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.