0
$\begingroup$

I have seen a few proofs specify for finite wells, step functions, and harmonic oscillators, that $V$ must be real for $H$ to be Hermitian. Why is that?

If we're solving the Schrodinger equation, we have $$i\hbar \frac{\partial}{\partial t}\psi=-\frac{\hbar^2}{2m}\nabla^2\psi+V\psi$$ and we know that $\psi$ being complex will have a complex spatial derivative. So for a real energy value to be observed, could the complex part cancel out with the complex part of $V$? That would still be hermitian, right?

Why does a electric Potential have to be real, but not a Potential in quantum mechanics? says that potentials in QM can be complex, but I want to know why we'd talk about complex potentials when they're not physically achievable and they cause operators to be non-hermitian and hence useless.

$\endgroup$
0
$\begingroup$

If $V$ is just a function of position then the requirement of being hermitian is equivalent to the requirement of reality. This is not true for different systems: suppose an electron is subject to an uniform magnetic field. The interaction with it's spin with the magnetic field will break the degeneracy associated with the spin states: spin up states will have a different energy than spin down states. This is effectively a 2 state system now, where we can represent the ground state as $$ |1\rangle = \begin{pmatrix} 1\\ 0 \end{pmatrix} $$ and the excited state as $$ |2\rangle = \begin{pmatrix} 0\\ 1\end{pmatrix}. $$ Suppose now we introduce another potential $V$ to this system. The most general form of $V$ will be represented by a 2x2 matrix $$ V = \begin{pmatrix} V_{11} & V_{12} \\ V_{21} & V_{22} \end{pmatrix}. $$ As you can easily deduce, the hermitian requirement ($V^\dagger = (V^*)^T=V$) allows for off-diagonal complex elements. And complex potentials may be achievable, one example I can think of is the Rabi oscillations potential that models a time-dependent 2 state system.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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