Let A,B be finite-dimensional Hermitian operators and show that $[A,B]= q~ i~I$ cannot hold for $q$ a real constant and $I$ the identity matrix. *
One of several ideas that has not worked: I guess by definition
$AB-BA = q~i~I = q~i\sum_j |j\rangle\langle j| $
Neither AB nor BA is Hermitian, since AB Hermitian iff $[A,B]=0.$
So let $|\psi\rangle,\lambda_1$ be an eigenvector, eigenvalue associated with AB and $\lambda_2, \langle \phi|$ with $BA$. Then
$\langle \phi|AB|\psi\rangle-\langle \phi|BA|\psi\rangle = \lambda_1\langle \phi|\psi\rangle -\lambda_2 \langle \phi|\psi\rangle =q~i \sum \langle\phi|j \rangle\langle j|\psi\rangle$
This only seems to allow us to say that $\lambda_1-\lambda_2$ is imaginary. I seem to be pretty far off course, any suggestions appreciated.
*The problem is from Commins, Quantum Mechanics (Cambridge, 2014)
Best Answer
Hint: What is the trace of a commutator?