let be the Hamiltonian H=f(xp) if we consider canonical quantization so
f(−ixddx−i2)ϕ(x)=Enϕ(x)
here 'f' is a real valued function so i believe that f(xp) is an Hermitian operator, however
How could i solve for a generic 'f' the eqution above ? , in case 'f' is a POlynomial i have no problem to solve it the same in case f is an exponential but for a generic 'f' how could i use operator theory to get the eigenvalue spectrum ?¿
Answer
xp is not Hermitian, but (xp+px)/2 is. Thus you need to use the latter as an argument.
To solve the time-independent Schroedinger equatione, first solve it for H=(xp+px)/2, and note that the eigenvectors don't change when taking a function of this operator.
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