If the Hamiltonian is given by H(x,p)=[tex]p^2+p[/tex] then is it Hermitian?(adsbygoogle = window.adsbygoogle || []).push({});

I'm guessing it's not, because quantum-mechanically this leads to:

[tex]H=-h^2 \frac{d^2}{dx^2}-ih\frac{d}{dx}[/tex]

and this operator is not Hermitian (indeed, for the Sturm-Liouville operator [tex]O=p(x)\frac{d^2}{dx^2}+k(x)\frac{d}{dx}+q(x)[/tex] to be Hermitian, p'(x) must equal k(x)). Actually, I'm no longer sure anymore, because the hermicity of the Sturm-Liouville operator assumes real coefficients, so maybe H is indeed Hermitian?

However, I just want to be sure because it's been awhile since I looked at quantum mechanics. Can the eigenvalue problem in quantum mechanics look like this:

[tex]H(x)\psi(x)=g(x)E\psi(x) [/tex]

for an arbitrary positive function g(x) (a weight function), and a Hermitian Hamiltonian H(x)?

Or does it have to be of this form instead:

[tex]H(x)\psi(x)=E\psi(x) [/tex]

?

Edit:

Never mind. P^2+P is Hermitian - the imaginary number involved with [tex]-i\frac{d}{dx} [/tex] allows P to be Hermitian by itself. I should have just calculated whether it was Hermitian directly, instead of trying to determine if it's Hermitian by comparing it to the Sturm-Liouville operator, which is the most general Hermitian operator for real 2nd-order linear operators.

Also, although [tex]H(x)\psi(x)=g(x)E\psi(x) [/tex] would be interesting, I don't think it's useful in constructing a quantum mechanics, because if you operate with H twice, then you don't get [tex]H(x)H(x)\psi(x)=(g(x)E)^2\psi(x) [/tex], so you don't get a simple expression for time translation [tex]e^(-iHt) [/tex] in the energy states.

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# Hermitian operators in schrodinger eqn.

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