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Hermitian operators in schrodinger eqn.

  1. Sep 6, 2009 #1
    If the Hamiltonian is given by H(x,p)=[tex]p^2+p[/tex] then is it Hermitian?

    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]



    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.
    Last edited: Sep 7, 2009
  2. jcsd
  3. Sep 7, 2009 #2


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    You answered the first part yourself. In general, if A and B are hermitian, then so are An+Bm for any integers n and m. This is easy to verify.

    The "eigenvalue equation" that you're asking about isn't an eigenvalue equation. Just look at the definition of an eigenvector. You should also think about what you get when you look for solutions to the Schrödinger equation of the form [itex]\psi(x,t)=T(t)u(x)[/itex]. This breaks the Schrödinger equation into two pieces, one of which is the eigenvalue equation for the Hamiltonian.
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