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Is Hamiltonian operator a Hermitian operator?

  1. Jul 16, 2010 #1
    Hi, there. It should be yes, but I'm very confused now.

    Consider a simple one-dimensional system with only one particle with mass of m. Let the potential field be 0, that's V(r) = 0. So the Hamiltonian operator of this system is:

    H = -hbar^2/(2m) * d^2/dx^2
    \hat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}

    To obtain the eigenvalue and eigenfunction of H, one should solve the following equation:

    H f = E * f
    \hat{H}f = Ef

    That leads:

    d^2f/dx^2 + 2mE/hbar^2 * f = 0
    \frac{d^2f}{dx^2} + \frac{2mE}{\hbar^2}f = 0

    However, there is always a solution f for any value of E, even imaginary number, such as E = i, E = 1+i... But a Hermitian operator can never have an imaginary eigenvalue.

    I don't know what's wrong. Can you help me? Thank you!
    Last edited: Jul 16, 2010
  2. jcsd
  3. Jul 16, 2010 #2
    Not all solutions [itex]f[/itex] are considered acceptable. We often speak about "physical solutions" and "non-physical solutions". I'm not sure if there's a clear definition for what a physical is, but you'll develop some intuition for it eventually.

    For example, when there is no potential term, all functions

    \psi(x) = e^{Ax}

    are solutions to the Schrödinger equation with energy [tex]E=-\frac{\hbar^2 A^2}{2m}[/tex], where [itex]A\in\mathbb{C}[/itex]. But solutions with [itex]\textrm{Re}(A)\neq 0[/itex] are not considered physical, because if [itex]\textrm{Re}(A)>0[/itex] then [itex]\psi(x)[/itex] diverges (it's amplitude diverges to infinity) as [itex]x\to\infty[/itex]. If instead [itex]\textrm{Re}(A)<0[/itex], then the limit [itex]x\to-\infty[/itex] causes similar divergence. This is why physical solutions always have [itex]\textrm{Re}(A)=0[/itex], and hence the solutions are

    \psi(x) = e^{ikx/\hbar},\quad\quad \big(A=\frac{ik}{\hbar}\big)

    where [itex]k\in\mathbb{R}[/itex]. The energy [tex]E=\frac{k^2}{2m}[/tex] is real then.

    Another example is the infinitely deep potential well. In that case we demand boundary conditions [itex]\psi(x)=0[/itex] for [itex]x=\pm R[/itex]. You will see that if you set wrong kind of eigenvalue to the SE, (for example a eigenvalue with imaginary component), then you cannot get the boundary values satisfied. The boundary values will force the eigenvalues to become real.
  4. Jul 16, 2010 #3


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    Science Advisor

    Welcome to PF, vact!

    Yes the Hamiltonian must be Hermitian. But as you succeeded in demonstrating, this is not something which you can derive from the Schrödinger equation itself.

    It's a separate condition which you have to enforce. In your example there's actually another condition you're not enforcing as well - which jostpuur indicates - the eigenfunction cannot diverge at infinity. And more-or-less equivalently, it has to be normalizable and square-integrable. (however if one keeps this in mind, plane-wave expansions are okay to use and are used all the time in fact)

    So for these reasons you have to reject the solutions with complex eigenvalues.
  5. Jul 16, 2010 #4
    Hi, jostpuur and alxm. Thank you for your answers.

    Yes, I've realized that the wavefunction was unacceptable in physics if Re(E) < 0 or Im(E) is not 0. But in my opinion, all concepts of eigenvalue, eigenfunction and Hermitian operator are defined in mathematics. So one can always solve the eigenvalue equation, which is [tex] \hat Hf = Ef [/tex], to get E and f in mathematics.

    However, I noticed that [tex] \hat H [/tex] with those functions which were unacceptable in physics, such as f = exp(Ax), where A > 0, was unable to be justified with the definition of Hermitian, because [tex]\int {g^* \hat Hfd\tau }[/tex] and [tex]\int {f (\hat Hg)^*d\tau }[/tex] have no finite value.

    Someone told me that these concepts were defined on some kinds of set of functions, maybe called "function space". Problems such as divergence can be avoided there. But I don't know the detail. I think this question is more mathematical. I'd better post it in a relevant forum. Of course I would very appreciate you for any detail about this topic.

    Thank you!
  6. Jul 17, 2010 #5

    just as a notice:

    Sometimes we are interested in the complex eigenvalues of the not hermitian Hamilton operator, because those correspond to the poles of the function you obtain by fourier transform the time evolution operator.

    By getting this complex eigenvalues, you directly get the lifetime of a quantum mechanical resonance states if you have the irreversible tunneling from a discrete to a continuum state ;)

    You see, we often use not hermitian operators but they are only a kind of tool to access some theories, just a way of describing.

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