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Stationary schrodinger equation (basic problems regarding the basis being used)

  1. Apr 30, 2009 #1
    i can't seem to understand something very basic about the stationary equation (a "simple" eigenvalue problem):


    H - hamiltonian operator
    Y - an eigenstate or an eigenfunction of the hamiltonian
    E - the eigenvalue of the eigenstate

    as far as i understand, the hamilotian operator (H) is represented in the energy basis, and the eigenstates (Y) therefore form a basis that span the hilbert space of the system.

    now this is where i get fairly confused. if the equation above is multiplied from the left by <x|, a bra representing one of the position states that form a position basis, then the equation is written as
    BUT, it is also said that the hamiltonian is represented in the energy basis, so why is "x" being the variable in use here?
    it seems as if the operator is in one basis, but the state is in a different one, but i know that i am wrong i just can't figure this out

    i hope someone can help me here...
  2. jcsd
  3. Apr 30, 2009 #2


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    If you multiply on the left by <x| you get:

    [tex] <x|H|Y> = E<x|Y> = E Y(x) [/tex]

    Note H can't be pulled out of the bracket since it's an operator. Now, H is an operator on the Hilbert space, not directly on the space of wavefunctions, but of course these spaces are really the same, so there should be a way of acting H on the wavefunction, and there is. Inserting a 1 in the form [itex]\int dx' |x'><x'|[/itex] we find:

    [tex] <x|H|Y> = \int dx' <x|H|x'><x'|Y> = \int dx' H(x,x') Y(x') [/tex]

    where I've defined [itex]H(x,x') = <x|H|x'>[/itex]. If you work this out, you'll find something like:

    [tex] H(x,x') = -\frac{\hbar^2}{2m} \delta''(x-x') + V(x) \delta(x-x') [/tex]

    And then inserting this into the integral you find:

    [tex] <x|H|Y> = -\frac{\hbar^2}{2m} Y''(x) + V(x) Y(x)[/tex]

    which gives you the Schrodinger equation in the position basis.
  4. May 1, 2009 #3
    oh very nice, this makes it a bit clearer, though i still have one more question.
    i have covered the missing steps (finding the H acting on the position basis) though one step isn't clear to me. from what i know, the inner product of the two position basis vectors is:
    <x|x'> = DiracDelta(x-x') (replacing the kronecker delta, moving from a discrete to a continuous space)

    but in order to reach the final equation of H(x,x') which you have written down, this needs to be true:
    <x|H|x'> = H<x|x'>
    why is this true? i have tried entering a complete set in between, but i couldn't get the result above.

    just to see if i actually get it now (hopefully), the state of the system is represented in the position basis, and the operator originally acts on the same hilbert space only under a different basis, and the idea is to find it's representation in the position base instead (did i get it right?)
    Last edited: May 1, 2009
  5. May 1, 2009 #4


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    Again, you need to be careful about the distinction between states and wavefunctions. An equation like:

    [tex] <x|H|x'> = H <x|x'> [/tex]

    makes no sense: the LHS is a number (depending on x and x') while the RHS is an operator, ie, a matrix on the infinite dimensional Hilbert space (also depending on x and x').

    Now, what you said at the end of your last post is true. Specifically, we can represent the Hilbert space in a number of ways, the most common of which is the position basis, where there is a one-to-one correspondence between states and (square integrable) complex valued functions on the real numbers. In this basis, we can also represent Hilbert space operators (like H), typically by differential operators. For example, we know from experience that the Hamiltonian has the representation:

    [tex] H |\psi> \leftrightarrow \left(-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x) \right) \psi(x) [/tex]

    (note the thing in paranetheses on the RHS is not H, but a differential operator that represents it in the position basis). What this equation says is that if the state [itex]|\psi>[/itex] is represented in the position basis by the function [itex]\psi(x)[/itex], then the state on the LHS of the above equation is represented by the function on the RHS. To put the last sentence in equations:

    [tex] <x|\psi> = \psi(x) \Rightarrow <x|H |\psi> = \left(-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x) \right) \psi(x) [/tex]

    In the case we're concerned with, since [itex]<x|x'> = \delta(x-x')[/itex] (in other words, the wavefunction of the state [itex]|x'>[/itex] is [itex]\delta(x-x')[/itex]), we have:

    [tex] <x|H|x'> = \left(-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x) \right) \delta(x-x') [/tex]

    which is the formula I gave above.

    You could also derive this from the more basic definition of the Hamiltonian (I'll put hats on things that are Hilbert space operators):

    [tex] \hat{H} = \frac{\hat{p}^2}{2m} + V(\hat{x}) [/tex]

    With a little work, using the commutation relations [itex] [\hat{x}, \hat{p}] = i \hbar [/itex] and the fact that [itex]|x'>[/itex] is an eigenstate of the operator [itex]\hat{x}[/itex] with eigenvalue x', you can derive the same formula without using the position basis at all.
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