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TISE in the position representation- basic question

  1. Jan 25, 2013 #1
    We were told in lectures that the time independent Schrodinger equation can be applied to wavefunctions, i.e. [itex]\frac{hbar^2}{2m}[/itex][itex]\frac{d^2U}{dx^2}[/itex]+V(x)U=EU where U is the wavefunction bra x ket psi. I don't understand why this is valid, as wavefunctions are probability amplitudes, and operators can't operate on mere numbers. Could someone explain how this result is derived? In other words, how do you apply the TISE in the position representation?

    Thanks in advance for your help! :smile:
  2. jcsd
  3. Jan 25, 2013 #2


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    Wave functions are probability amplitudes, as such they are functions of space (and time). Operators in the position representation operate on functions of space (and time), so I don't see what's the problem.

    You apply it like you would apply a (second) derivative and multiplication to any function from regular calculus.
  4. Jan 25, 2013 #3
    Thanks for your help.
    To rephrase my question:
    H/En>=En/En> (where /x> is ket x)
    Multiplying by bra x, <x/H/En>=En<x/En>.
    I would like to know why <x/H/En>=H<x/En>.
  5. Jan 25, 2013 #4


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    Abstractly, in terms of bras and kets, H|ψ> = E|ψ>. If we want to write this in the x representation, we multiply on the left by <x|, and also insert the identity operator, in the form I = ∫|x>d3x<x|

    ∫<x|H|x'>d3x'<x'|ψ> = E<x|ψ>

    Then <x|ψ> is the wavefunction ψ(x), and <x|H|x'> is the Hamiltonian expressed as an operator acting on functions of x. For example, the V term is <x|V|x'> = V(x) δ3(x - x').
  6. Jan 25, 2013 #5


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    Well, you can convince yourself this way:
    [itex]<x| \dfrac{p^2}{2m} |\Psi>[/itex]
    [itex]= <x| \dfrac{p^2}{2m} |k><k|\Psi>[/itex]
    [itex]= <x| \dfrac{h^2 k^2}{2m} |k><k|\Psi>[/itex]
    [itex]= \dfrac{h^2 k^2}{2m} <x|k><k|\Psi>[/itex]
    [itex]= \dfrac{h^2 k^2}{2m} \dfrac{1}{2\pi} e^{i kx} <k|\Psi>[/itex]
    [itex]= - \dfrac{h^2}{2m} \dfrac{\partial^2}{\partial x^2} \dfrac{1}{2\pi} e^{i kx} <k|\Psi>[/itex]
    [itex]= - \dfrac{h^2}{2m} \dfrac{\partial^2}{\partial x^2} <x|k><k|\Psi>[/itex]
    [itex]= - \dfrac{h^2}{2m} \dfrac{\partial^2}{\partial x^2} <x|\Psi>[/itex]
  7. Jan 27, 2013 #6
    When operator act on a ket state, it produces an eigenvalue which is merely number and commutes with everything. Also, operators have their own positional form, such as momentum operator which we usually know as the partial derivative with respect to space. Written like that, I think it would be more reasonable to act on a function of space.

    In addition, wavefunction came up first, and bra-ket are introduced later to be more general rather than specific to positional. Therefore, physicists definitely have ensured that they are equivalent.
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