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Eigenfunction of Hamiltonian

  1. Oct 15, 2014 #1
    Could someone please explain Hψ = Eψ? I understand that H = Hamiltonian operator and ψ is a wavefunction, but how is H different from E? I am confused. I am trying to understand "Hψ = Eψ" approach
  2. jcsd
  3. Oct 15, 2014 #2


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    In this context H is the Hamiltonian while E is an energy. The difference between them is that H is a operator while E is a number. So, what that equation is saying is H, an operator, will take ##\psi## and return a number E times ##\psi##. In general, an operator, like H, will take a function like ##\psi## and give you another function, for example, ##\phi## so we expect ##H\psi=\phi##. Now, a priori, there's no reason to expect ##\phi## is related to ##\psi## in any particular way (other than being what you get after acting H on it), but the Eigen-function equation is forcing ##\phi## to be a multiple of ##\psi## i.e. ##\phi=E\psi##. This is perhaps easiest to understand in terms of vectors and linear transformations on vectors, see here: http://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors
    You are doing exactly an analogous thing here except instead of acting on vectors in 3-D with a linear transformation (i.e. a matrix), we are acting on functions (also a vector space, but not the regular vectors we are familiar with) with linear operators (e.g. a differential operator).
  4. Oct 15, 2014 #3


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    Hi terp,
    1) By Hψ = Eψ, the time-independent schrodinger equation, we can get the E given that we know H and ψ (which H acts on). H is a matrix; E is an eigenvalue of H; ψ is an eigenvector of H.
    2) How is H different from E? H and E is different things. H is a matrix while E is a scalar. It's like E is an apple while H is an apple tree.
    Also, we can have
    Hψ1 = E1ψ1
    Hψ2 = E2ψ2, where E1 is not equal to E2. Why is that? It is because the ψ H acts on is different.

    I hope my suggestion helps you.
  5. Oct 15, 2014 #4
    Wait, E is a constant?
  6. Oct 16, 2014 #5
    Hψ = Eψ

    is an eigenvalue problem
    you can read about it here


    H is the eigenvector and E is it's corresponding eigenvalue. eigenvalues are constants. for each eigenvalue you can find a corresponding eigenvector.

    again, think of it like a eigenvalue problem and that should be clear.
  7. Oct 16, 2014 #6
    you just need to read eigenvalue problem. that seems like where you are having trouble.

    in Hψ = Eψ

    ψ do NOT cancel out. if you like, you can write it in a different notation

    H|ψ> = E|ψ>

    where H should actually be written as "H hat" to make a distinction that it is an operator.

    so, Hamiltonian operator, H, is acting on your wave function, ψ, and the result is the same wave function, ψ, in the same space with some constant, E, multiplied to it.
  8. Oct 16, 2014 #7


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    I think you made a typo. ##\psi## is the eigenvector, not H.
  9. Oct 16, 2014 #8


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    Yes, E is a number, a constant. H is an operator. They're different kinds of things.
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