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Eigen Value and Eigen Function

  1. Jun 18, 2010 #1
    I know that the eigen equation is given by [tex]A\psi =\lambda\psi[/tex] where A is an operator and [tex]\psi[/tex] is an eigen function and [tex]\lambda[/tex] is an eigen value.
    For the particle in an infinite square well of width L the wavefunction is given by [tex]\psi (x)=\sqrt{\frac{2}{L}}sin(\frac{n\pi x}{L})[/tex] and the energy is given by [tex]E=\frac{n^2h^2}{8ml^2}[/tex]
    I want to know why [tex]\psi[/tex] is called eigen function and E is called the eigen value.
     
  2. jcsd
  3. Jun 18, 2010 #2

    alxm

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    From German eigen, meaning "distinct", "particular", "[one's] own".

    It comes by analogy to linear algebra. If a vector [tex]\vec{v}[/tex] is such that when you multiply a matrix A, by it, you get back the same matrix multiplied by some scalar (lambda):
    [tex]\mathbf{A}\vec{v} = \lambda\mathbf{A}[/tex]
    Then [tex]\vec{v}[/tex] is an eigenvector of A, and [tex]\lambda[/tex] the corresponding eigenvalue. Any linear algebra textbook will tell you all about it.

    This concept can be generalized to linear operators (matrix multiplication being a linear operation). For a linear operator [tex]\hat{O}[/tex], and a function f, if:
    [tex]\hat{O}f = \lambda f[/tex]

    Then f is an eigenfunction of [tex]\hat{O}[/tex], and [tex]\lambda[/tex] is again the corresponding eigenvalue. If you have a given operator and some boundary conditions, and wish to find the eigenfunctions and eigenvalues for it, then that's called an "eigenvalue problem".

    For instance, the differential operator [tex]\hat{D}^n = \frac{d^n}{d x^n}[/tex] is a linear operator.
    So differential equations, such as [tex]\frac{df}{d x} - \lambda f(x) = 0[/tex] are eigenvalue problems.
     
  4. Jun 18, 2010 #3

    tiny-tim

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    hi roshan2004! :smile:

    (have a psi: ψ and a lambda: λ and a pi: π and a square-root: √ and a curly d: ∂ :wink:)

    your wavefunction should also have an eiωt factor

    the (energy) operator A is ih∂/∂t (where i'm writing h for h-bar = h/2π),

    so the (energy) eigenvalue is hω

    (which just happens to be h(nπ/L)2/2m = h2n2/8mL2

    see http://en.wikipedia.org/wiki/Square_well" [Broken] :wink:)​
     
    Last edited by a moderator: May 4, 2017
  5. Jun 18, 2010 #4
    Thanks, but in this case I couldnot find operator here, which is the operator in this case?
     
  6. Jun 18, 2010 #5

    alxm

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    The Hamiltonian. The time-independent Schrödinger equation is simply the eigenvalue problem for the Hamiltonian.
    Also, since V = 0 and you only have one dimension in the infinite square well, the Hamiltonian in that case is just the second-order differential operator (multiplied by [tex]-\hbar/2m[/tex])
     
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