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Homework Help: Use the variation method with trial Wavefunction (Szabo and Oslund ex 1.18)

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Poll closed Sep 10, 2011.
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  1. Aug 10, 2011 #1
    1. The problem statement, all variables and given/known data

    The Schrodinger equation (in atomic units) of an electron moving in one dimension under the influence of the potential -delta(x) [dirac delta function] is:


    use the variation method with the trial function psi'=Ne-a.x2 to show that -pi-1 is an upper bound to the exact ground state energy (which is -0.5). You will need the integral:

    int(-inf..inf)dx. x2me-a.x2=(2m)!pi1/2/22mm!.am+1/2

    2. Relevant equations

    a) 1D Shrodinger eq: (-1/2.d2/dx2-delta(x)).phi=E.psi
    b) trial wave function: psi'=Ne-a.x2
    c) Useful integral: int(-inf..inf)dx. x2me-a.x2=(2m)!pi1/2/22mm!.am+1/2

    d) variation method: E0<= <psi'|H|psi'>/|<psi'|psi'>|2

    3. The attempt at a solution

    You basically need to plug in the wave function psi' (equation b) into the variation method (equation d) using the hamiltonian in equation a). I do this but do not arrive at -1/pi. Instead I get the answer: a-(2a)1/2/pi1/2

    Can anyone find the error in my logic/arithmetic below?

    First determine the normalization constant N in psi' by computing: <psi'|psi'> = 1 and solving for N.

    This gives: N2int(-inf..inf)dxe-2ax2 using equation c) for this integral with m = 0 I obtain:

    therefore N2=(2a)1/2/pi1/2

    The hamiltonian has two terms the kinetic energy (Ke) and the potential energy (V) term.

    <psi'|H|psi'> = <psi'|Ke|psi'> + <psi'|V|psi'>

    where Ke = -1/2d2/dx2
    and V = -delta(x)

    First compute the Ke term in the integral <psi'|H|psi'>

    Ke integral = <psi'|-1/2d2/dx2|psi'>

    The first derivative of psi' is:
    d/dx psi' = N(-2ax.e-ax2)

    Therefore the second derivative is:
    d2/dx2 psi' = N( (2ax)2 -2a) e-ax2

    using this the Ke integral becomes:

    =-1/2 N2int(-inf..inf).dx ( (2ax)2 -2a) e-2ax2

    the first term in the brackets corresponds to m = 1 in the useful integral (equation c) and the second term in the brackets corresponds to m = 0 in the useful integral (equation c)

    So the Ke integral becomes (after cancelling terms and rearranging):

    <psi'|Ke|psi'> = (1/2)N2((2a)1/2pi1/2)

    The potential energy integral is the following:

    <psi'|V|psi'> = <psi'|-delta(x)|psi'>

    = -N2 ea.0


    so the whole term < psi'|H|psi'> is:

    (1/2)N2((2a)1/2pi1/2) -N2

    now all that remains is to plug in N2 which we computed at the beginning:

    this reduces the above to:


    this is not -1/pi given in the question..... so can anyone tell me what algebra/logic error have I committed?

    Thanks in advance.
  2. jcsd
  3. Aug 10, 2011 #2


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    You need to check your calculation for the kinetic energy. I found
    [tex]\left\langle \psi' \left\lvert -\frac{1}{2}\frac{d^2}{dx^2} \right\rvert \psi' \right\rangle = \frac{1}{2}N^2\sqrt{\frac{a\pi}{2}}[/tex]
    You're going to end up with an upper bound for the energy that depends on a. What you want to do is then find the value of a that minimizes the upper bound. At that value of a, the upper bound is [itex]-1/\pi[/itex].
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