# Use the variation method with trial Wavefunction (Szabo and Oslund ex 1.18)

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1. Aug 10, 2011

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:

(-1/2.d2/dx2-delta(x)).psi=E.psi

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:

N2pi1/2/(2a)1/2=1
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

=-N2

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:

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

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

$$\left\langle \psi' \left\lvert -\frac{1}{2}\frac{d^2}{dx^2} \right\rvert \psi' \right\rangle = \frac{1}{2}N^2\sqrt{\frac{a\pi}{2}}$$
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 $-1/\pi$.