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

[gradstudentNZ]

Homework Statement

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. x^2me^-a.x2=(2m)!pi^1/2/2^2mm!.a^m+1/2

Homework 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. x^2me^-a.x2=(2m)!pi^1/2/2^2mm!.a^m+1/2

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

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/pi^1/2

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:

N²pi^1/2/(2a)^1/2=1
therefore N²=(2a)^1/2/pi^1/2


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)^1/2pi^1/2)


The potential energy integral is the following:

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

= -N² e^a.0

=-N²


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


(1/2)N²((2a)^1/2pi^1/2) -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/pi^1/2

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

Thanks in advance.

 

[vela]

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].