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## Homework Statement

Consider a one-dimentional particle in a box with infinite potential walls at x=0 and x=L. Employ the variational method with the trial wave function Ψ

_{T}(x) = sin(ax+b) and variational parameters a,b>0 to estimate the ground state energy by minimising the expression

[tex]E_{T}= \frac{\left \langle \Psi _{T}(x) |H |\Psi _{T}(x) \right \rangle}{\left \langle \Psi _{T}(x) |\Psi _{T}(x) \right \rangle}[/tex]

## Homework Equations

## The Attempt at a Solution

I calculate the denominator first

[tex]\left \langle \Psi _{T}(x) |\Psi _{T}(x) \right \rangle = \int_{0}^{L} sin^{2}(ax+b) dx[/tex]

I use the trig function

[tex]sin^{2}A = \frac{1}{2} - cos (2A)[/tex]

then integrate to give

[tex] = \frac{L}{2} - \frac{1}{2a}sin(2aL+2b)+\frac{1}{2a}sin(2b)[/tex]

then the numerator

[tex]\left \langle \Psi _{T}(x) |H |\Psi _{T}(x) \right \rangle = \frac{(-i\hbar)^{2}}{2m}\int_{0}^{L}\Psi ^{*}\frac{d^{2}}{dx^{2}} sin(ax+b)dx[/tex]

[tex]= \frac{\hbar^{2}a^{2}}{2m}\int_{0}^{L}sin^{2}(ax+b)dx[/tex]

which is the same integral as before so gets the same result, therefore we have

[tex]E_{T}= \frac{\hbar^{2}a^{2}}{2m}\frac{\frac{L}{2} - \frac{1}{2a}sin(2aL+2b)+\frac{1}{2a}sin(2b)}{\frac{L}{2} - \frac{1}{2a}sin(2aL+2b)+\frac{1}{2a}sin(2b)}[/tex]

which gives

[tex]E_{T}= \frac{\hbar^{2}a^{2}}{2m}[/tex]

this is the bit i'm stuck on, I need to minimise this. if i differentiate with respect to a and set to zero I get a=0.

if there is a potential term it's fine i can do that, but with this question there isnt, any ideas how to minimise it?

as always thanks for any help.