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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 ΨT(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.