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Here's the problem:

1. The problem statement, all variables and given/known data

Quantum particle moving in 1D. Potential energy function is [tex]V(x) = C|x|^{3}[/tex]. Using the variational method, find an approx. ground-state wave function for the particle.

3. The attempt at a solution

Using [tex]\psi = Ae^{-ax^{2}}[/tex], I find that

[tex]A = \left(\frac{2a}{\pi}\right)^{1/4}[/tex]

Then,

[tex]<H> = <T> + <V> = -\frac{\hbar^2}{2m}\frac{d^{2}}{dx^{2}}+C|x^{3}|[/tex]

[tex]<T> = -\frac{\hbar^2}{2m}A^{2}\left( -2a\int_{-\infty}^{+\infty}e^{-2ax^{2}} dx + 4a^{2}\int_{-\infty}^{+\infty}x^{2}e^{-2ax^{2}} dx \right)[/tex]

which I work out to be

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

For the potential energy component, I state that

[tex]<V> = CA^{2}\int_{-\infty}^{+\infty} e^{-ax^{2}} |x|^{3} e^{-ax^{2}} dx = 2CA^{2} \int_{0}^{+\infty} x^{3}e^{-2ax^{2}} dx[/tex]

which I find to be

[tex]<V> = \frac{CA^{2}}{4a^{2}} = ... = \frac{C}{2a\sqrt{2a\pi}}[/tex]

And thus

[tex]<H> = \frac{\hbar^{2}a}{2m} + \frac{C}{2a\sqrt{2a\pi}}[/tex]

Minimising (diff. H wrt. a and set to zero) I obtain

[tex]\left(\frac{3Cm}{2\hbar^{2}\sqrt{2\pi}}\right)^{2/5}[/tex],

but this (substituted back into the expression for <H> above) produces a rather ugly looking expression, which makes me suspicious that I have made a mistake somewhere down the line.

Can anyone see where (or if) I am going wrong?

Cheers!

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# Homework Help: Variation method (Quantum Mech.)

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