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Variation method (Quantum Mech.)

  • Thread starter T-7
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T-7
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Hi,

Here's the problem:

1. Homework Statement

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!
 
Last edited:

Answers and Replies

Dr Hooley intended you to do the assessed work on your own. ;)
 
T-7
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Dr Hooley intended you to do the assessed work on your own. ;)
Whilst we are intended to work through the exercises ourselves, as far as I know we're encouraged to discuss all the problems, and the lecturer has been willing to take questions himself on any of the set problems. I've not reproduced my working in detail here for anyone to copy, nor am I asking for detailed working in reply. It's against this forum's policy to do your homework for you.

However, if you're under the impression that 'starred' Qs can't be discussed at all, I'll ask my tutor for clarification. Consider the thread closed.
 
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