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Variational Principle

  • Thread starter cepheid
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cepheid
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1. Homework Statement

Find the best bound state on Egs for the one-dimensional harmonic oscillator using a trial wave function of the form

[tex] \psi(x) = \frac{A}{x^2 + b^2} [/tex]

where A is determined by normalization and b is an adjustable parameter.


2. Homework Equations

The variational principle

[tex] \langle \psi |H| \psi \rangle \geq E_{gs} [/tex]


3. The Attempt at a Solution

This problem has some really nasty integrals. E.g., just to normalize the wavefunction, you need to calculate:

[tex] \int_{-\infty}^{\infty} \frac{dx}{(x^2 + b^2)^2} [/tex]

I was able to find most of them using residue theory. It was a last resort...the only method I could think of! I had to dig through my old complex analysis notes to remind myself of the technique using a semicircular contour of radius r and then letting r go to infinity. The integral of f(z) over the arc is zero, leaving the integral on the real axis, which is the part I want to evaluate. Anyway, I'm actually pretty sure of my answers, which are as follows

The best value of b is:

[tex] b^2 = \frac{\hbar}{m\omega \sqrt{2}} [/tex]

[tex] \langle \psi |H| \psi \rangle = \frac{\hbar \omega}{\sqrt{2}} \geq E_{gs} [/tex]

My question is, in a subsequent problem, Griffiths asks us to generalize this to a trial wavefunction of the form

[tex] \psi(x) = \frac{A}{(x^2 + b^2)^n} [/tex]

Huh? Did I miss something?! I had enough trouble with the integrals the first time around!! I don't think contour integration is going to help me. Am I missing some more obvious method for evaluating these integrals?
 
Last edited:

Answers and Replies

pam
455
1
That first integral is more easily done with the substitution x=tan y, and some trig.
 
cepheid
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hmm...you're right that does work rather nicely. Not too sure about the general case though.
 
kdv
336
1
1. Homework Statement

Find the best bound state on Egs for the one-dimensional harmonic oscillator using a trial wave function of the form

[tex] \psi(x) = \frac{A}{x^2 + b^2} [/tex]

where A is determined by normalization and b is an adjustable parameter.


2. Homework Equations

The variational principle

[tex] \langle \psi |H| \psi \rangle \geq E_{gs} [/tex]


3. The Attempt at a Solution

This problem has some really nasty integrals. E.g., just to normalize the wavefunction, you need to calculate:

[tex] \int_{-\infty}^{\infty} \frac{dx}{(x^2 + b^2)^2} [/tex]

I was able to find most of them using residue theory. It was a last resort...the only method I could think of! I had to dig through my old complex analysis notes to remind myself of the technique using a semicircular contour of radius r and then letting r go to infinity. The integral of f(z) over the arc is zero, leaving the integral on the real axis, which is the part I want to evaluate. Anyway, I'm actually pretty sure of my answers, which are as follows

The best value of b is:

[tex] b^2 = \frac{\hbar}{m\omega \sqrt{2}} [/tex]

[tex] \langle \psi |H| \psi \rangle = \frac{\hbar \omega}{\sqrt{2}} \geq E_{gs} [/tex]

My question is, in a subsequent problem, Griffiths asks us to generalize this to a trial wavefunction of the form

[tex] \psi(x) = \frac{A}{(x^2 + b^2)^n} [/tex]

Huh? Did I miss something?! I had enough trouble with the integrals the first time around!! I don't think contour integration is going to help me. Am I missing some more obvious method for evaluating these integrals?
note that

[tex] \frac{1}{(x^2+b^2)^3} = -\frac{1}{2} \frac{d}{db^2} ( \frac{1}{(x^2+b^2)^2}) [/tex]

You can generalize to an arbitrary n. Therefore, if you know how to integrate with 1/(b^2+x^2), you can integrate any power n. Your result will contain the nth derivative which you can probably find explicitly (it depends, what is the result of the integral with 1/(x^2+b^2) ?)
 
Not to grave dig here too bad, but I don't think the OP came to the right solution and I didn't want someone else to be confused.

The ground state energy of a harmonic oscillator is known to be

[tex] E_{gs}=\frac{1}{2} \hbar \omega >
\frac{1}{\sqrt{2}} \hbar \omega [/tex]

The variational principle can only find a lower bound on the energy, meaning that the approximation must be greater than or equal to the actual energy of the system.

Working out the problem myself, I got a little different value of b. I believe this is where the OP went wrong.
 
Last edited:

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