# Variational Principle

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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

$$\psi(x) = \frac{A}{x^2 + b^2}$$

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

2. Homework Equations

The variational principle

$$\langle \psi |H| \psi \rangle \geq E_{gs}$$

3. The Attempt at a Solution

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

$$\int_{-\infty}^{\infty} \frac{dx}{(x^2 + b^2)^2}$$

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:

$$b^2 = \frac{\hbar}{m\omega \sqrt{2}}$$

$$\langle \psi |H| \psi \rangle = \frac{\hbar \omega}{\sqrt{2}} \geq E_{gs}$$

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

$$\psi(x) = \frac{A}{(x^2 + b^2)^n}$$

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:

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pam
455
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That first integral is more easily done with the substitution x=tan y, and some trig.

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Gold Member
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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

$$\psi(x) = \frac{A}{x^2 + b^2}$$

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

2. Homework Equations

The variational principle

$$\langle \psi |H| \psi \rangle \geq E_{gs}$$

3. The Attempt at a Solution

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

$$\int_{-\infty}^{\infty} \frac{dx}{(x^2 + b^2)^2}$$

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:

$$b^2 = \frac{\hbar}{m\omega \sqrt{2}}$$

$$\langle \psi |H| \psi \rangle = \frac{\hbar \omega}{\sqrt{2}} \geq E_{gs}$$

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

$$\psi(x) = \frac{A}{(x^2 + b^2)^n}$$

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

$$\frac{1}{(x^2+b^2)^3} = -\frac{1}{2} \frac{d}{db^2} ( \frac{1}{(x^2+b^2)^2})$$

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

$$E_{gs}=\frac{1}{2} \hbar \omega > \frac{1}{\sqrt{2}} \hbar \omega$$

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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