Perturbation Theory energy shift

[raintrek]

Homework Statement

I'm trying to calculate the energy shift given an electron in a 1D harmonic potential has a wavefunction

$$\Psi_{0}(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}exp\left(\frac{-m\omega x^{2}}{2\hbar}\right)$$

The shift in $$E_{0} = \frac{\hbar\omega}{2} = 2eV$$

due to the time independent perturbation
$$V(x) = V_{0}cos\frac{\pi x}{L}$$
where $$V_{0}=1eV, L = 5x10^{-10}m$$.

I'm told that $$\int^{\infty}_{-\infty}e^{-a^{2}x^{2}}cos(bx)dx = \frac{\sqrt{\pi}}{a}e^{-b^{2}/(4a^{2})}$$

Homework Equations

The Attempt at a Solution

OK, here's what I have:

$$\Delta E_{0} = V_{00} = \int^{\infty}_{-\infty}\Psi_{0}^{*}V\Psi_{0} dV$$

$$= V_{0}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2} \int^{\infty}_{-\infty}exp\left(\frac{-m\omega x^{2}}{\hbar}\right)cos\left(\frac{\pi x}{L}\right)dx$$

Which, using the integration provided leads to

$$V_{0}exp\left(\frac{-\pi^{2}\hbar}{4L^{2}m\omega}\right)$$

Now the exponential I have = 1 using the values provided, leading me to $$\Delta E_{0} = V_{0} = 1eV$$

HOWEVER, the answers I've been provided with show that:

$$\Delta E_{0} = \sqrt{2} exp\left(\frac{\pi^{2}\hbar}{2L^{2}m\omega}\right)$$

The exponential still goes to one, leaving delta E at 1.41 eV, but I can't see for the life of me how it's been arrived at... Help!

 

[Jhero]

Your approach is correct, but there are a few small mistakes in your calculations. First, the integration provided is missing a factor of 2 in the exponent, so it should be:

\int^{\infty}_{-\infty}e^{-a^{2}x^{2}}cos(bx)dx = \frac{\sqrt{\pi}}{a}e^{-b^{2}/(4a^{2})}

Next, when you substitute in the values for V_0 and L, you forgot to include the factor of the wavefunction in front of the integral. It should be:

\Delta E_0 = V_0 \left(\frac{m\omega}{\pi\hbar}\right)^{1/2} \int^{\infty}_{-\infty}exp\left(\frac{-m\omega x^{2}}{\hbar}\right)cos\left(\frac{\pi x}{L}\right)dx

= V_0 \left(\frac{m\omega}{\pi\hbar}\right)^{1/2} \left(\frac{\sqrt{\pi}}{2L}\right) exp\left(\frac{-\pi^{2}\hbar}{4L^{2}m\omega}\right)

This gives the correct answer of \Delta E_0 = \sqrt{2} V_0 = \sqrt{2} eV.

I hope this helps clarify any confusion. Keep up the good work!