Quantum Mechanics (position uncertainty)

[roam]

Homework Statement

I need some help with the following problem:

http://img827.imageshack.us/img827/4061/prob1y.jpg [Broken]

Homework Equations

For some particle in state Ψ the expectation value of x is given by:

$\left\langle x \right\rangle = \int^{+\infty}_{-\infty} x |\Psi(x, t)|^2 \ dx$

The wave function in ground state:

$\psi_0 (x)= A e^{\frac{-m \omega}{\hbar}x^2}$

The Attempt at a Solution

Let α = mω/2ħ, so

$\left\langle x \right\rangle = \int^{+\infty}_{-\infty} x A^2 (e^{-\alpha x^2})^2 \ dx$

$= A^2 \int^{+\infty}_{-\infty} x (e^{-2 \alpha x^2}) \ dx$

$=A^2 \left[ \frac{-e^{-2 \alpha x^2}}{4 \alpha} \right]^{+\infty}_{-\infty} \ dx$

But since e^∞=∞ and e^-∞=0 (i.e. the limits as x approaches ±∞) we get:

<x>=A² ∞ = ∞

So, what have I done wrong here?

Also for <x²> we have the following

$\left\langle x^2 \right\rangle = \int^{+\infty}_{-\infty} x^2 A^2 (e^{-\alpha x^2})^2 \ dx$

But similarly here I will encounter the same problem. So how can I evaluate the <x> and <x²> without getting infinity?

Clearly they can't equal to infinity since we must obtain:

$\Delta x = \sqrt{\left\langle x^2 \right\rangle -\left\langle x \right\rangle^2} = \sqrt{\frac{\hbar}{2m \omega}}$

Any help is greatly appreciated.

 

[klawlor419]

Check the evaluation of your upper and lower limits on your integrals again. ;)

 

[roam]

Check the evaluation of your upper and lower limits on your integrals again. ;)

The upper limit is +∞ and the lowe limit is -∞. Here's what I did:

$A^2 \left( \frac{e^{-2 \alpha \infty^2}}{4 \alpha} - \frac{e^{2 \alpha \infty^2}}{4 \alpha} \right)$

$= A^2 \left( \frac{e^{-\infty^2}}{4 \alpha} - \frac{e^{\infty^2}}{4 \alpha} \right)$

$= A^2 (0- 0) = 0$

Edit: this time I get 0. So <x>=0?

 

[klawlor419]

Yes! Not infinity as you were getting before. It makes sense to, a simple ground state harmonic oscillator spends most of the time close to origin.

 

[roam]

Okay, but what about <x²>? The integral $\int^{+\infty}_{-\infty} x^2 A^2 (e^{-\alpha x^2})^2 \ dx$ turns out to be extremely messy. I'm just wondering if I am on the correct path?

If I integrate that I get this:

$\left[ \frac{\sqrt{\pi/2} \ erf(\sqrt{2} \sqrt{\alpha} x)}{8 \alpha^{3/2}} - \frac{x e^{-2\alpha x^2}}{4 \alpha} \right]^{+\infty}_{-\infty}$

But I must get <x²>=ħ/2mω so that the position uncertainty would be the square root of that.

 

[klawlor419]

You are the right path, if you evaluate the integrals you should get the right value.