# Homework Help: Expectation Value For a Given Wave Function

1. May 16, 2010

### CanIExplore

1. The problem statement, all variables and given/known data
Find the expectation value of x (Find <x>) given the wave function:
$$\psi(x)$$=[sqrt(m*alpha)/h_bar]e^[(-m*alpha*|x|)/(h_bar)^2]
This wave function represents the single bound state for the delta-function potential.
It's the solution to the shrodinger equation given the potential V(x)=-alpha*\delta(x)

3. The attempt at a solution

So this amounts to evaluating the integral of x|\psi(x)|^2 from negative infinity to positive infinity. Because of the absolute value of x, I split the integral into two parts. The first integral has limits from negative infinity to 0 and the second integral has limits from 0 to infinity. The first integral is of the form xe^cx and the second is xe^(-cx) where c is a constant. This is evaluated simply using integration by parts which gives me [-(x/c)e^(-cx)] - [(1/c)^2]e^(-cx) for the second integral. But when I plug in my limits, from 0 to infinity you'll notice I get negative infinity minus (1/c)^2. This doesn't make much sense to me. Since we're talking about a BOUND state, the expectation value of the position should be finite shouldn't it? Actually if we measure any physical quantity of a particle, it must always be finite. I think, or else it wouldn't make much physical sense. So I might be doing something wrong in the math or maybe I'm missing something else. Could someone please help me understand the results or even correct me if I've done something wrong because I can't seem to see it. Thanks

2. May 16, 2010

### NanakiXIII

Let me see if I understand correctly. Your problem is that

$$\left[ x e^{-c x} \right]_0^{\infty}$$

yields an infinite result? I think you should rethink the value of

$$\lim_{x->\infty} x e^{-c x}$$.

3. May 16, 2010

### jdwood983

Hmm, I get

$$\frac{m\alpha}{\hbar}\int_{-\infty}^\infty x\exp\left[-\frac{m\alpha}{\hbar} |x|\right]dx=\frac{m\alpha}{\hbar}\left(\int_{-\infty}^0x\exp\left[\frac{m\alpha}{\hbar}x\right]dx+\int_0^\infty x\exp\left[-\frac{m\alpha}{\hbar}x\right]dx\right)$$

$$=2\frac{m\alpha}{\hbar}\int_0^\infty x\exp\left[-\frac{m\alpha}{\hbar}x\right]dx=-\frac{2\hbar}{m^2\alpha^2}\left(m\alpha x\exp\left[-\frac{m\alpha}{\hbar}x\right]+\hbar\right)_{x=0}^{x=\infty}=2\frac{m\alpha}{\hbar}\cdot\frac{\hbar^2}{m^2\alpha^2}$$

$$\frac{m\alpha}{\hbar}\int_{-\infty}^\infty x\exp\left[-\frac{m\alpha}{\hbar} |x|\right]dx=\frac{2\hbar}{m\alpha}$$

4. May 17, 2010

### CanIExplore

You're absolutely right, I see my mistake.

I realized that this problem is far simpler since xe^(-|x|) is an odd function the integral just comes out to be zero. Thanks for the replies you guys.