# <p> Operator on Probability Density in X-Space

1. Oct 29, 2013

### Legaldose

1. The problem statement, all variables and given/known data
Consider a particle whose wave function is:

$$\Psi(x)=\left\{\begin{array}{ccc} 2\alpha^{3/2}xe^{-\alpha x} & \text{if} & x> 0\\ 0 & \text{if} & x\leq 0 \end{array}\right.$$

Calculate <p> using the $$\hat{p}$$ operator on probability density in x space.

2. Relevant equations

$$\hat{p}=-i \hbar \frac{\partial}{\partial x}$$

3. The attempt at a solution
So I'm pretty sure I'm missing something, I figured since you take the derivative then do the integral, they will "cancel" each other, then you can just evaluate -ih|Psi|^2 from 0 to infinity, in which case you get 0. Here is my work:

$$<p> =\int_{0}^{\infty}\hat{p}|\Psi(x)|^{2}dx=\int_{0}^{\infty}-i \hbar 4\alpha^{3} \frac{\partial}{\partial x} x^{2} e^{-2 \alpha x}dx=-i \hbar 4\alpha^{3}\int_{0}^{\infty}\frac{\partial}{\partial x} x^{2} e^{-2 \alpha x}dx$$

Then when you take the derivative, then integrate it over that range:

$$=-i \hbar 4\alpha^{3} x^{2} e^{-2 \alpha x}|_{0}^{\infty}=0$$

And when you evaluate this it goes to 0, did I do something wrong? TBH, I did miss this lecture, and unfortunately my professor forgot to record it like he usually does, so I had to take notes from his PowerPoint, as well as a friend. If it matters, I have already found <p>, but only after I did the Fourier Transform from the position space. It turned out to be:

$$<p> =\frac{-\alpha \hbar}{3\pi}$$

If I did it correctly. So can anyone give me insight on what I did wrong? Any help is appreciated.

2. Oct 29, 2013

### Simon Bridge

$$\langle \hat{p}\rangle = \int_0^\infty \psi^\star (x)\hat{p}\psi (x) dx$$ ... spot the difference.

3. Oct 29, 2013

### Legaldose

Oh, I see my problem, thank you, so I need to apply the p operator to psi before multiplying by the conjugate?

4. Oct 29, 2013

### Simon Bridge

That's the one - the relation you used only works for operators like position.
In general, the operator goes in between the wavefunctions.

In Dirac notation: <Q>=<ψ|Q|ψ>

5. Oct 29, 2013

### vanhees71

Yes, and the complete derivation is as follows
$$\langle A \rangle =\langle \psi|\mathbf{A}| \psi \rangle= \int_{\mathbb{R}} \mathrm{d} x \langle \psi|x \rangle \langle x|\mathbf{A}|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x \; \psi^*(x) \hat{A} \psi(x).$$
Note that there is a subtle difference in my use of symbols. While $\mathbf{A}$ is the operator representing the observable $A$ in abstract (basis independent) Hilbert space, $\hat{A}$ denotes the representation of the operator in position representation. Your equation for $\hat{p}$ is of course correct.