Calculate d<p> / dt

1. May 18, 2013

Cogswell

1. The problem statement, all variables and given/known data
Calculate $\dfrac{d <p>}{dt}$

Answer: $\left< -\dfrac{\partial V}{\partial x} \right>$

2. Relevant equations

Schrodinger equation: $i \hbar \dfrac{\partial \Psi}{\partial t} = -\dfrac{\hbar ^2}{2m} \frac{\partial ^2 \Psi}{\partial x^2} + V \Psi$

3. The attempt at a solution

Here's what I did:

$\displaystyle \dfrac{\partial}{\partial t} \int^{\infty}_{- \infty} \Psi ^* \left( \dfrac{\hbar}{i} \dfrac{\partial}{\partial x} \right) \Psi dx$

$\displaystyle \dfrac{\hbar}{i} \int^{\infty}_{- \infty} \dfrac{\partial}{\partial t} \Psi ^* \dfrac{\partial \Psi}{\partial x} dx$

$\displaystyle \dfrac{\hbar}{i} \int^{\infty}_{- \infty} \left[ \dfrac{\partial \Psi ^*}{\partial t} \dfrac{\partial \Psi}{\partial x} + \Psi^* \dfrac{\partial}{\partial t} \dfrac{\partial \Psi}{\partial x} \right] dx$ (Differentiation by Product rule)

From the Schrodinger equation we get that: $\dfrac{\partial \Psi}{\partial t} = \dfrac{i \hbar}{2m} \dfrac{\partial ^2 \Psi}{\partial x^2} - \dfrac{i}{\hbar} V \Psi$

And it's conjugate: $\dfrac{\partial \Psi ^*}{\partial t} = -\dfrac{i \hbar}{2m} \dfrac{\partial ^2 \Psi^*}{\partial x^2} + \dfrac{i}{\hbar} V \Psi^*$

Putting those into my integral I get:

$\displaystyle \dfrac{\hbar}{i} \int^{\infty}_{- \infty} \left[ \left( -\dfrac{i \hbar}{2m} \frac{\partial ^2 \Psi^*}{\partial x^2} + \dfrac{i}{\hbar} V \Psi^* \right) \dfrac{\partial \Psi}{\partial x} + \Psi^* \dfrac{\partial}{\partial x} \left( \dfrac{i \hbar}{2m} \frac{\partial ^2 \Psi}{\partial x^2} - \dfrac{i}{\hbar} V \Psi \right) \right] dx$

Expanding out everything:

$\displaystyle \dfrac{\hbar}{i} \int^{\infty}_{- \infty} \left[ -\dfrac{i \hbar}{2m} \frac{\partial ^2 \Psi^*}{\partial x^2} \dfrac{\partial \Psi}{\partial x} + \dfrac{i}{\hbar} V \Psi^* \dfrac{\partial \Psi}{\partial x} + \dfrac{i \hbar}{2m} \frac{\partial ^3 \Psi}{\partial x^3} \Psi^* - \dfrac{i}{\hbar} \Psi^* \dfrac{\partial}{\partial x} (V \Psi) \right] dx$

$\displaystyle \dfrac{\hbar}{i} \int^{\infty}_{- \infty} \left[ -\dfrac{i \hbar}{2m} \frac{\partial ^2 \Psi^*}{\partial x^2} \dfrac{\partial \Psi}{\partial x} + \dfrac{i}{\hbar} V \Psi^* \dfrac{\partial \Psi}{\partial x} + \dfrac{i \hbar}{2m} \frac{\partial ^3 \Psi}{\partial x^3} \Psi^* - \dfrac{i}{\hbar} \dfrac{\partial V}{\partial x} \Psi \Psi * - \dfrac{i}{\hbar} \dfrac{\partial \Psi}{\partial x} V \Psi ^* \right] dx$

$\displaystyle \dfrac{\hbar}{i} \int^{\infty}_{- \infty} \left[ \underbrace{-\dfrac{i \hbar}{2m} \frac{\partial ^2 \Psi^*}{\partial x^2} \dfrac{\partial \Psi}{\partial x}}_1 + \underbrace{\dfrac{i \hbar}{2m} \frac{\partial ^3 \Psi}{\partial x^3} \Psi^*}_2 - \underbrace{\dfrac{i}{\hbar} \dfrac{\partial V}{\partial x} \Psi \Psi *}_3 \right] dx$

I'm stuck at this point. I'm presuming there's a way to cancel out each of the integrals? I know the last integral is the one I want but I do not know how to cancel out the first 2.

2. May 18, 2013

Simon Bridge

I'd have kept the operator notation for longer, vis...

$\renewcommand{H}{\hat{H}} \renewcommand{p}{\hat{p}} \renewcommand{\expn}[1]{\left \langle #1 \right \rangle} \renewcommand{dt}[1]{\frac{d #1}{dt}} \renewcommand{dx}[1]{\frac{d #1}{dx}} \renewcommand{intf}[1]{\int_{-\infty}^\infty #1 \; dx} \renewcommand{ddx}[1]{ \frac{d^2 #1}{dx^2} }$ Need to show:$$\text{(1)... }\dt{}\expn{\hat{p}} = \expn{-\dx{V}}$$ (i.e. Newton's second law...)
- expand: $$\begin{array}{rl} i\hbar \dt{}\expn{\p} & = i\hbar \dt{} \intf {\Psi^\star p\Psi} \\ &= \intf { \left ( i\hbar \dt{} \Psi^\star \right ) \hat{p} \Psi + \Psi^\star \left ( i\hbar\dt{}(\hat{p}\Psi ) \right )} \; \text{ ...(2)} \end{array}$$ - from the Schrodinger equation: $$\text{(3)... }i\hbar\dt{} \Psi = \H\Psi\\ \text{(4)... } \dt{}\expn{\p} = \frac{1}{i\hbar}\intf{ \H \Psi^\star \p\Psi + \Psi^\star \H\p\Psi}$$... hence, need to show that $$\text{(5)... } \H \Psi^\star \p\Psi + \Psi^\star \H\p\Psi = -i\hbar\Psi^\star \dx{V} \Psi$$ - which, I think, is pretty much where you are up to ;)
(caveat: do not rely on me to get the math right - check!)

- note that $$\text{(6)... }\p\H = - i\hbar \dx{} \left ( -\frac{\hbar^2}{2m}\ddx{}+V \right )$$ ... gives you third-order differentiation in x as well as the dV/dx you need.

So how do you change the order of the operations?

Last edited: May 18, 2013
3. May 19, 2013

vela

Staff Emeritus
If you want to continue on from where you got to, use integration by parts to move $\frac{\partial}{\partial x}$ between $\Psi$ and $\Psi^*$.

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