Solving an Ensemble of Particles Wave Function Problem

[leoflindall]

b]1. Homework Statement [/b]

I have the soloution to this question, but am confused as to what has been done between each step between lines 2,3,4. Can anyone explain how they have been simplified (espicially what happened to the operator) and what the value of the intergral is? I think I am just missing something here...

Any help would be greatly appreciated!

Leo

The Question and solution are as follows,

Suppose, an ensemble of particles of mass, M, is prepared in a state as below

The Wave function = $$\sqrt{\frac{2}{L}}$$ cos ($$\frac{\Pi x}{L}$$ Between L/2 and -L/2, and is 0 otherwise.

Evaluate the expectation value {H} $$\psi$$ for an energy measurement on an en-
semble of particles prepared in $$\psi$$ (x)


The Solution

The soloution is shown in the reply below, for somereason, it wouldn't write it all out here. Exscuse the poor latex. PLease also note that h= h bar and the intergral is between limits of L/2 and -L/2


Cheers Guys!

 

[leoflindall]

{H} = $$\frac{2}{L}$$$$\int$$cos( $$\frac{x\Pi}{L}$$ )( $$\frac{P^2}{2M} )cos(\frac{x\Pi}{L}$$ .dx

= $$\frac{1}{ML}\int$$cos($$\frac{x\Pi}{L}$$)(-ihd/dx)cos($$\frac{x\Pi}{L}$$) .dx

= $$\frac{h^{2}}{ML}$$ ( $$\frac{\Pi^{2}}{L^{2}}$$ ) $$\int$$ cos$$^{2}$$ ( $$\frac{\Pi x}{L}$$ ) .dx

= $$\frac{h^{2} \Pi^{2}}{2ML^{2}}$$

 

[jdwood983]

In quantum mechanics momentum is an operator defined by

$$\mathbf{p}=i\hbar\nabla\rightarrow p=i\hbar\frac{\partial}{\partial x}$$

for a 1D problem. Since you have $p^2$, this gives you

$$\mathbf{p}^2=-\hbar^2\nabla^2\rightarrow p^2=-\hbar^2\frac{\partial^2}{\partial x^2}$$

So between lines 1 and 2, you have that (and you were missing a squared term)

$$\langle H\rangle=-\frac{\hbar^2}{ML}\int\cos\left[\frac{\pi x}{L}\right]\cdot\frac{\partial^2}{\partial x^2}\cos\left[\frac{\pi x}{L}\right]dx$$

You should know that the second derivative of cosine gives you

$$\frac{\partial^2}{\partial x^2}\cos[\alpha x]=-\alpha\frac{\partial}{\partial x}\sin[\alpha x]=-\alpha^2\cos[\alpha x]$$

Using this between 2 and 3, you get

$$\langle H\rangle=\frac{\hbar^2}{ML}\,\frac{\pi^2}{L^2}\int\cos^2\left[\frac{\pi x}{L}\right]dx$$

Performing this integral gives

$$\langle H\rangle=\frac{\hbar^2}{ML}\,\frac{\pi^2}{L^2}\cdot\left(\frac{\sin\left[2\pi x/L\right]}{4\cdot\pi/L}+\frac{x}{2}\right)$$

Which, evaluating at the limits of +L/2 and -L/2, we get

$$\langle H\rangle=\frac{\hbar^2}{ML}\,\frac{\pi^2}{L^2}\cdot\frac{L}{2}$$

Which reduces to

$$\langle H\rangle=\frac{\pi^2\hbar^2}{2ML^2}$$

 

[leoflindall]

That makes perfect sense! I think it was a stupid error on my part, missing out the squared term on the operator threw me off.

Thanks for your help!

Leo