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

 

[paranormal]

Dear Leo,

Thank you for your question about the solution to the ensemble of particles wave function problem. It seems like you are having trouble understanding the simplification that occurs between lines 2, 3, and 4. I will explain the steps that were taken and what they mean.

In line 2, the expectation value of the Hamiltonian operator is being evaluated. This operator represents the total energy of the system, and it is denoted by the symbol H. The wave function is represented by the symbol \psi, which describes the probability amplitude of finding a particle at a certain position.

In line 3, the Hamiltonian operator is applied to the wave function. This means that the operator acts on the wave function, resulting in a new expression. In this case, the operator is being multiplied by the wave function, which is why it appears next to it in the equation.

In line 4, the integral is being evaluated. An integral is a mathematical operation that represents the area under a curve. In this case, the integral is being used to calculate the expectation value of the Hamiltonian operator. The integral is being taken over the limits of L/2 and -L/2, which represent the boundaries of the system.

The value of the integral is the expectation value of the Hamiltonian operator, which represents the average energy of the ensemble of particles. This value is important because it gives us an idea of the energy distribution of the particles in the system.

I hope this explanation helps clarify the steps that were taken in the solution. If you have any further questions, please do not hesitate to ask. Keep up the good work in your studies!

Best regards,