No 2D momentum average for a particle in a box?

1. Nov 28, 2013

Duave

Hello Everyone, Happy Thanksgiving. I have physics issues. I went around the entire boundless universe in the last two days, and found out the Schrodinger forgot to develop the math for the momentum expectation value for a 2-D particle in a box. It's nowhere to be found. I am a chemistry major, and so I need help from the physics department to engineer the math to complete quantum mechanics. Can anyone help me with my marvelous discovery below? It's series of equations used as an attempt to solve the problem that I posted. Everyone's help would be greatly appreciated. Thank you.

1. The problem statement, all variables and given/known data
Suppose that a particle of mass m is confined to a rectangular region between 0<x<L and 0<y<L/2 by an infinitely high potential energy function.

Calculate the expectation value p$^{2}$, the square of the magnitude of the momentum

2. Relevant equations

3. The attempt at a solution

p^2(x)=∫(2/L) sin(n*pi*x/L)(ih[d/dx](2/L) sin(n*pi*x/L)dx
p^2(y)=∫(4/L) sin(2n*pi*y/L)(ih[d/dx](4/L) sin(2n*pi*y/L)dy
<p^2> =p^2(x) +p^2(y)
<p^2> =1/2 +1
<p^2> =3/2

Last edited: Nov 28, 2013
2. Nov 28, 2013

bp_psy

The basic idea is good but the results of your integrals are wrong.
On the other hand this problem is usually done without doing the integrals.
Think about the relation between <p^2> and <E> is, considering that the Hamiltonian is H=p^2/2m inside the box. Try doing it for the 1D case then go for the 2D case.

3. Nov 28, 2013

Duave

@bp_psy,

Thank you for your response. The image that I posted is of the one-dimensional case.

https://scontent-a.xx.fbcdn.net/hphotos-ash3/541843_10151745284505919_675157014_n.jpg

It's all over the internet, and I still had to ask about the 2D case. I don't understand "how to take it to the next level".

1. The problem statement (with respect to your response)

I'm not sure if I can just solve P^2 by simply:

P^2(x)/2m = E

3. The attempt at a solution (with respect to your response)

P^2(x)= 2m*E
E=$\frac{h^2}{8mL^2}$*[n^2(i) +4*n^2(ii)]

hence,

P^2(x)= 2m*$\frac{h^2}{8mL^2}$*[n^2(i) +n^2(ii)]

hence,

P^2(y)= 2m*$\frac{h^2}{8mL^2}$*[n^2(i) +4*n^2(ii)]

P^2= P^2(x) + P^2(y)

P^2= {2m*$\frac{h^2}{8mL^2}$*[n^2(i) +n^2(ii)]} + {2m*$\frac{h^2}{8mL^2}$*[n^2(i) +4*n^2(ii)]}

P^2= 2m*E

{2m*$\frac{h^2}{8mL^2}$*[n^2(i) +n^2(ii)]} + {2m*$\frac{h^2}{8mL^2}$*[n^2(i) +4*n^2(ii)]} = $\frac{h^2}{8mL^2}$*[n^2(i) +4*n^2(ii)]

I'M LOST..........

4. Nov 28, 2013

bp_psy

You seem to get the basic idea but you should always mark when you are talking about expectations <p(x)>^2 is not the same as p(x)^2 and E is not the same as <E>.
No each direction is associated with some of the energy not the whole energy.So just one of the ns, n_x for p_x and n_y for p_y

In this case <p>^2/2m=(1/2m)(<p(x)>^2+<p(y)>^2)=<E>=<E_x>+<E_y>. Each component of the momentum is associated with some of the energy. The energy associated with each direction is the same as for the the 1D case You are also missing some ∏ factors and you forget to simplify some m factors.

5. Nov 29, 2013

B-80

This is how to proceed with calculating any expectation value (this has nothing to do with physics by the way, it's elementary probability theory). And you're basically right, except you're trying to calculate the average of p^2, yet you're only applying p once in the integral. Make sure you apply p two times to get the average of p^2.

I'm not sure what you're doing in line 3, but just compute the integral you have there in line 2 with the momentum operator applied twice over the region where the wave function is not zero.