# Force exerted by a particle in a box on the boundary

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1. Dec 26, 2017

### Pushoam

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

The energy eigen - value of a particle in a box is given by $E_n = \frac { n^2 h^2}{8mL^2}$ .

Now, applying classical mechanics , $\frac { p^2}{2m} = \frac { n^2 h^2}{8mL^2}$ .

$p \propto \frac { 1} L$ ,

So, the force is proportional to $\frac { 1} L$.

So, the answer is option (c).

2. Dec 26, 2017

### TSny

Can you add some explanation of getting from $p \propto \frac { 1} L$ to "the force is proportional to $\frac { 1} L$"?

Imagine the length of the box is slowly increased by an infinitesimal amount $\delta L$. By how much does the energy of the system change? How do you account for this change in energy?

3. Dec 27, 2017

### Pushoam

$\vec F = \frac { d\vec p }{dt}$

$p \propto \frac1{ L}$ and L is independent of t. So, $F \propto \frac1{ L}$ .

But, this approach should not be right as it is classical, should it?
In quantum mechanics, do we talk of force?

The eigen energy is the total energy of the particle, right?

It is not K.E. or P.E. of the particle. Right?

$dE = \frac { C} L dL$

$\frac { dE} {dL } = \frac { C} L$

Where C is an appropriate constant.

Is $F = \frac { dE} {dL }$ ?

Then, isn't this energy only potential energy instead of total energy?

4. Dec 27, 2017

### TSny

I think you can get the answer with a hybrid classical/quantum argument where you use a kinetic-theory-of-gases approach to relate the force to the momentum. But, it would be better to get the answer without using the classical "hand waving".

Not often in my experience. But here we are.

Right.

I don't follow this. Use your expression for En given in the first post.

Yes.

No, you want E to be the total energy. The idea is that the loss of total energy of the system (as it expands) equals the work done by the system on the environment.

5. Dec 27, 2017

### Pushoam

$dE = \frac { C} {L^3} dL$

$\frac { dE} {dL } = \frac { C} {L^3}$

Where C is an appropriate constant.

dE is the work done on the surrounding

dE = F. dl

So, F = $\frac { dE}{dl}$ .

So, the correct option is (b).

6. Dec 27, 2017

To take a completely classical approach, for gaseous particles, $PV=nRT$, and in this case we are considering the pressure or force in one direction. For a single particle at a given energy, the force should indeed be inversely proportional to the length of the box. That is a result of the number of collisions with the wall is inversely proportional to the length. Any QM result should agree with this. When considering the Q.M. wave in momentum space, it the particle doesn't have a precise location, but simply has an average momentum at the wall that causes a force there. I agree with the OP's original answer.

7. Dec 27, 2017

### TSny

Looks good to me.

8. Dec 27, 2017

### TSny

But the force depends on more than just the rate of collisions.

9. Dec 27, 2017

My expertise at Q.M. is rather limited. I'm going to need to study this further. :)

10. Dec 27, 2017

### Pushoam

According to the Bhor's Correspondence principle, for large n, quantum results tend towards the classical results.
So, for large n,
$\frac { p^2}{2m} = \frac { n^2 h^2}{8mL^2}$ .

$p \propto \frac { 1} L$ ,

So, the force is proportional to $\frac { 1} L$.

So, the answer is option (c).

I got this idea right now. So, I posted it to get it checked.

11. Dec 27, 2017

You get a minus sign when you take $dE/dL$. You need to go to a higher quantum number $n$ to keep the particle at the same energy as you expand the box. The energy isn't staying the same at a given $n$ when you expand the box, and any energy you input is not getting expended as work done to the surroundings. I think any results that are obtained by interpreting $dE/dL$ are incorrect here, but again, this is outside of my area of expertise, and I will need to study it further.

12. Dec 27, 2017

### Pushoam

The particle will not be at the same energy as I expand the wall. Its energy will decrease. The decrease in energy is workdone by the particle on the boundary walls.

Why do I need to go to a higher quantum no. n for this purpose?

13. Dec 27, 2017

### Pushoam

No energy is being given to the particle from outside agent. The particle loses its own energy by increasing walls separation.

14. Dec 27, 2017

Your post 5 gets you a different answer. That's why I'm giving this one further study. I'm inclined to believe your first answer (c), but at present, I'm not completely satisfied. $\\$ Editing: A google of this seems to show that $1/L^3$ is the correct answer, so that I'm going to need to try to digest it in detail. And the answer seems to be, you need to push on it to compress it, when it stays in the same quantum state $n$, and its energy increases. Thereby, the force is proportional to $1/L^3$. Thank you @TSny . These Q.M. problems can be quite educational. :)

Last edited: Dec 27, 2017
15. Dec 27, 2017

### TSny

Yes, $p \propto \frac { 1} L$. (You can also get that from de Broglie's $p = h/\lambda$ and the fact that $\lambda =2L$ for the ground state.)

But how does this imply that the force is proportional to $\frac { 1} L$?

16. Dec 27, 2017

### TSny

For a classical particle bouncing back and forth in the box, the average force is the change in momentum per collision times the number of collisions per unit time.

The change in momentum per collision is 2p. The rate of collisions at one end of the box is v/(2L) = p/(2mL).
So, classically, F = 2p⋅p/(2mL) = p2/(mL).

Using the de Broglie relation p = h/λ = h/(2L), you get F = h2/(4mL3).

This hand waving argument happens to give the same result as dE/dL.

17. Dec 27, 2017

### Pushoam

F = $\frac {dp }{dt}$
Since L is independent of t, it is size of box, I thought of taking $\frac 1 L$ out of differentiation and I didn't think of other factors (thinking that the question asks only for L - dependence of force).
This wrong attitude towards other factors led to the wrong answer.

18. Dec 27, 2017

### Pushoam

Why do you call it a hand waving attitude?

19. Dec 27, 2017