Force exerted by a particle in a box on the boundary

[Pushoam]

Homework Statement

Homework Equations

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).

 

[TSny]

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 $\frac { 1} L$.

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?

 

[Pushoam]

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?

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

$p \propto \frac1{ L}$ 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?

 

[TSny]

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

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

But, this approach should not be right as it is classical, should it?

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".

In quantum mechanics, do we talk of force?

Not often in my experience. But here we are.

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

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

Right.

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

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

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

Yes.

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

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.

 

[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).

 

[Charles Link]

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.

 

[TSny]

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

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

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).

Looks good to me.

 

[TSny]

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.

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

 

[Charles Link]

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

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

 

[Pushoam]

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).

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.

 

[Charles Link]

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

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.

 

[Pushoam]

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 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?

 

[Pushoam]

any energy you input is not getting expended as work done to the surroundings

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

 

[Charles Link]

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?

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. :)

 

[TSny]

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 $\frac { 1} L$.

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 $\frac { 1} L$?

 

[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.

 

[Pushoam]

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

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.

 

[Pushoam]

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

Why do you call it a hand waving attitude?

 

[Charles Link]

Why do you call it a hand waving attitude?

It's called a "handwaving" approach or argument. I think the phrase comes from when a magician waves his hands over the hat, and makes the rabbit disappear, etc. I use the phrase quite commonly also. It generally applies when the mathematics is not completely rigorous.