Force on Walls of Rectangular Potential Box by Particle Inside

[maverick280857]

Hi everyone..

I'm trying to prove the following proposition:

The force exerted on the wall perpendicular to the x-axis by a particle of mass m contained in a rectangular box of dimensions a, b, c is given by the negative of the expectation of the derivative of the Hamiltonian wrt a:

F = -\left\langle \frac{\partial \hat{H}}{\partial a}\right\rangle

But I can't see why I should differentiate with respect to the size of the box. Any ideas? This proposition is given in Landau/Lifgarbagez.

Thanks in advance.

 

[Vanadium 50]

I love this problem.

Think about this like a Halliday and Resnick problem. I have a pressure on the walls, so when i push them in, I need to apply a force. If I apply the force F through an infinitesimal distance dx, I do work dE. So F = dE/dx. Now it's simply a matter of recognizing that I can't change x without changing a, and then quantizing the system.

 

[maverick280857]

Thanks for your reply Vanadium.

Now it's simply a matter of recognizing that I can't change x without changing a, and then quantizing the system.

Thats what I am trying to understand more deeply. x is just a coordinate here and classically I would have to differentiate potential energy with respect to the coordinate to get the force. Here however, I am differentiating wrt to the size of the box. What do you mean by changing x and then quantizing the system? Could you please elaborate.

 

[Vanadium 50]

If I squeeze in on the box, I make it smaller, no?

"Then quantizing the system" means "solve using QM".

 

[Vanadium 50]

Now that I think about it more, maybe this will help.

Suppose someone gave you a box, and asked you what the pressure was. You don't (yet) know the contents of the box, but suspect it's some number of particles bouncing around inside (like a gas). You'd measure the force on the walls, probably by squeezing it and measuring the resistance - or, equivalently, by seeing how much work you had to do on it.

So measuring the pressure is equivalent to measuring dE/dx. (In fact, pressure has dimensions of energy density)

Now, I tell you what's in the box - a single particle. Given your knowledge of QM, you can calculate dE/dx from that. Now you have all the pieces.

 

[maverick280857]

I'm a dunce, I still don't get the idea behind differentiating it wrt to the size of the box. Is a more rigorous 'derivation' of this equation possible?

 

[Vanadium 50]

Does it help to replace the "box" with a piston?

 

[Minich]

Any ideas? This proposition is given in Landau/Lifgarbagez.

The problem is not so simple. Landau never liked to think about fundamentals of QM.
1. You should think about adiabaticity. Is it possible to have 90⁰ shift of phase between F and C (length of box) for sinusoidal C change in time?
2. Has the box exact 0 position? Is then the momentum of the box equal to infinity? What is then?
3. Elimination of center of mass motion is the most unsolved (it is more unsolved than solved, remember Messbauer effect and Nobel prize, Landau criterium for superfluidity) problem in QM.

 

[Vanadium 50]

If the OP is having trouble with the problem as is, it's probably not going to help him to immediately launch into subtleties.

 

[maverick280857]

See, I understand that in CM, the force is equal to minus one times the partial derivative of potential energy with respect to position x. Can you tell me how it generalizes to this form in Quantum Mechanics and why am I differentiating with respect to the size of the box?

Vanadium, I get your point about having to change a before I can quantize the system. I think I have an intuitive feel for the expression, but I don't see how I can derive it...any ideas in this direction are particularly welcome.

 

[Vanadium 50]

It's a piston. Push on it, and it gets smaller. That's why you are differentiating it with respect to the size of the box.

 

[maverick280857]

It's a piston. Push on it, and it gets smaller. That's why you are differentiating it with respect to the size of the box.

Hmm okay, thanks Vanadium.