Hi Everyone,

Quick QM problem here.

Classic particle-in-a-box scenario. So, in 1D

[math]\psi = -\frac{\hbar}{2m} \frac{d^2\psi}{dx^2} + V\psi = E\psi[/math]

and energy is

[math]E = \frac{\pi^2 \hbar^2}{2ma^2}[/math]

Now, is it right to consider the force on the sides (e.g at x=0 and x=a) to be [imath]\frac{d\psi}{dx}[/imath]? I'm trying to show that the force is [math]F = \frac{\pi^2 \hbar^2}{ma^3}[/math] and that's all I can see which would do it... I've never come across this before.

## QM problem; particle in a box

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- massivefoot
**Posts:**39**Joined:**Mon Nov 05, 2007 8:19 am UTC

### Re: QM problem; particle in a box

Well firstly that's wrong and you can see so quite easily. \frac{d\psi}{dx} changes sign under \psi \mapsto -\psi . But these are the same physical state (they differ only by a phase), so they must exert the same force on the walls of the box.

- massivefoot
**Posts:**39**Joined:**Mon Nov 05, 2007 8:19 am UTC

### Re: QM problem; particle in a box

What you need to do is this: (hopefully I'll get the latex right this time...)

Recall that a classical system in a potential at position [imath]x[/imath] experiences a force [imath]F = -V^{\prime}(x)[/imath]

Now, the box itself is classical (force is a classical concept, so it had better be...), and the energy of the system is [imath]V(a) = E(a)[/imath], where [imath]E[/imath] is the energy of the particle. So the force on each wall is [imath]{1 \over 2} dE/da[/imath].

Recall that a classical system in a potential at position [imath]x[/imath] experiences a force [imath]F = -V^{\prime}(x)[/imath]

Now, the box itself is classical (force is a classical concept, so it had better be...), and the energy of the system is [imath]V(a) = E(a)[/imath], where [imath]E[/imath] is the energy of the particle. So the force on each wall is [imath]{1 \over 2} dE/da[/imath].

### Re: QM problem; particle in a box

This is what was confusing me... I realise that [imath]F = - \frac{dV}{dx}[/imath]... but by definition V=0 inside the box. I don't see how you can say E(a) = V(a).

### Re: QM problem; particle in a box

The force is infinite at the walls of the box, and zero elsewhere. It's not a very physical situation when it comes to forces.

massivefoot, no. The kinetic energy does not come into force computations.

Potentials, in many ways, are more fundamental than forces- and so it's a lot easier to get used to thinking in potentials than to try to convert potentials to forces to get a handle on QM problems. You will be able to get some use out of considering forces at a reflection wall (V=V0 when x>0, V=0 when x<0), or a Hydrogen atom- but in both those situations your intuition from how forces act can direct you to the wrong place, so be careful.

[edit]Oh, on the sides not of the sides. That's what I get for not reading carefully!

massivefoot, no. The kinetic energy does not come into force computations.

Potentials, in many ways, are more fundamental than forces- and so it's a lot easier to get used to thinking in potentials than to try to convert potentials to forces to get a handle on QM problems. You will be able to get some use out of considering forces at a reflection wall (V=V0 when x>0, V=0 when x<0), or a Hydrogen atom- but in both those situations your intuition from how forces act can direct you to the wrong place, so be careful.

[edit]Oh, on the sides not of the sides. That's what I get for not reading carefully!

Last edited by Vaniver on Mon May 18, 2009 5:40 pm UTC, edited 1 time in total.

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- ThinkerEmeritus
**Posts:**416**Joined:**Sat Jan 19, 2008 11:32 pm UTC

### Re: QM problem; particle in a box

seladore wrote:This is what was confusing me... I realise that [imath]F = - \frac{dV}{dx}[/imath]... but by definition V=0 inside the box. I don't see how you can say E(a) = V(a).

F is supposed to be a force exerted on the wall, and V is the potential energy due to forces on the particle inside the box. So F and V refer to different forces and are not directly related to each other.

What we need is the potential energy produced by the force on the wall. Visualize F as an exterior force and detach the walls from the top and bottom of the cube. Then if the wall moves inward a little bit, the work done on the particle-wall system is

F dx. That work must go into the system somehow, and in fact it changes the energy E(a) by accelerating the particle a little bit with each collision with the moving wall. Thus E(a) is the "potential energy" corresponding to the force on the wall. Hence it is really true that F = dE/da.

[There is no factor of 1/2 in this formula. The quantity a is the total width of the well and is therefore the distance moved by either one of the walls.]

"An expert is a person who has already made all possible mistakes." -- paraphrase of a statement by Niels Bohr

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### Re: QM problem; particle in a box

I think the question is asking what force the particle exerts on the sides of the box, in which case, yes, it's just -dE/da.

Think of the particle in a box as well, a black box. It doesn't matter what form the energy of takes while it's in the box "potential", "kinetic", whatever. All that matters is the energy of the system, as a function of its external properties (in this case, a, the size of the box).

So massivefoot is right, except for the factor of 1/2 at the end.

Think of the particle in a box as well, a black box. It doesn't matter what form the energy of takes while it's in the box "potential", "kinetic", whatever. All that matters is the energy of the system, as a function of its external properties (in this case, a, the size of the box).

So massivefoot is right, except for the factor of 1/2 at the end.

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- massivefoot
**Posts:**39**Joined:**Mon Nov 05, 2007 8:19 am UTC

### Re: QM problem; particle in a box

ThinkerEmeritus, I think we may have two slightly different physical situations in mind.

What I'm thinking is that the box is interacting only with the particle inside it, which is remaining in an energy eigenstate (which in the square well case has zero momentum expectation value) therefore the forces acting on the walls must be of equal magnitude and in opposite directions (otherwise the box's centre of mass would begin moving, which would not conserve momentum).

What you seem to be describing is that one wall is pinned in position, and the other wall moves. I believe this accounts for our difference of a factor of 1/2. Please let me know if it doesn't!

What I'm thinking is that the box is interacting only with the particle inside it, which is remaining in an energy eigenstate (which in the square well case has zero momentum expectation value) therefore the forces acting on the walls must be of equal magnitude and in opposite directions (otherwise the box's centre of mass would begin moving, which would not conserve momentum).

What you seem to be describing is that one wall is pinned in position, and the other wall moves. I believe this accounts for our difference of a factor of 1/2. Please let me know if it doesn't!

### Re: QM problem; particle in a box

If you have water in a tank, does the force it exerts on one of the sides depend on whether someone is holding the opposite side?

Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?

- ThinkerEmeritus
**Posts:**416**Joined:**Sat Jan 19, 2008 11:32 pm UTC

### Re: QM problem; particle in a box

massivefoot wrote:ThinkerEmeritus, I think we may have two slightly different physical situations in mind.

What I'm thinking is that the box is interacting only with the particle inside it, which is remaining in an energy eigenstate (which in the square well case has zero momentum expectation value) therefore the forces acting on the walls must be of equal magnitude and in opposite directions (otherwise the box's centre of mass would begin moving, which would not conserve momentum).

What you seem to be describing is that one wall is pinned in position, and the other wall moves. I believe this accounts for our difference of a factor of 1/2. Please let me know if it doesn't!

If you let both walls move, you have forces acting at each end, which gives a factor of 2 in the work formula. However,a in the energy formula is the width of the well, so da is the change of width of well, and if each side moves by dx, dx = da/2. The factors of two cancel out.

Please note I have been ignoring the sign all along since there seemed to be general agreement about it. I probably shouldn't have done that. The work has a minus sign I haven't included, and the minus sign in F = -dU/da cancels the one in the work.

"An expert is a person who has already made all possible mistakes." -- paraphrase of a statement by Niels Bohr

Seen on a bumper sticker: "My other vehicle is a Krebs cycle".

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