Calculate the number of states for a particle in a box

[Higgsono]

The multiplicity of states for a particle in a box is proportional to the product of the volume of the box and the surface area of momentum space.

$$ \Omega = V_{volume}V_{momentum}$$

The surface area in momentum space is given by the equation:

$$p^{2}_{x}+ {p}^{2}_{y}+{p}^{2}_{z} = \frac{U}{2m}$$

But this seem rediculous to me. It says that, "the number of possible directions a particle can travel in is proportional to the total energy of the particle". I don't understand this, why would the particle have more directions to choose from when it have higher energy? I know what the mathematics says, but the physics of it I don't understand.

 

[BvU]

It says that, "the number of possible directions a particle can travel in is proportional to the total energy of the particle".

No it doesn't. The states are steady-state solutions and there is no traveling involved.

 

[Higgsono]

No it doesn't. The states are steady-state solutions and there is no traveling involved.

Doesn't answer my question.

What do you mean? Being in a moment state means it has a certain momentum.

 

[BvU]

Doesn't answer my question

I'm sorry about that.

Being in a moment state means it has a certain momentum

Yes, but you mix up a classical mindset with terms used in quantum mechanics. The expectation values for the momentum are all zero for the momentum states.

I know what the mathematics says, but the physics of it I don't understand.

Stick to the math.

 

[BvU]

why would the particle have more directions to choose from when it have higher energy?

I can almost imagine a particle being there in that box and making is choices ...

In 1D the density of states is a constant. You can imagine dots on a half-line. A line segment length $l$ starting at 0 has $l$ dots (eigenstates). Increase $l$ with ${\sf d}l$ and you get ${\sf d}l$ more eigenstates.

In 2D same - imagine dots on a grid. In a circle with radius $r$ on that grid there are $\pi r^2$ possible eigenstates. An increase of $r$ with ${\sf d}r$ gets you $2\pi r\;{\sf d}r$ more of them.

In 3D you get a sphere and with radius $r$ on that 3D grid there are ${4\over 3}\pi r^3$ possible eigenstates. An increase of $r$ with ${\sf d}r$ gets you $4\pi r^2\;{\sf d}r$ more grid points.

But you know the math.

 

[BvU]

And you are definitely not the only one who wonders about this. cf @JohnnyGui here
(warning:154 posts ! )
( I hope JG doesn't mind my referring to the thread - I remember it well )

And wondering about things is a good habit for physicists !

 

[Higgsono]

Thanks for your answers! Maybe some of the confusion lies in my assumption that we are dealing with classical particles. In that case, intuition tells me that there should be a continnum of directions regardless of the energy of the particle.

 

[BvU]

Well, the name Quantum Mechanics itself already suggest quantization effects ...

The transition from QM to classical (usually by means of $\hbar\downarrow 0$) requires a delicate approach, not always intuitive. And for the particle in a box I don't even know.

 

[JohnnyGui]

And you are definitely not the only one who wonders about this. cf @JohnnyGui here
(warning:154 posts ! )
( I hope JG doesn't mind my referring to the thread - I remember it well )

And wondering about things is a good habit for physicists !

I don't mind it at all . I remember having difficulty understanding some aspects of the derivation of the continuous MB distribution, and I was too stubborn to truly nail it down in my head, hence the huge thread.