# I 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.

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#### BvU

Homework Helper
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 travelling involved.

#### Higgsono

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

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

#### BvU

Homework Helper
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

Homework Helper
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

Homework Helper
(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

Homework Helper
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

(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.

BvU

"Calculate the number of states for a particle in a box"

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