I Calculating the number of energy states using momentum space

JohnnyGui

A question came up about deducing the number of possible energy states within a certain momentum $p$ using momentum space.
To make my question easier to understand, I deliberately chose $p$ and not a particular increment $dp$ and I assume a 2 dimensional momentum space with coordinates $x$ and $y$. The concerning particle thus only has translational kinetic energy in these 2 coordinates.

A particle within a box of volume $V$ can have the same momentum $p$ in different directions within that box. In a 2D momentum space this momentum $p$ is therefore given by a circle with radius $p$.
From what I understand, the number of possible energy states $N_s$ in this 2D case is then deduced from the area of the circle multiplied by the number of energy states in the $x$ and $y$ coordinates:
$$N_s = \frac{L_x \cdot p_x}{h} \cdot \frac{L_y \cdot p_y}{h} \cdot \pi$$
Where $L$ is the length of the box in a certain dimension (given by subscript $x$ or $y$).

Here's my question regarding this formula:
I can see that the formula assumes that the density of energy states is homogenous over the circular p-space because it is merely multiplying the number of energy states in the $x$ dimension by the number of energy states in the $y$ dimension. However, I don't understand why this is the case, because from what I know, the number of possible energy states in a certain direction is proportional to the length of the box in that very same direction. If a certain momentum has a combined $x$ and $y$ direction, shouldn’t the number of possible energy states within that momentum vector be dependent on the length of the box in that same direction and not by the $x$ and $y$ coordinates seperataly?

JohnnyGui

Perhaps a better alternative way to formulate my question is like this:

Why is the number of possible energy states independent of the shape of the container? Why is it merely dependent on the number of states in only 3 perpendicular container dimensions while a momentum vector can be directed at any direction within the container?
Shouldn't the length of the container in that same direction as the momentum vector also determine the number of energy states in that direction?

Last edited:

BvU

Homework Helper
Why is the number of possible energy states independent of the shape of the container?
says who ?

In both the directions the number density for a rectangular box is dependent on the length. Work it out: there is a lower bound (dependent on length) and no upper bound. Lower p can only occur in one direction. Your circle is an ellipse in $nx, ny$ coordinates.

Everyone assumes a square box (eq 25) here is an exception). For e.g a circle you get something quite different

JohnnyGui

says who ?

In both the directions the number density for a rectangular box is dependent on the length. Work it out: there is a lower bound (dependent on length) and no upper bound. Lower p can only occur in one direction. Your circle is an ellipse in nx,nynx,nynx, ny coordinates.
Does this imply that, for a momentum vector which is a combination of these $x,y$ coordinates, the number of states within that momentum vector is dependent on the length of the container in that same vector direction? I have illustrated my question (in 2D momentum space) to show what I mean:

Attachments

• 17.6 KB Views: 383

BvU

Homework Helper
I don't see any other shape here than a cube with side length L (nice book, though!)

I do see $k_x = {n\pi\over L}$ $n$ = 1, 2, ,3, ... so for a rectangular box you get $k_x = {n\pi\over L_x}$ etc. And with that
Does this imply that, for a momentum vector which is a combination of these $x,y$ coordinates, the number of states within that momentum vector is dependent on the length of the container in that same vector direction?
Correct.

So in your picture the steps in the x-direction are smaller than in the y-direction.

by the way:
$p = \hbar k= \displaystyle {hk\over 2\pi}$ so don't forget the 2.

and:
you make life difficult using mixed notation, as in $\displaystyle {L_yp_y\over h} = p_y\ \ \ \ \$ ...
better write something like:
$$n_{x, {\rm max}}= {2L_x h\over |p|_{\rm max}}$$ etcetera.

 small mistake (see below). Should be $$n_{x, {\rm max}}= {2L_x |p|_{\rm max}\over h}$$
(hey, how do I get red $\LaTeX$ ?
So you get a red ellipse in n-space (3D: ellipsoid) instead of a red circle. (for counting, we usually make use of the n-space).

In p-space you do have a circle, but there the grid point density differs per Cartesian axis.

Last edited:

JohnnyGui

Correct.So in your picture the steps in the x-direction are smaller than in the y-direction.
Let me restate my question to make sure it came across clearly. Does this mean that the number of states in $p_e$ shown in my picture is dependent on the cross-sectional container length $L_e$ shown in the picture (the diagonal light blue line)?

I don't see any other shape here than a cube with side length L (nice book, though!)
A quote in the link says the following: The semiconductor is assumed a cube with side L. This assumption does not affect the result since the density of states per unit volume should not depend on the actual size or shape of the semiconductor.

nx,max=2Lxh|p|maxnx,max=2Lxh|p|max
Apologies, but I can't seem to understand how this formula is derived from my initial formula, even after implementing the factor of 2. I thought that the total number of states $n_t$ within a 3D spherical momentum space is $n_t = \frac{V \cdot 4\pi p^3}{3h^3}$ and that for 1 dimension (e.g. the x-coordinate) it would be: $n_{x,max} = \frac{L_x \cdot p_{max}}{h}$ (the factor of 2 is added when there are 2 possible spins, in the case of electrons, according to the link)

BvU

Homework Helper
Does this mean that the number of states in $p_e$ shown in my picture is dependent on the cross-sectional container length $L_e$ shown in the picture (the diagonal light blue line)?
You are mixing up p-space with x-space. $L_e$ lives in a different world than $p_e$.

Take a case where $L_x << L_y$ and draw the points that have $|p| \le \text {some value}$

The semiconductor is assumed a cube with side L. This assumption does not affect the result since the density of states per unit volume should not depend on the actual size or shape of the semiconductor.
crucial here is the 'per unit volume' (see his 2.4.6 where the $L^3$ divides out, and the application to a (rectangular!) box in example 2.3)

$n_{x, {\rm max}}= {2L_x h\over |p|_{\rm max}}$ follows from his 2.4.2: $k_x = {n_x \pi \over L}$ combined with $p=\hbar k$.

Last edited:

JohnnyGui

You are mixing up p-space with x-space. LeLeL_e lives in a different world than pepep_e.
Yes, I am indeed aware that they are in different worlds. But my question is about the relationship between these 2 worlds formula-wise (relationship between $n$ and $L$ according to the mentioned formula). The number of states in momentum vector $p_x$ is dependent on the length of the box in the x-coordinate $L_x$, since the momentum $p_x$ is also directed in the $x$ direction, right? In that case, why isn't the number of states in momentum vector $p_e$ dependent on the length of the box $L_e$?

crucial here is the 'per unit volume' (see his 2.4.6 where the L3L3L^3 divides out, and the application to a (rectangular!) box in example 2.3)
Ah, that's what I missed. In that case, does the number of states differ "per unit momentum" depending on which direction the momentum is directed at?

nx,max=2Lxh|p|maxnx,max=2Lxh|p|maxn_{x, {\rm max}}= {2L_x h\over |p|_{\rm max}} follows from his 2.4.2: kx=nxπLkx=nxπL k_x = {n_x \pi \over L} combined with p=ℏkp=ℏkp=\hbar k.
I'm sorry but perhaps I'm missing something very obvious here. If I substitue $k_x$ with $\frac{p_x}{\hbar}$, then I still get $n_x = \frac{2p_x \cdot L_x}{h}$. Why are $p$ and $h$ parameters switched in your case compared to mine?

BvU

Homework Helper
Does this mean that the number of states in $p_e$ shown in my picture is dependent on the cross-sectional container length $L_e$ shown in the picture (the diagonal light blue line)?
Ah, maybe I get it: For a given direction of $p_e$ you have
the number of states in the x-direction = $\displaystyle{2L_x p_{e,x}\over h}$
and in the y-direction = $\displaystyle{2L_y p_{e,y}\over h}$ .
So in n-space you get $n_e = \sqrt{n_x^2+n_y^2} = \displaystyle{2L_e p_e\over h }$.

does the number of states differ "per unit momentum" depending on which direction the momentum is directed at?
Yes. You have an expression.

switched in your case compared to mine
Can you point it out ? I don't know where that occurs.

JohnnyGui

Ah, maybe I get it: For a given direction of pepep_e you have
the number of states in the x-direction = 2Lxpe,xh2Lxpe,xh\displaystyle{2L_x p_{e,x}\over h}
and in the y-direction = 2Lype,yh2Lype,yh\displaystyle{2L_y p_{e,y}\over h} .
So in n-space you get ne=√n2x+n2y=2Lepehne=nx2+ny2=2Lepehn_e = \sqrt{n_x^2+n_y^2} = \displaystyle{2L_e p_e\over h }.
Yes, this is indeed what I was wondering. However, shouldn't the number of states in the $x$ and $y$ projection of $p_e$ in that case be dependent on the projection of length $L_e$ in those coordinates ($L_{e,x}$ and $L_{e,y}$), not the full $L_x$ and $L_y$ of the container? After all, the shape of the container could be so irregular that $L_e$ does not have any relationship with the $x$ and $y$ dimensions of the container.

For example, the number of states of $p_{e,y}$ would be $\frac{L_{e,y} p_{e,y}}{h} = n_{e,y}$

Can you point it out ? I don't know where that occurs.
You formulated it as $n_{x, {\rm max}}= {2L_x h\over |p|_{\rm max}}$ whereas I formulated it as $n_x = \frac{2p_x \cdot L_x}{h}$.

Last edited:
BvU

BvU

Homework Helper
shouldn't the number of states in the $x$ and $y$ projection of pepep_e in that case be dependent on the projection of length $L_e$ in those coordinates ($L_{e,x}$ and $L_{e,y}$), not the full LxLxL_x and LyLyL_y of the container
No. The number of states is dependent on the projection of $p_e$ only. The distance between allowed states for $p_{e, x}$ depends on $L_x$ - idem y.

Why are p and h parameters switched in your case compared to mine?
Oops, big small mistake I missed, even when you pointed it out... You're perfectly correct. Sorry about that, ehmmm...
I edited the first occurrence

JohnnyGui

No. The number of states is dependent on the projection of pepep_e only. The distance between allowed states for pe,xpe,xp_{e, x} depends on LxLxL_x - idem y.
Ah, this is what I can't seem to grasp. According to your statement this means that:
$$n_e = \sqrt{\frac{2L_x \cdot p_{e,x}}{h}^2 + \frac{2L_y \cdot p_{e,y,}}{h}^2} = \frac{2L_e \cdot p_e}{h}$$
However, the length of the container in the $L_e$ dimension can be any size, regardless of how large $L_x$ and $L_y$ are, which can lead to the equation falling apart (in the case of a weird shaped container for example). The same goes for if the particle is near one of the walls of the container, in which case the $L_e$ length of the container would change as well. Is there a way to explain why these cases don't matter?

Oops, big small mistake I missed, even when you pointed it out... You're perfectly correct. Sorry about that, ehmmm...I edited the first occurrence
No problem at all, thanks for verifying it

BvU

Homework Helper
the length of the container in the $L_e$ dimension can be any size
How so ? It's always between Lx and ly.

Note that x and y are fully independent: we solve for each one completely separately.

JohnnyGui

How so ? It's always between Lx and ly.

Note that x and y are fully independent: we solve for each one completely separately.
Something like this for example:

However, since you said $L_e$ should be always between $L_x$ and $L_y$, does this mean that the largest dimensions of an irregular shaped container are chosen for the calculation?
If the answer is yes, doesn't the number of states in a momentum vector also depend on the direction of a momentum vector and the location of the concerning particle within the container? For example, momentum $p_y$ in this case is confined within length $L_{y2}$ and not length $L_y$ of the container.

Attachments

• 27.7 KB Views: 335

BvU

Homework Helper
Something like this for example
You'll have a hard time finding solutions for the Schroedinger equation in this funny case !

JohnnyGui

You'll have a hard time finding solutions for the Schroedinger equation in this funny case !

Does this mean that the formula is only valid for symmetrically boxed containers, since the number of states within a momentum vector does depend on the length dimension in which the momentum is directed at, such as in the case of my last irregular shaped container?

BvU

Homework Helper
Bear in mind that these boxes are highly artificial. They are only used to unearth features that scale nicely (e.g. density per volume). The direction of a momentum isn't all that relevant.

The number of states with $|p| \le$ a given momentum depends on direction also in a symmetrically boxed container. We've been through that, haven't we ?

JohnnyGui

The number of states with |p|≤|p|≤|p| \le a given momentum depends on direction also in a symmetrically boxed container. We've been through that, haven't we ?
Yes we have. But what I find very peculiar is that $L_e$ is represented by $L_x$ and $L_y$ instead of its projections, even in a symmetrically boxed container. Let's put the particle at the very upper left corner within the symmetrically boxed container (in my first post). In that case, $L_e$ would be very short. How can the equation for $n_e$ represented by the constants $L_x$ and $L_y$ then still hold for a changing $L_e$ that changes with particle position?

BvU

Homework Helper
The origin of all spaces is in the 'center'. Don't mix up n, p and x space

JohnnyGui

The origin of all spaces is in the 'center'. Don't mix up n, p and x space
So no matter where the particle is positioned in the container, it is always considered to be in the center, even in x-space? What would be the siginificance of a container in that case then be?

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving