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?

 

[BvU]

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 ?

Here is an example: https://ecee.colorado.edu/~bart/book/book/chapter2/ch2_4.htm

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:

Link for a larger version

 

[BvU]

Here is an example: https://ecee.colorado.edu/~bart/book/book/chapter2/ch2_4.htm

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.

[edit] 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.

 

[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]

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 $p=\hbar k$.

 

[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]

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 $n_x = \frac{2p_x \cdot L_x}{h}$.

 

[BvU]

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]

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.

 

[BvU]

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]

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]

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?

 

[BvU]

where the particle is positioned in the container

is not determined in quantum mechanics. You only have a probability density from the wave function.

it is always considered to be in the center

Not the particle, the origin for counting the number of possible states for a given |p|

 

[JohnnyGui]

Not the particle, the origin for counting the number of possible states for a given |p|

So if I understand correctly, the particle can be positioned anywhere within the container but the origin of its momentum is assumed to be in the center of the container?

 

[PeterDonis]

the origin of its momentum

What does "the origin of its momentum" mean? You seem to be confusing position space with momentum space. The origin of momentum is in momentum space, not position space; it isn't anywhere in the container, because the container is in position space.

 

[PeterDonis]

a changing $L_e$ that changes with particle position?

$L_e$ doesn't change with particle position. It's determined by the container, not by the particle.

 

[JohnnyGui]

What does "the origin of its momentum" mean? You seem to be confusing position space with momentum space. The origin of momentum is in momentum space, not position space; it isn't anywhere in the container, because the container is in position space.

Doesn't a momentum vector have a direction in spatial direction as well since it's a function of velocity, with the particle's position as the spatial origin? Wouldn't it have a corresponding spatial container length in that same spatial direction?

$L_e$ doesn't change with particle position. It's determined by the container, not by the particle.

Is there a way to explain why taking my above question into account?

 

[PeterDonis]

Doesn't a momentum vector have a direction in spatial direction as well

It does, but not the way you are thinking it does. A better way of putting it would be that momentum space has directions that correspond to the directions in position space; but momentum vectors are still vectors in momentum space, not position space.

with the particle's position as the spatial origin?

No. The particle's position is not represented anywhere in momentum space. The "origin" of momentum space is the "zero momentum vector"--the state of having exactly zero momentum. It does not correspond to any position at all. A momentum vector is a vector in momentum space, not position space; you need to take a step back and think very carefully about what that means.

Is there a way to explain why taking my above question into account?

I don't see what the issue is: the container has a size and shape, and that size and shape is the same no matter where any particles are located within the container. That seems too obvious to even need mentioning, so I have a hard time seeing why you are having trouble with it. And the container's size and shape is what determines $L_e$.

 

[JohnnyGui]

I don't see what the issue is: the container has a size and shape, and that size and shape is the same no matter where any particles are located within the container. That seems too obvious to even need mentioning, so I have a hard time seeing why you are having trouble with it. And the container's size and shape is what determines LeLeL_e.

It's because I'm seeing that the number of states within a particular momentum vector is a function of the contaner's length in same the spatial direction as that of the concerning momentum vector, just like $L_e$ and $p_e$ as shown in the illustration. It makes me think they are tied together.

If the momentum vector has another spatial direction, then another dimension length of the container (different from $L_e$), that has the same spatial direction as that momentum vector, would be chosen to calculate the number of states in that momentum. Is this incorrect?

 

[BvU]

It makes me think they are tied together

They are not. $x$ and $y$ (and an possible $z$) are completely separate for the solution of the Schroedinger equation. It's only when you start counting states that satisfy a particular criterion involving $x$ and $y$ they have to be combined ( in p space and in n space)

advice: make two drawings, one in p space and one in n space.

 

[BvU]

Other idea: look at the wave functions for the lower values of $n$. Contour plots, expectation values for $x$ and $p$, and the (half the time surprising) probability densities at such location, etc.

 

[JohnnyGui]

Other idea: look at the wave functions for the lower values of nnn. Contour plots, expectation values for xxx and ppp, and the (half the time surprising) probability densities at such location, etc.

Thanks, I'll see what I can get out of this.

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

Just to make sure I understand this correctly, based on this explanation and your mentioned formula:
$$n_e = \frac{2L_e \cdot p_e}{h}$$
Does this mean that for a particle at any location within the container, to calculate the number of states within a momentum in a random spatial direction, $L_e$ would be the spatial length of the container in that same spatial direction but intersecting the center of the container?

 

[BvU]

Forget about the location of the particle in the container. There is no such thing in QM.

 

[PeterDonis]

for a particle at any location within the container

If the particle is in a momentum eigenstate, which is what you are considering, then it has no definite location. It can't, because of the uncertainty principle: position and momentum are non-commuting observables. So any reasoning you do in which you think of the particle as having a location is not valid.

 

[vanhees71]

Forget about the location of the particle in the container. There is no such thing in QM.

To the contrary! In this container (i.e., the one with rigid boundary conditions) the position is well defined as a self-adjoint operator, but momentum is not. There are thus also no momentum eigenstates. Nevertheless the energy is a well-definied observable since the Hamiltonian is a self-adjoint operator. Just use the search function in these forums. We've discussed this at length once. The original question in the OP is thus pointless in this case.

The wave numbers of the energy eigenstates are just labeling these energy eigenstates, but they are not to be associated as "momentum components" of the particle in the rigid box.

It's the opposite in the "toroidal box", which is often used in QFT to solve notorious problems with the infinite-volume limit, there you give periodic boundary conditions rather than rigid boundary conditions. Then, of course, no position observable exists anymore, because you cannot define a self-adjoint position operator on this Hilbert space anymore, but then the momentum observables are well defined as self-adjoint operators.

 

[JohnnyGui]

Forget about the location of the particle in the container. There is no such thing in QM.

I see. In that case, is the remainder of my statement then correct based on your explanation and formula?

"To calculate the number of states within a momentum in a random spatial direction, $Le$ would be the spatial length of the container in that same spatial direction intersecting the center of the container"

 

[BvU]

Try to move away from this picture. Look at the wave function. Don't think 'marble in box' is the same as what we are dealing with here. You have solved the wave equation in a specific context and found states that satisfy the equation and the boundary conditions; now continue in this wave picture.

@vanhees71: I have always learned that the position may well be an operator, but that you can not point at a position in the box and say: that's where the particle is located right now. All there is, is a probability density. My estimate is that the poster is hung on a mixup of p, x and n space in a Schroedinger picture and can't move on.

 

[PeterDonis]

the position is well defined as a self-adjoint operator

But the particle is not in an eigenstate of this operator, so it has no definite position.

 

[JohnnyGui]

My estimate is that the poster is hung on a mixup of p, x and n space in a Schroedinger picture and can't move on.

What I'm noticing is that I merely want to understand which length $L$ of the container I have to fill in the formula $n = \frac{2L p}{h}$ for a diagonal momentum $p$ in any spatial direction that has both an $x$ and $y$ coordinate. Is it always $L = \sqrt{L_x^2 + L_y^2}$ regardless of where the concerning momentum is spatially directed at??

 

[vanhees71]

@vanhees71: I have always learned that the position may well be an operator, but that you can not point at a position in the box and say: that's where the particle is located right now. All there is, is a probability density. My estimate is that the poster is hung on a mixup of p, x and n space in a Schroedinger picture and can't move on.

I'm always careful enough to say a self-adjoint operator is representing an observable in the formalism of QT, it's not the observable itself, but that's semantics.

Math is always clear, and an self-adjoint operator is a self-adjoint operator by definition, and there's no exception for (admittedly academic and oversimplified) models like the box with rigid boundary conditions. An operator has a domain and a co-domain, and the definition of a self-adjoint operator implies that the co-domain must be the same as its domain.

For the rigid 1D box the Hilbert space is $\mathrm{L}^2([0,L])$. The position operator is then defined as in infinite space by $\hat{x} \psi(x)=x \psi(x)$. It's obviously Hermitean, i.e., $\langle \psi|\hat{x} \phi \rangle=\langle \hat{x} \psi|\phi \rangle$ for all wave functions, for which $\hat{x} \psi(x)$ and $\hat{x} \phi(x)$ are again square-integrable. For the rigid boundary conditions, $\psi(0)=\psi(L)=0$, obviously also $\hat{x}\psi(x)$ is again fulfilling these boundary conditions, and thus the so defined position operator is self-adjoint.

The putative momentum operator $\hat{p} \psi(x)=-\mathrm{i} \partial_x \psi(x)$ is also Hermitean (which you can easily check by calculating the scalar products explicitly), but it's not self-adjoint. If this were the case, the eigenvectors which are $\sin(k x)$ and $\cos(k x)$ with $k$ chosen for both cases such as to fulfill the boundary conditions, should have the same co-domain as the domain, but that's not the case, because the derivative of the eigenvectors does not fulfill the boundary conditions and thus is outside of the Hilbert space.

What's, however, self-adjoint is the Hamiltonian, i.e., $\hat{H}=-\hbar^2 \partial_x^2/(2m)$, and thus you have a well-defined position opreator and a well defined Hamiltonian, and that's enough to justify this nice example for eigen-value problems. Nevertheless the corresponding eigenstates are eigenstates of the Hamiltonian not of the only Hermitean momentum operator, and there is no true momentum observable for this example.

 

[vanhees71]

But the particle is not in an eigenstate of this operator, so it has no definite position.

It cannot be in an eigenstate of the position operator, because the position eigenstates are distributions, not square-integrable functions. That's not different from the infinite-volume case.

 

[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 }.

I learned something new today:

There is no such thing as "number of states within a specific momentum direction". The number of states is proportional to the area in momentum space. Since a specific momentum direction (such as my mentioned $p_e$) does not have a defined momentum-space area, calculating the number of states in a specific momentum direction is not possible.

Please correct me if I'm wrong on this and why.

 

[vanhees71]

I still have no clue what you are after! Again: There is no momentum for the rigid-boundary box and thus it's nonsensical to look for momentum-level densities. For the periodic-boundary box it's an obvious and very important finding that the number of single-particle momentum states in a phase-space volume is given by
$$\mathrm{d}^3 p \frac{V}{(2 \pi \hbar)^3}.$$
For more details, see

https://th.physik.uni-frankfurt.de/~hees/publ/kolkata.pdf

Sect. 1.2 and Sect. 1.8.

 

[JohnnyGui]

I still have no clue what you are after! Again: There is no momentum for the rigid-boundary box and thus it's nonsensical to look for momentum-level densities. For the periodic-boundary box it's an obvious and very important finding that the number of single-particle momentum states in a phase-space volume is given by
$$\mathrm{d}^3 p \frac{V}{(2 \pi \hbar)^3}.$$
For more details, see

https://th.physik.uni-frankfurt.de/~hees/publ/kolkata.pdf

Sect. 1.2 and Sect. 1.8.

From what I understand your formula gives the total number of states of all the possible momentums within an area of $d^3p$ space in the case of a periodic-boundary box, correct? Since there's no momentum in a rigid-boundary box, what I want to know is if it's possible to calculate the number of states within just one momentum at a certain direction in p-space in the case of periodic boundary conditions.

Looking at your mentioned formula and from what I read previously, I can see that this is not possible because a single momentum does not have a defined volume in p-space; the number of states should be proportional to the volume in p-space. Is this correct?

 

[vanhees71]

Yes, the reason to introduce a finite volume and periodic boundary conditions often is to have a calculational tool to make sense of troublesome features of continuous eigenvalues of unbound operators in Hilbert space. It's a kind of regularization procedure. A highly non-trivial example is Haag's theorem in relativistic QFT, which is only due to using the continuous momentum spectrum (or "infinite-volume limit"). Then the trick with the finite volume and periodic boundary conditions to keep well-defined momenta helps a lot.

 

[JohnnyGui]

Yes, the reason to introduce a finite volume and periodic boundary conditions often is to have a calculational tool to make sense of troublesome features of continuous eigenvalues of unbound operators in Hilbert space. It's a kind of regularization procedure. A highly non-trivial example is Haag's theorem in relativistic QFT, which is only due to using the continuous momentum spectrum (or "infinite-volume limit"). Then the trick with the finite volume and periodic boundary conditions to keep well-defined momenta helps a lot.

Thanks. This has helped me remove the thought that one could calculate the number of states within a single momentum in one direction in p-space.

 

[JohnnyGui]

Something came up when reading further about the number of states that I can't seem to grasp.

The allowed kinetic energies of a particle in a 1-dimensional container is formulated as $E_k = \frac{h^2 \cdot n^2}{8mL^2}$. I read that in this case the integer values of $n$ does not have a limit and can have any integer value, thus there is no limit for the allowed kinetic energies of the particle in this 1 dimensional case.
In the case of a 3D cube container, the allowed kinetic energies of a particle is formulated as
$$E_k(x,y,z) = \frac{h^2(n_x^2+n_y^2+n_z^2)}{8mL^2}$$
It is said in this 3D case that each $n$ does have a limit; the total number of combinations of $n_x$ , $n_y$ and $n_z$ must be at max the volume of an eighth of a sphere with radius $R = \sqrt{n_x^2+n_y^2+n_z^2}$

I understand the derivation of all this. But what is the reason that in a 3D scenario, the integer values of $n$ must have a limit unlike in the case of a 1 dimensional value? What is the limiting cause here that doesn't play a role in the 1 dimensional case?

 

[JohnnyGui]

I might have figured out what the answer is to my above question.

There is no limit for $n$ and thus also not for $E_n$ for a particle in the 3D scenario. It all boils down to the probability based on the Boltzmann statistics for a particle to have a certain kinetic energy $E_n$.

Please correct me if I'm wrong on this.

 

[JohnnyGui]

Guys, I have a question about the approximation of counting the number of quantum states for up until a given $p$. I understand that:
$$\sqrt{n_x^2+n_y^2+n_z^2} = \frac{p \cdot 2L}{h}$$
The parameter $\sqrt{n_x^2+n_y^2+n_z^2}$ is then used as a radius to calculate the volume of an eighth of a sphere in n-space to get the number of quantum states all up to momentum $p$. However, because the quantum numbers $n_x, n_y, n_z$ are integer numbers, using this method is considered an approximation.

Is it true that the accuracy of this approximation decreases when lower momentums is used for $p$? The smaller the used momentum is, the less qantum states there are. This makes the integer "grid dots" in n-space under the calculated n-sphere volume larger. The n-sphere volume would in that case not represent the number of states accurately.

If this is true, why is it considered sufficiently accurate to calculate the number of states for a verly small increment of $dp$ using this technique?

 

[BvU]

Because there are normally such an incredibly high number of states that are occupied.

 

[JohnnyGui]

Because there are normally such an incredibly high number of states that are occupied.

Even in an infinitesimally small $dp$?

 

[BvU]

It's the density of states as a function of E (or $|\vec p|$) that is of interest here. Not how grainy it is for infinitesimal $dp$.