Calculating the number of energy states using momentum space

[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 $dp$.

 

[JohnnyGui]

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 $dp$.

But isn't the density of states deduced from $dp$ through $\frac{dN}{dp}$ which maintains the inaccuracy further?

Also, probability calculations using formulas such as the Maxwell-Boltzmann distribution is based on the number of states within an infinitesimally small increment like $dv$.

 

[BvU]

Look at the numbers in your link, e.g. in example 2.3 and figure 2.4

 

[JohnnyGui]

Look at the numbers in your link, e.g. in example 2.3 and figure 2.4

Sorry, I'm not sure which link and example you're talking about. Can't find any example 2.3 or figure 2.4 in the link I gave in post #45.

 

[BvU]

Post #4, way back when. It works out your whole conundrum ...

 

[JohnnyGui]

Post #4, way back when. It works out your whole conundrum ...

Sorry for the very late reply. Just checked the link and noticed that the number of states density actually increases as energy $E$ increases. This would mean that the approximation of the number of states per $dE$ by using n-space geometry would be increasingly more accurate. The "least" relative accurate approximation of the number of states per $dE$ is at very low levels of energy $E$ since the quantum states are relatively low there.

Please correct me if I'm wrong.

Also, one other thing, I read that the numer of states density $N'$ in terms of energy (without the factor of 2 for 2 possible spins):
$$N' = \frac{V \cdot 4\pi \cdot \sqrt{2} \cdot m^{1.5}}{h^3} \cdot \sqrt{E}$$
I also read that the number of states density in terms of momentum is:
$$N' = \frac{V \cdot 4\pi \cdot p^2}{h^3}$$
I can't seem to derive them from one another.

 

[BvU]

From (2.4.3) with $p= \hbar k$: $$
N = 2 \times{1\over 8} \times \left(L\over \pi\right)^3 \times {4\over 3}\pi k^3 ={{4\over 3}\pi V p^3\over h^3 }\Rightarrow dN ={4\pi V p^2\over h^3 }\ dp
$$ With $\quad E= \displaystyle {{ p^2\over 2m} \Rightarrow dE = { p\over m} dp \Rightarrow { dp\over dE } = { m\over p} }\quad$ you use $\quad\displaystyle {{dN\over dE } = { dN\over dp} {dp\over dE}}\quad$ to get $$
dN ={4\pi V p^2\over h^3 }\ {m\over p}\ dE = {4\pi\; V \sqrt{2mE} \over h^3 } m \ dE $$

 

[JohnnyGui]

From (2.4.3) with $p= \hbar k$: $$
N = 2 \times{1\over 8} \times \left(L\over \pi\right)^3 \times {4\over 3}\pi k^3 ={{4\over 3}\pi V p^3\over h^3 }\Rightarrow dN ={4\pi V p^2\over h^3 }\ dp
$$ With $\quad E= \displaystyle {{ p^2\over 2m} \Rightarrow dE = { p\over m} dp \Rightarrow { dp\over dE } = { m\over p} }\quad$ you use $\quad\displaystyle {{dN\over dE } = { dN\over dp} {dp\over dE}}\quad$ to get $$
dN ={4\pi V p^2\over h^3 }\ {m\over p}\ dE = {4\pi\; V \sqrt{2mE} \over h^3 } m \ dE $$

Thanks a lot, I wrongly assumed it is merely done by writing $\sqrt{E}$ in terms of momentum thinking this would transform the derivative somehow in the number of states density per unit momentum.

Is my statement in my previous post before the question about the formula more or less correct?

 

[BvU]

I'd say yes.

 

[JohnnyGui]

I'd say yes.

Thanks for verifying. I noticed something peculiar that I hope you could help me with.

I know that for each increment $dp$, a shell containing a certain number of quantum states gets added to an 8th of a sphere in n-space, increasing its radius. I concluded that the radius of that 8th sphere in n-space in terms of momentum is:
$$R=\frac{p \cdot 2L}{h} = (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}$$
I want to calculate the thickness $ΔR$ of each n-shell that gets added to the 8th n-sphere when each increment $dp$ is added to a certain momentum $p$. According to the above formula, this should be:
$$ΔR = \frac{(p+dp) \cdot 2L}{h} - \frac{p \cdot 2L}{h} = \frac{p+dp}{p}\cdot (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}- (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}$$
According to this formula, each n-shell that gets added to the n-sphere decreases in thickness as each $dp$ is added to a larger value of $p$.
I find this very weird, because the derivative $\frac{dN}{dp}$ actually shows that the number of quantum states per $dp$ increases exponentially as $p$ gets larger. How can an exponentially increasing number of quantum states per $dp$ fit into an n-shell with decreasing thickness per $dp$? I'm aware that each added n-shell also increases in surface but that does not compensate enough for the decreasing thickness to make the number of quantum states in each n-shell get exponentially higher. For an exponentially increasing number of quantum states per $dp$, I would expect n-shells of at least a fixed thickness.

How is this possible? Is there something wrong in my calculation?

 

[BvU]

You already had $$ dN ={4\pi V p^2\over h^3 }\ dp $$ You know how to differentiate $$R = \frac{ 2Lp}{h}\Rightarrow dR = \frac{ 2L}{h} \;dp$$ the volume in there (between $R+dR$ and $R$ in n-space) increases with $R^2$ -- as discussed.

All clean and consistent. Why not move on to the next chapter ?