Determining the Number of Points in the n-Space

[Dario56]

TL;DR Summary

Determining the Number of Points in the n-Space for Certain Fermi Energy

Electron gas is a collection of non - interacting electrons. If these electrons are confined to certain volume (for example, cube of metal), their behavior can be described by the wavefunction which is a solution to the particle in a box problem in quantum mechanics. Allowed energy states for any electron is given by: $$ \epsilon = \frac {h^2} {8mL^2} ({n_x}^2 + {n_y}^2 + {n_z}^2) = \frac {h^2} {8mL^2} n^2 $$

where $n_x$, $n_y$ and $n_z$ are quantum numbers which determine allowed momentum components of the electron: $$p_x = \frac {hn_x}{2L}, p_y = \frac {hn_y}{2L}, p_z = \frac {hn_z}{2L}$$

Allowed states of the electrons can be visualized in the n-space, where the states can be shown graphically in the $n_x, n_y, n_z$ coordinate system. Every coordinate in the n-space actually represents two possible states for certain energy level because there are two electrons with opposite spin states for certain energy level. If we know maximum occupied energy level (Fermi energy), we can calculate maximum quantum number $n_{max}$ from the first equation.

To determine the number of points in the n-space for known Fermi energy, we can first count the ''volume'' of the one electron state in the n-space (particular $n_x, n_y, n_z$ coordinate). To do so, we note that the difference between two ''neighboring'' quantum numbers is equal to one (quantum numbers are positive integers) which means that the ''volume'' of any cube composed of neighboring points in the n-space is equal to 1. Each cube is enclosed by eight points in the n-space, but since any point is shared by the 8 neighboring cubes, there is actually only one point per each cube since only 1/8 of each point belongs to the individual cube. This means that the ''volume'' of the one electron state in the n-space is equal to 1.

When the ''volume'' per electron state in the n-space is known, we can calculate the total number of points in the n-space (for certain Fermi energy, $E_f$) by taking the total volume in the n-space and dividing it by the ''volume'' per electron state (which is equal to 1, so total number of points equals the total volume).

What I don't understand is that the total volume taken is equal to the 1/8 volume of the sphere with the ''radius'' equal to $n_{max}$: $$V_f = \frac {1} {8}\frac {4 \pi n_{max}^3} {3}$$ According to my understanding, it doesn't really make sense to take the sphere as the total volume because we determined the volume per state as being equal to the volume of the cube with sides having a length 1. Because of this, we should know the total volume of all the cubes for certain $E_f$ and since the volume per state is equal to 1, total number of points is equal to the total volume as mentioned previously.

If this is the case, why is volume of the sphere taken as the paramount when the total volume of the cubes should be taken?

 

[vanhees71]

This is because you take the volume as given by "rigid boundary conditions" of the single-electron wave function, i.e., your $n_x$, $n_y$, and $n_z$ are only the positive integers.

 

[Dario56]

This is because you take the volume as given by "rigid boundary conditions" of the single-electron wave function, i.e., your $n_x$, $n_y$, and $n_z$ are only the positive integers.

It is true that $n_{x}, n_{y}, n_{z}$ are positive integers and that because of it we should only look at the first octant. However, what does that have to do with my question?

Maybe there is a misunderstanding about what my question is, so I did some editing to make it more clear.

 

[vanhees71]

Then I don't understand your question. I thought, it's about why to use only the octant of the sphere. Of course the quantum-mechanically correct single-particle phase-space measure is
$$\frac{\mathrm{d}^3 x \mathrm{d}^3 p}{(2 \pi \hbar)^3}$$
when taking the thermodynamic limit, i.e., volume to infinity keeping the particle density fixed.

Strictly speaking the use of the rigid boundary conditions is a bit unclean from a mathematical point of view, because with these you don't have proper self-adjoint momentum operators. A better alternative is to use periodic boundary conditions. The final result, however, is the same in the thermodynamic limit. It's a good exercise to prove this!

 

[Dario56]

Then I don't understand your question. I thought, it's about why to use only the octant of the sphere. Of course the quantum-mechanically correct single-particle phase-space measure is
$$\frac{\mathrm{d}^3 x \mathrm{d}^3 p}{(2 \pi \hbar)^3}$$
when taking the thermodynamic limit, i.e., volume to infinity keeping the particle density fixed.

Strictly speaking the use of the rigid boundary conditions is a bit unclean from a mathematical point of view, because with these you don't have proper self-adjoint momentum operators. A better alternative is to use periodic boundary conditions. The final result, however, is the same in the thermodynamic limit. It's a good exercise to prove this!

Okay, so my question is essentially, how can we use 1/8 of sphere's volume as an total volume from which we calculate the number of points for certain $E_f$ if we defined volume per state as an CUBE with a side equal to 1? How can sum of the volume's of cubes be equal to the volume of the sphere?

According to my understanding, we should use the total volume of the cubes inside $E_f$ in the n-space not the sphere.

 

[vanhees71]

Ok, so I indeed misunderstood your question, and the counting in this case is indeed correct, because you ask for the number of microstates for a given Fermi energy $E_F$ (microcanonical ensemble at $T=0$) for an electron gas in a cavity using the rigid boundary conditions to define the volume.

Indeed one can show that the single-particle energy operator,
$$\hat{H}=-\frac{\hbar^2}{2m} \vec{\nabla}^2$$
is a proper self-adjoint operator on the Hilbert space of square-integrable functions on the cube $[0,L]^3$, and the eigenstates are labeled by three quantum numbers $n_1$, $n_2$, $n_3$ with $n_j \in \mathbb{N}$. The energy eigenfunctions are
$$u_{n_1,n_2,n_3}(\vec{x})=\sqrt{\frac{8}{L}} \sin(\pi n_x x/L) \sin(\pi n_y y/L) \sin(\pi n_z y/L).$$
There is no well-defined momentum observable for the electrons in this case!

So what you can count is the number of one-particle states with $E<E_f$, i.e., how many dots in the grid $(n_1,n_2,n_3) \in \mathbb{N}^3$. Since the energy eigenvalues are
$$E_{n_1,n_2,n_3}=\frac{\hbar^2 \pi^2}{2m L^2}(n_x^2+n_y^2+n_z^2)$$
you have
$$n_x^2+n_y^2+n_z^2 < \frac{2m L^2 E_F}{\hbar^2 \pi^2}.$$
This means the said number is roughly given by the volume of the octand of the sphere with the corresponding radius,
$$N=\frac{g}{8} \frac{4 \pi}{3} \left (\frac{2m L^2 E_F}{\hbar^2 \pi^2} \right)^{3/2},$$
where $g=2s+1$ is the spin degeneracy factor (for electrons $s=1/2$, i.e., g=2).

 

[Dario56]

Ok, so I indeed misunderstood your question, and the counting in this case is indeed correct, because you ask for the number of microstates for a given Fermi energy $E_F$ (microcanonical ensemble at $T=0$) for an electron gas in a cavity using the rigid boundary conditions to define the volume.

Indeed one can show that the single-particle energy operator,
$$\hat{H}=-\frac{\hbar^2}{2m} \vec{\nabla}^2$$
is a proper self-adjoint operator on the Hilbert space of square-integrable functions on the cube $[0,L]^3$, and the eigenstates are labeled by three quantum numbers $n_1$, $n_2$, $n_3$ with $n_j \in \mathbb{N}$. The energy eigenfunctions are
$$u_{n_1,n_2,n_3}(\vec{x})=\sqrt{\frac{8}{L}} \sin(\pi n_x x/L) \sin(\pi n_y y/L) \sin(\pi n_z y/L).$$
There is no well-defined momentum observable for the electrons in this case!

So what you can count is the number of one-particle states with $E<E_f$, i.e., how many dots in the grid $(n_1,n_2,n_3) \in \mathbb{N}^3$. Since the energy eigenvalues are
$$E_{n_1,n_2,n_3}=\frac{\hbar^2 \pi^2}{2m L^2}(n_x^2+n_y^2+n_z^2)$$
you have
$$n_x^2+n_y^2+n_z^2 < \frac{2m L^2 E_F}{\hbar^2 \pi^2}.$$
This means the said number is roughly given by the volume of the octand of the sphere with the corresponding radius,
$$N=\frac{g}{8} \frac{4 \pi}{3} \left (\frac{2m L^2 E_F}{\hbar^2 \pi^2} \right)^{3/2},$$
where $g=2s+1$ is the spin degeneracy factor (for electrons $s=1/2$, i.e., g=2).

Thank you for the detailed answer. What you derived in the end is the number of electron states if we know $E_f$.

I am familiar with this formula and your derivation is clear. However, it didn't answer my question.

This means the said number is roughly given by the volume of the octand of the sphere with the corresponding radius

As I said in the main post, why is that the sphere defines the number of states when we defined volume per state as an CUBE with a side equal to 1? How can sum of the volumes of cubes be equal to the volume of the sphere?

 

[vanhees71]

But I derived it in my posting:

$$n_x^2 + n_y^2 + n_z^2 \leq \frac{2m L^2 E_F}{\hbar^2 \pi^2}.$$
This is the description of a sphere (ball) with the right-hand side being the square of its radius, if the $n_x$, $n_y$, and $n_z$ were real numbers. Now first of all it's restricted by the demand that all the $n_j>0$, i.e., that restrict it to only the corresponding octant of the sphere. If the radius is not too small, you can just take the volume of this sphere to count the number of points $(n_x,n_y,n_z)$ within this octant of the sphere to a good approximation.

 

[Dario56]

But I derived it in my posting:

$$n_x^2 + n_y^2 + n_z^2 \leq \frac{2m L^2 E_F}{\hbar^2 \pi^2}.$$
This is the description of a sphere (ball) with the right-hand side being the square of its radius, if the $n_x$, $n_y$, and $n_z$ were real numbers. Now first of all it's restricted by the demand that all the $n_j>0$, i.e., that restrict it to only the corresponding octant of the sphere. If the radius is not too small, you can just take the volume of this sphere to count the number of points $(n_x,n_y,n_z)$ within this octant of the sphere to a good approximation.

I agree that equation mentioned is the equation of the ball with the radius equal to the expression on the right-hand side, but that isn't what this question is about since we defined the volume per state as an cube with the side length equal to 1. How can sum of the volumes of the cube be equal to the volume of the ball?

As far as I understand, idea of counting all points defined by $E_f$ is to take the whole volume in the n-space defined by the $E_f$ and to divide this volume by the volume per point. If we defined volume per point as an cube, how can we use volume of the ball as an total volume since volume of the ball isn't a multiple of volume per point (since we defined volume per point to be a cube and sum cube volumes can never be equal to to the volume of the ball)?

If the radius is not too small, you can just take the volume of this sphere to count the number of points (nx,ny,nz) within this octant of the sphere to a good approximation.

Maybe this part of your answer is a key to my question. Volume of the ball is just an approximation which is valid if $n$ is big. However, I am not sure why does this approximation hold?

 

[vanhees71]

It's of course an approximation. However, the usual order of magnitude of the largest $n_j$ for a given macroscopic amount of gas is huge. So you can approximate the sum of the volumes of the cubes (with volume 1) inside the octant of the sphere by the corresponding integral, which is of course the volume of this octant of the sphere.

 

[Dario56]

It's of course an approximation. However, the usual order of magnitude of the largest $n_j$ for a given macroscopic amount of gas is huge. So you can approximate the sum of the volumes of the cubes (with volume 1) inside the octant of the sphere by the corresponding integral, which is of course the volume of this octant of the sphere.

This now boils down to a math problem.

Important question is what is in fact the exact volume we are approximating with the sphere? It must be a multiple of the cube volume, but I am really not sure how does such object look in the n-space. Do you know?

So, you say because $n_{max}$ is huge, sum of volumes of all the cubes (volume of the concerned object) is basically the same as the volume of the octant of the sphere with the radius of $n_{max}$? I see you mention some integral used to approximate concerned volume. Can you write which integral are you referring to?

 

[Dario56]

@vanhees71 I tagged you just in case you weren't notified with my answer because I noticed that when it takes me more time to answer to the thread that often I don't get any answers back. It seems that notification doesn't arrive in that case.

 

[PeterDonis]

we defined the volume per state as an cube with the side length equal to 1.

Where is this defined?

 

[Dario56]

Where is this defined?

I explained how is this defined in the main post. There is one point per volume of the cube of the unit volume in the n-space. We need this because we determine the number of the points by the total volume in the n-space.

 

[hutchphd]

So using these units, what is the radius of a "typical" Fermi sphere? The fractional error in volume estimate making the contimuum approximation is de minimis.

 

[PeterDonis]

There is one point per volume of the cube of the unit volume in the n-space.

Ok. Now: Which cubes are included in the total sum you are computing?

 

[Dario56]

It's of course an approximation. However, the usual order of magnitude of the largest $n_j$ for a given macroscopic amount of gas is huge. So you can approximate the sum of the volumes of the cubes (with volume 1) inside the octant of the sphere by the corresponding integral, which is of course the volume of this octant of the sphere.

I must say that know this answer has sense because in fact we previously solved it (in the deleted post) when we discussed calculation of the internal energy of Fermi gas.

Sum of the electron energy over all quantum numbers can adequately be approximated by the integral as the difference between two consecutive quantum numbers (which is equal to 1) is very small compared to the total number of the electron states. This is the same as in the numerical integration, where more correct results are obtained when more points are taken.

By the same reasoning, total volume of the Fermi sphere must approximately be equal to the sum of the unit volume cubes enclosed by the sphere since volume of each cube is very small compared to the total volume of the sphere. It is like the numerical integration where volume element is finite, but very small.