Volume in K space occupied per allowed state

[amjad-sh]

An allowed state of a molecule in a gas that is in a box of length L can be represented by a point in 3 dimensional K-space, and these points are uniformly distributed.In each direction points are separated by a distance π/L. A single point in K-space occupies a volume (π/L)^3.

The number of allowed states with wave vector whose magnitude lies between k and k+dk is described by the function g(k)dk, where g(k) is the density of states.This number is then given by:
g(k)dk= (volume of k-space of one octant of a spherical shell)/ (volume in k-space occupied per allowed state)

Then g(k)dk=\frac{1/8\times 4πk^{2}dk} {2π^2}

My question is : Can we say that dk=(π/L)^3 ?

 

[EmilyRuck]

An allowed state of a molecule in a gas that is in a box of length L can be represented by a point in 3 dimensional K-space, and these points are uniformly distributed.In each direction points are separated by a distance π/L. A single point in K-space occupies a volume (π/L)^3.

Ok, and so first of all you can say now that the density of the allowed states in the K-space is

\displaystyle \frac{1}{\left( \displaystyle \frac{\pi}{L} \right)^3}

The number of allowed states with wave vector whose magnitude lies between k and k+dk is described by the function g(k)dk, where g(k) is the density of states.This number is then given by:
g(k)dk= (volume of k-space of one octant of a spherical shell)/ (volume in k-space occupied per allowed state)

that is (volume of k-space of one octant of a spherical shell)*(density of allowed states in K-space.)

My question is : Can we say that dk=(π/L)^3 ?

No, absolutely, for several reasons:
- dk is an infinitesimal, while (π/L)^3 is a finite quantity, if L is finite;
- this computation is made to find the number of states contained in the spherical shell (considering just one octant) when k is incremented by the infinitesimal quantity dk. What you obtain, g(k)dk, is not a density: it is instead a number of states. Every time you increment k by a quantity dk, you have to consider g(k)dk more states.
k is the modulus of the wavevector \mathbf{k}, that is | \mathbf{k}| = k. So, if you want to know the number of allowed states between two values of k you can integrate

N_a = \displaystyle \int_{k_1}^{k_2} g(k) dk

(g(k)dk is the number of states in a shell of infinitesimal thickness dk; N_a is instead the number of states in a shell of thickness k_2 - k_1).

Nonetheless, usually you want to integrate not in the K-space, but with respect to energy E. There is a relation between energy and k, so that E = E(k). If you want to know the number N_b of states between two energy values E_1 and E_2, you can integrate the density of states in dE. You first need to know g(k)dk (and it is the reason why you computed this above); then, you have to express it in terms of E instead of k (and usually you can, if you know the relation between E and k). You will find h(E)dE:

N_b = \displaystyle \int_{E_1}^{E_2} h(E) dE

and you have N_b. So, g(k)dk is just a step to find the density of states with respect to energy E and to obtain h(E)dE. dk is simply an infinitesimal increment of the k variable, which will be related to dE (used for integration in the E variable).
Hope it will help.

 

[amjad-sh]

Thanks a lot for your answer.
but even if (π/L) is not infinitesimal, isn't it what separates between the k's in the k-space? I mean the difference between a k-state and another k-state can't be less than (π/L). So if I do not consider it as infinitesimal, the dk will not make sense then, isn't this right?

 

[EmilyRuck]

Thanks a lot for your answer.

You're welcome :).

but even if (π/L) is not infinitesimal, isn't it what separates between the k's in the k-space?

Yes, it is. If \mathbf{k} is a vector with components k_x, k_y, k_z, the minimum spacing between two adjacent values of k_x (or k_y, or k_z) is \pi / L. So, each of the acceptable values of \mathbf{k} result to be "alone" inside a volume (\pi / L)^3 (and then you obtain the density of states in K-space as the reciprocal of this volume).

I mean the difference between a k-state and another k-state can't be less than (π/L). So if I do not consider it as infinitesimal, the dk will not make sense then, isn't this right?

I can understand your doubt. Unfortunately, not every mathematical operator has a physical meaning which is strictly related to your problem. Here, dk has above all a mathematical meaning: if you have to integrate in the k variable (and you must do it, before handling energy), the integral needs a differential, which is dk.
Roughly speaking, every integral must in fact be expressed as

\displaystyle \int_a^b f(x) dx

where f(x) is the function you want to integrate and dx is the differential of the variable x. Take a look at Calculus notes about integration in one variable and differentials, to be more confident with this.

As we said before, k = |\mathbf{k}| and the physical meaning of dk is an infinitesimal increment in |\mathbf{k}|. Evaluating the quantity g(k)dk means to count "how may states are contained between a sphere in the K-space of radius k and a sphere of radius k + dk" (that is: the first sphere, whose radius is being incremented of a quantity dk). Mathematically speaking, dk is an infinitesimal quantity and is needed to correctly write the integral.

 

[amjad-sh]

the physical meaning of dkdkdk is an infinitesimal increment in |k||k||\mathbf{k}|.

Here where I get confused, as dk means nothing physically.

 

[EmilyRuck]

Here where I get confused, as dk means nothing physically.

1) If you read again my answer, I didn't say that dk has not a physical meaning. I said instead that dk is not related to the spacing between the states
Be careful: the \mathbf{k}-space is a vector space. Your variable \mathbf{k} = k_x \mathbf{x} + k_y \mathbf{y} + k_z \mathbf{z}, is in fact a vector. It is defined by its components along the \mathbf{x}, \mathbf{y} and \mathbf{z} axes. The space between the allowed states is \pi / L for each component of \mathbf{k}, as you stated in your first post. So, the density of the allowed states in this vector space is (L / \pi)^3.

2) When you deal with g(k), you are no more dealing with the vector \mathbf{k}, but just with its modulus: the real number k = | \mathbf{k} | = \sqrt{k_x^2 + k_y^2 + k_z^2}. You take a sphere of radius
k = | \mathbf{k} | = \sqrt{k_x^2 + k_y^2 + k_z^2}, then you take a sphere of radius k + dk = | \mathbf{k} | = \sqrt{k_x^2 + k_y^2 + k_z^2} + dk and you want to count how many states of the vector space are present in the shell between the two spheres. And correctly you make the multiplication of your first post. Now you are only interested in the radii k and k + dk of the spheres, and you are no more interested in the \mathbf{k} vectors.
It is convenient for you to say that the number of the states in the shell is not just a number n, but a quantity g(k)dk, that is: (\mathrm{density\ per\ unit\ length}) \times \mathrm{length}.
The density per unit length is g(k), density per unit length of k: note that this variable k is new for you. It is the modulus of the vectors \mathbf{k}, not the vector \mathbf{k} itself. The length is dk: it is a linear length, because k is a real variable and it can be represented as an horizontal axis, just like x. You were dealing with spheres, but because you are only interested in their radius, you are basically dealing with a real number. You now can forget the spheres: you simply have a function g(k) to integrate, just like in Calculus you had a function f(x).
Why not considering a finite length \Delta k? You can. No problem. But this in general doesn't allow you to solve an integral:

\displaystyle \int_{k = k_1}^{k = k_2} g(k) \Delta k

is not correct. In order to solve an integral (and you will have to solve it, when dealing with energy) you need an infinitesimal length dk:

\displaystyle \int_{k = k_1}^{k = k_2} g(k) dk

is correct and practicable.

So, the physical meaning of dk is the following. The states in \mathbf{k}-space are fixed and you know them pretty well. In order to solve this integral, now you must wonder: how many states are enclosed between two spherical shells whose radii are respectively k and k + dk? The outer sphere must have a radius infinitesimally longer than the first, in order to write g(k) dk. That is: how many states I enclose if I consider a real (not vector) variable k and I increase it by an infinitesimal quantity? This is how to proceed in Calculus courses when you deal with derivatives and integrals.

So, if you hoped dk has a practical, evident meaning, the answer is no: it doesn't. But you can draw two concentric spheres in the \mathbf{k} space, reduce the outer one to almost overlap the inner one, and measure the distance between them. dk is roughly that distance.

You are interested in k and not in \mathbf{k} because energy is a function of k only (k^2 to be more clear). Now you can write k as a function of E and dk in terms of k and dE and solve the integral, to find the density of states with respect to energy.

 

[amjad-sh]

Thanks a lot!
g(k)dk gives the number of states between k and k+dk, but this number is not necessarily to be really exist, the real number of states can be obtained by integrating g(k)dk from K1(which is the magnitude of the first k state) to K2 (which is the magnitude of the next available K-state).

 

[EmilyRuck]

Thanks a lot!

g(k)dk gives the number of states between k and k+dk, but this number is not necessarily to be really exist

Yes, exactly!

the real number of states can be obtained by integrating g(k)dk from K1(which is the magnitude of the first k state) to K2 (which is the magnitude of the next available K-state).

I understand what you are meaning and yes, it is correct. Obviously, very often you are not interested in counting just the number of states between two available modulus values k_1 and k_2: instead, you will be interested in counting the total number of states, so between the smallest value of modulus and the largest value of modulus in your k-space. But your way of thinking is correct and yes, this is the way the integral let's you count the states!