# K-space and Brillouin zone

1. Oct 18, 2008

### carbon9

Hi,

I'm following the book of Kaxiras on solid state physics and I'm a bit confused about Brillouin zone and solving Schroedinger equation in BZ. Please let me to write the logical statements I've understood (maybe correct ot not):

1. Crystals are made up of atoms located periodically in 3-D. This space is real space.

2. There is an abstract space called reciprocal space whose lattice vectors are defined in terms of real space vectors. The points in reciprocal space are k-vectors. k-vectors represent momenta of electrons.

3. In periodic potential, electron wavefunctions can be expressed as plane waves as:

http://img145.imageshack.us/img145/5051/94472449is2.th.jpg [Broken]http://g.imageshack.us/thpix.php [Broken]

4. The Schroedinger equation for each electron for each k is given as:

http://img379.imageshack.us/img379/7445/50770285gk9.th.jpg [Broken]http://g.imageshack.us/thpix.php [Broken]

It is said that "Solving this last equation determines uk(r), which with the factor exp(i.k.r) makes up the solution to the original single particle equation."

First question is: After we have got the solution of the equation, we get a wavefunction Psi k (r) where k is a subscript. Does this mean that: "The probability of finding an electron at r which has a momentum of k is |Psi k (r)|^2"?

The second question is: Do we have to solve the above equation for a set of k-vectors in the first Brillouin zone and then will we sum all the wavefunctions to get the actual wavefunction in the real space?

Regards,

Last edited by a moderator: May 3, 2017
2. Oct 20, 2008

### Cthugha

Well, almost. It means more or less, that if you have a look at an electron with (pseudo)momentum k, the probability to find it at r is given as you described it.

It is not a distribution, which describes the probability, that some randomly picked electron will have (pseudo)momentum k and be at the position r.

What do you mean exactly by mentioning "the actual wavefunction in real space"? The wavefunction you described is already the actual wavefunction for a single electron with (pseudo)momentum k in the used approximation. If you instead want to know the wavefunction for n electrons, it gets a bit more complicated as the wave function needs to be antisymmetric. A first approach to this problem is the usage of Slater determinants. However, this method does not work in any possible situation, for example there are problems in systems with electronic correlations.

3. Oct 20, 2008

### carbon9

Hi Cthugha,

Thanks for your answers. I tried to mean Psi (r) (without subscript k) to denote the function that represents all possible values of k in Brillouin zone.

In this concept, please let me to as again "he second question is: Do we have to solve the above equation for a set of k-vectors in the first Brillouin zone and then will we sum all the wavefunctions to get the actual wavefunction in the real space?"

Regards,

4. Oct 20, 2008

### Cthugha

Well, to be honest, I am still a bit puzzled and it is still unclear to me what "the actual wavefunction in real space" might mean.

At first: The function you mentioned is the actual wavefunction for a single electron with wave vector k in real space.

Maybe you think you need a wave function, which is independent of k, so let me address this topic:

Have a look at the simple case of a 3d potential well of infinite height. Here the wave function for some given quantum numbers will look like this:

\psi_{n_x,n_y,n_z}(x,y,z) \prop sin(k_x x) sin(k_y y) sin(k_z z)

with

k_x=\frac{n_x \pi}{a_x}

and so on for y and z.

Now you can easily see, that the choice of of a certain wave vector k equals choosing certain quantum numbers n_x, n_y and n_z. In solids it is roughly speaking the same. So just as you do not sum over wave functions for different quantum numbers in potential well problems or do not sum over wave functions for different orbitals in atomic problems to get just one wave function, you do not need to sum over wave functions for different wave vectors to get one wave function in solids. You will just get a set of eigenfunctions for different quantum numbers.

edit: Tex does not seem to work at the moment, so I removed the [/tex] and [tex] tags and hope you can read the math anyway.

Last edited: Oct 20, 2008
5. Oct 21, 2008

### carbon9

Hi again Cthugha,

Thank you for the answers. I could now understood the facts clearer I think.

Finally, we have a set of eigenfunctions for different k's. How can we go to the electron density function (rho(r)) that is dependent of r now? Can we write it as

rho(r)=sum(up to n) [sum (for all k in BZ) {|Psi n k (r)|^2}]

where Psi n k (r) is the nth solution to eigenfunction problem with momentum k?

Is it right?

Regards,

6. Oct 21, 2008

### Cthugha

The approach you used has been used in some first approaches to theoretical solid state physics, but it has some drawbacks, where its predictions do not meet reality. Summing up all the probability densities for n electrons means more or less, that all electrons are independent of each other and the wave function for the n-electron system is just the product of the single wave functions. However you will notice, that this simple product will give a symmetric wavefunction. As electrons are fermions, you will need an antisymmetric wavefunction instead. So the question is, how to antisymmetrize the wave function. The straight forward method is to take all wave functions and write them into a (n*n)-matrix:

\psi_1(r_1) \psi_2(r_1)....\psi_n(r_1)

and so on on the horizontal lines and

\psi_1(r_1) \psi_1(r_2).... \psi_1(r_n)

and so on on the vertical lines. The determinant of this matrix, the so called Slater determinant will give an antisymmetrized wave function. However, this is almost uncomputable in any real solid state system due to n being extremely huge.

A more modern approach (density functional theory) was developed in the sixties (and got a Nobel prize in 1998). According to the Hohenberg-Kohn-theorem the ground state of an system of N electrons has an electron density N(r), which is unambiguous.

Now the trick is to use the Born-Oppenheimer approximation, which just treats the electrons quantum mechanically and considers the atoms as static, which gives an effective potential for the electrons to move in. Now one uses an ansatz of n single electron wave functions, which are n solutions of the Schrödinger equation in such an effective potential (so now n single functions and not a slater determinant - pretty much easier). Now one gets some electron density. The interesting point is that the effective potential itself depends on the electron density. It has in principle three parts: one for the interaction of the electrons with the atoms, one for the electrostatic electron-electron interaction and one exchange parts, which takes the n-body effects into account. As the effective potential depends on the electron density, the calculated electron density changes the effective potential used before, so the Schrödinger equation changes as well. Now one starts again with this changed Schrödinger equation and solves it again, gets a new electron density, which gives a new effective potential and so on. Hopefully, at some point, there will be a stable and self consistent solution, where the potential and the density do not change anymore. Then you have the electron density (and ground state energy).

So sorry to say that, but: getting correct analytical expressions, which can be calculated is almost impossible in solid state physics, even for problems, which look simple at first.

7. Oct 21, 2008

### carbon9

Hi Cthugha,

Again Thank you for the answers.

I understood somethings I think but is there a way to get electron density function from wavefunctions in density functional theory?

Regards,