What is the significance of k-values in a 2D quantum well system?

[Janeabc]

It is a classic 2DE quantum well problem. I do not understand few things. I haven't seen the solution anywhere on the new, so this is why i ask here.

It is a quantum well - Lx is it's dimension along x-axis Ly is it's dimension along the y-axis.There is a potential as in picture let's call it Uo.

I have found from post here that the k-value is quantized. If i understood it correctly, the Kz is quantized. I am not sure what does this means. Can actually the particle in the quantum well even move along the z-axis?

What are possible values of wave-vector Kz ?As i seen in my lecture there are bound condition - so as far as i understand, Kx and Ky are quantized as well.
Kx*Lx = 2*n*Pi
Ky*Ly = 2*n*Pi

Does it change something, that in the lecture is described not single quantum well, but a periodic potential - like in a GaAs based hetero-structure.

 

[Cthugha]

The picture and your description of it are somewhat misleading. Let us see whether we can clarify.

It is a quantum well - Lx is it's dimension along x-axis Ly is it's dimension along the y-axis.There is a potential as in picture let's call it Uo.

The term quantum well is pretty much reserved for structures which have confinement in exactly one direction and no confinement in other directions. The k-value is then quantized in the direction of the confinement.

Looking at your picture, it definitely does not show a quantum well. The picture is also somewhat misleading. On the left and right it says "bulk", but you see what looks like a surface. Either this is really a surface (not really probable) or that is just a means to better illustrate where the bulk lies on the x- and y-axes and the bulk goes on in z-direction. In that case the structure shown is rather a quantum wire with confinement in x- and y- direction (and correspondingly quantized k-values in x- and y-direction), but not in z-direction.

If it is supposed to picture confinement in all three directions, you have a quantum, dot or quantum box with confinement and quantized k-values in all three dimensions.

I have found from post here that the k-value is quantized. If i understood it correctly, the Kz is quantized. I am not sure what does this means. Can actually the particle in the quantum well even move along the z-axis?

What are possible values of wave-vector Kz ?


As i seen in my lecture there are bound condition - so as far as i understand, Kx and Ky are quantized as well.
Kx*Lx = 2*n*Pi
Ky*Ly = 2*n*Pi

As said above it does not look like you have confinement in z-direction. However, generally speaking you will always get quantized values that look like Kx*Lx = 2*n*Pi in every direction where confinement is present and a continuum of possible values in directions without confinement.

Does it change something, that in the lecture is described not single quantum well, but a periodic potential - like in a GaAs based hetero-structure.

A single quantum well is already one example for a GaAs based heterostructure, but not for a periodic potential. Whether changes occur when introducing periodic potentials depends on what setting you have in mind exactly. For example one possible scheme for having periodic potentials lies in placing several quantum wells in some fixed distance to each other, so that you have bulk-QW-bulk-QW-bulk-QW-bulk...

In that case you can get for example tunneling of particles from one quantum well to the next that depends on the distance between quantum wells.

 

[Janeabc]

Thanks very much for answer . I uploaded another (i hope) better picture. It is a AlGaAs-GaAs-AlGaAs hetero-structure.

So K-z is quantified in this case, K-x and K-y have continual values. So the electron is free to move along x-axis and y-axis (using conduction level), but when he decided to move along the z-axis he bumps into potential barrier, so it is confined along z-axis.


On previous as well on this picture Lx and Lx are both infinite (or equal to the dimension of material).As electron can move freely along x-axis,

But another problem i have had is density of states in a quantum well. If K-x and K-y are both continual values the the density of states is infinite. I have read this in the post of yours.

In 2D one of the k-values is quantized. The k-space volume will be a circle around that fixed k-value. All states with the same energy will lie on the circumference of that circle. The density of states will be proportional to 2 Pi k.

 

[Cthugha]

But another problem i have had is density of states in a quantum well. If K-x and K-y are both continual values the the density of states is infinite. I have read this in the post of yours.

The value of the density of states given in that post is the density of states in momentum-space. Basically you look at a fixed kz and have a look at the energy of the states as a function of kx and ky. You will find that states with equal energy will then lie on the circumference of a circle around kx=ky=0 with kx^2+ky^2=constant. Obviously you can create many of these circles and each will correspond to a different energy. If you go to very large values of kx^2+ky^2 the circle will get rather large and the number of states will tend to infinity as kx^2+ky^2 tends towards infinity. However that situation corresponds to states with large energy. In any realistic setting only states with energy smaller than the confinement potential you have make sense physically. That density of states is obviously always finite.

Also, I would like to stress that the density of states given in that post was the one in momentum space. Typically more interesting is the density of states in energy.
Starting from what is given above, the density of states in k-space can be expressed as

$$g(k)dk=2\frac{2\pi\left|k\right|}{V_{2D}}dk$$

where V2d gives the total volume of momentum space (just (2 pi/L)^2).

To get to the more common representation of density of states in energy space you need to take the typical relation for energy:
$$E(k)\frac{\hbar^2 k^2}{2 m}$$
and solve for k:
$$k(E)\frac{\sqrt{2 m E}}{\hbar}.$$

You also need to replace dk in the density of states by dE:
$$\frac{dk(E)}{dE}=\frac{2 m}{2\hbar \sqrt{2 m E}}.$$

Plugging all that into your equation for the density of states and inserting the volume in k-space, you get:

$$g(E) dE=\frac{m}{\pi \hbar^2}dE.$$

Which is the famous result that the density of states does not depend on the energy for a single quantum well energy level (=single value of kz). As you have several energy levels in a QW, you will then get the famous picture of a constant DOS that increases steplike at several certain energies when a new energy level comes into play due to the next kz coming into play.

 

[Janeabc]

I got everything except V2d=(2*Pi/L)*2. What is L.

It only make sense if L is a x -dimension of the material, but if so, the states in x and y direction are quantized as well, Just Lx and Ly are much larger that Lz. So z quantization is the most important.

 

[Cthugha]

I got everything except V2d=(2*Pi/L)*2. What is L.

It only make sense if L is a x -dimension of the material, but if so, the states in x and y direction are quantized as well, Just Lx and Ly are much larger that Lz. So z quantization is the most important.

Oh, sorry. The L indeed refers to the real space extent of the solid in the direction without quantization. The total volume of your solid obviously determines the number of particles you have and therefore also has an influence on the density of states. However, you can of course also calculate a normalized DOS per unit volume.