PHYS 7440 Solutions 1
Folks seemed to do just fine on this set. Here are my solutions:
1. Marder 6.1
Product wavefunctions are great when they work. In this case, you are essentially asked to show that in the special case where the total hamiltonian is a sum of single particle hamiltonians:
and
that the total wavefunction is just a product of single particle wavefunctions i.e.,
where
and that this total wavefunction solves the total hamiltonian with an energy eigenvalue that is just the sum of single particle energys:
Really, you are being asked to show that separation of variables will solve this type of problem.
In many ways, you just need to plug in the product wavefunction and chug out the results for operating on it with the total hamiltonian. It's just some book keeping:
Think about the operation of one of the single-particle hamiltonians on the total wavefunction. Since the hamiltonian is a function of only one of the coordinates, it one operates on that particular single particle wavefunction in the product. Thus,
Therefore, we find:
Plug and chug.
In the second part of the problem, you are asked to consider the antisymmetrized sum of product wavefunctions:
This wavefunction represents an extension of the product wavefunction idea, to allow consistency with the Pauli exclusion principle. In other words, a simple product wavefunction is NOT antisymmetric when you exchange coordinates of any two electrons. However, the 'anti-symmetrized' sum of product wavefunctions IS!! For example, suppose that you have two electrons (1 and 2) and two single particle wavefunctions (say i and j). Then, a simple product wavefunction would be:
While this product wavefunction does satisfy the two-particle hamiltonian, it is not antisymmetric with respect to exchange as required by the Pauli exclusion principle i.e., any wavefunction for a pair of identical fermions must satisfy:
An example of an extension to the simple product, which DOES satisfy the antisymmetry is:
Each term is a product wavefunction (and therefore satisfies the Schrodinger equation) AND each term results in the same eigenenergy.
John Slater first pointed out the extension of this two-particle case to the general situation where you have N particles, N single-particle wavefunctions, and you want to produce a sum of product wavefunctions that satisfies the antisymmetry condition. In his recipe, you simply form the so-called 'Slater determinant': 1) Let the columns in some square matrix refer to the electron coordinate. 2) Let the rows refer to the different single particle functions. 3) Take the determinant to find the product wavefunction sum. 4) Multiply by a factor of to take into account the N! different terms (of equal energy) that such a determinant will generate.
Here is the case of three electrons in three distinct single particle functions:
Each of these terms yields the same energy because each of the single particle hamiltonians has the same single particle potential. Marder's (6.80) is the algebraic way of writing out the determinant.
2. Marder 6.2
Let's assume that we are looking at the square box problem, so that the single particle energies are:
In building the ground state energy, we just need to ask for the lowest energy single particle states available at each stage of filling and keep adding electrons until we are forced to move to higher energy states.
For the first two electrons, we can always go into the zero energy state. Next, there are six different quantum numbers that all lead to the same single particle energy:
Each of these states is doubly degenerate due to spin. Thus, we can fit 12 electrons into this set of energy levels. Along with the lowest energy, that's a total of 14 electrons. The 15th electron must go into a state with two of the quantum numbers non-zero. In a notation where we list the values of the x- y- and z-quantum numbers, we have:
/ Degeneracy / Energy(000) / 2 / 0
(100),(010),(001) / 2 each for 12 total /
(110),(101),(011) / 2 each for 24 total /
There is an obvious or natural energy in this problem, namely the energy of the first excited state. In terms of this energy, we can produce a plot of the total energy. Here is some Mathematica code for plotting:
Here is a table with the number of electrons added to the box, and the associated energy in units of the first excited state energy:
Here's a list plot of the energies:
3. Marder 6.4
The density of states is created in each case to allow us to convert summations to integrations. To convert a k-space summation to a k-space integration, you need to k-space density of states. To convert the k-space summation to an energy integration, you need the energy density of states. In each case, the issue is one of the counting of states. For 2D, just take the area around a given quantum state:
Thus, per unit area, we have a k-space density of states (including the spin degeneracy) of:
Therefore, we find that any k-space summation (normalized by the total system area, A) in a two-dimensional problem becomes:
If you would prefer to evaluate an integral over energy, then we need to find an energy density of states so that the total number of states in some energy range is related to the same total number of states in some k-vector range:
Therefore, using the 2D polar coordiates,
In other words, a two-dimensional system has a constant energy density of states. This result is to be compared to the 3D case, where we find a square root of energy increase in the density of states. This result can also be written in a more physically transparent way, by recalling that the units of energy density of states in 2D should be number of states per unit energy and per unit area. We can get the energy units by introducing two factors of the 2D Fermi wavevector:
Just as we did for the volume of the Fermi sphere in 3D, for a 2D system, we equate the area of the Fermi disk to the total number of electrons in the system:
Thus, the energy density of states in 2D can be more transparently written as an obvious number per unit area, per unit energy:
Very nice!
The arguments for 1D systems are similar. We find:
4. Marder 6.5
OK, you are asked to consider a very thin film of silver (Ag) that is a square patch of 100 microns on a side in the x- and y- directions, but only 4.1 angstroms (0.41nm) thick in the z-direction. Physically, this film would have a single layer of silver atoms. That's pretty thin! You are asked to consider a free electron gas model for the silver, to assume hard wall boundary conditions in the z-direction, and determine the depth of the Fermi sea of electrons i.e., find the energy difference between the lowest single particle electronic level and the highest occupied single particle level in the ground state.
I will assume that the coordinate system has one surface of the film at z=0 and the other at z=0.41nm. Then, the hard wall boundary condition forces us to use sine wave solutions. Assuming that we continue to use periodic boundary conditions in the x- and y-directions, we find that the single particle wavefunctions are:
The associated single particle eigenenergies are as usual:
The boundary conditions restrict the various k-components to be:
Therefore, due to the different boundary condition in the z-direction, the k-space of allowed quantum states is only a half space of positive z-components. Note also that the zero energy state is no longer available. The quantization condition in the z-direction eliminates the strictly zero energy states. However, in this new k-space, we still allow two electrons per quantum state and we can still define a typical volume around each state that is equal to :
This k-space density is very similar to the usual case considered in the text.
Finally, notice that the huge difference in x- and y-direction lengths compared to the z-direction. It is really useful to think about the k-space as being a set of planes, each parallel to the x- and y-components of the k-vector, where the quantum states are so closely spaced that it's reasonable to treat them as continuously distributed, while still thinking of the z-direction as discrete.
The remaining question is to determine the energy of the highest occupied single particle level compared to the lowest energy. To find this energy, we need to fill energy levels until we have created a charge neutral system. In the case of silver, there is one electron per atom free to move, and a typical volume density of (see Marder Table 6.1) 5.86x1022 atoms/cm3. For the sake of argument, let's assume that we can fit all of these electrons into k-space by using only the first plane of x- and y-components of the k-vector. If we're wrong, then we'll find out quickly.
In this case, we have effectively a two-dimensional problem. Then, assuming that all the electrons fit into states on the first plane of x- and y-states, we would have a Fermi wavevector (see the 2D problem above) of:
This wavevector tells us the radius of the 2D Fermi disk of occupied states. We would also have a Fermi energy given by:
It is clear that we will only have a Fermi disk in the first plane if the Fermi wavevector is short enough so that there is no need to use states in the next higher energy plane. In other words, only if:
We can use the actual numbers to test this inequality. We have:
Compare this number to the associated inverse film thickness:
Score!!
It turns out that the Fermi disk will indeed be small enough that we will not need to move to the second layer of states! Immediately, because we already know the relationship between the density of electrons and the 2D Fermi wavevector, we can write down the Fermi energy:
Therefore, the energy between the lowest state and the highest state is:
The bulk, 3D Fermi energy from Table 6.1 is 5.50eV. Thus, you can see that by reducing one of the dimensions to atomic scales, you squeeze the electronic energies up a bit. However, notice that even this extreme case shifts the important energy scale by only a small amount. Therefore, the idea that many ground state properties are insensitive to details of sample dimensions and boundary conditions is well supported.
b) The nastier case where the film is 0.82nm thick.
The same arguments as above hold, but now, due to the thicker film, it's clear that you're going to need to populate more than one layer of states. Bad luck, because now you need to figure out nasty details about how to simultaneously fill the multiple layers in k-space.
This is the place where you can really see the value of the energy density of states compared to the k-space density of states. Essentially, if we know the energy density of states, then we simply integrate out to the (presently unknown) Fermi energy, insist that the total number of states is equal to the number of electrons (divided by the degeneracy of each state), and use the known number of electrons to solve for the Fermi energy.
In this problem, each layer in k-space introduces an energy density of states that is like the 2D form we calculated above. Notice that the 2D energy density of states is just a constant value, so it is pretty easy to integrate. The only complication is that new additional states appear each time the energy reaches a threshold that allow population of the next plane in k-space. This threshold happens for each z-component quantum number. Since each plane of states has the same energy density of states, we just need to know how many planes are populated at a given energy level. The number of planes occupied is just the z-quantum number. In other words,
Since each sheet contributes the same density of states, we have:
Since this density of states is piece-wise continuous, the integral for the number of states is easy to evaluate:
It's ugly, but not too bad. Notice that each term has the correct units of areal density. Therefore, we just need to solve for the energy needed so that the correct number of particles per area are occupied. In other words, figure out the energy such that the integral above is equal to the apparent 2D density of electrons for the given thickness:
By looking at the integrated density of states above, it is clear that the first layer is the only layer we need up to a density given by:
Above this density, we will start to fill the second layer as well as the first. We fill only these two layers up to a density of:
Still not enough states... However, if you go to the third disk, you can accomodate a total density of:
That's enough states. Therefore, we really only need to worry about filling states on the first three k-space disks. Explicitly, we fill up to an energy given by:
Solving for the required energy yields the top energy of:
The depth of the Fermi sea is then 6.1eV.
Sol 1.12