...we will look at the origins of magnetism, particularly ferromagnetism in metals. We will look at the role of the Fermi energy, Pauli exclusion principle, the fermion nature of electrons. Electron-electron repulsion plays a key role. Why is that? How do electron spins get involved? What causes spins to align?
HW 8.
1. Band energy: The total band energy can be defined as the sum of the energies of all electrons in single-electron-states. That is, it is the sum of the energies of the occupied single-electron states. Considering a band of bandwidth B for which we make the simplifying assumption that the density of states, \(D_c\), does not depend on energy.
a) What is the total band energy in terms of E_f and E_c?
b) What is the total band energy of we set \(E_c = 0\) ?
c) What is the total band energy in terms of n?
d) What is the relationship between n and Ef?
e) How many total states are there on the band
Suppose that just, for the fun of it, we divide the density of states into two parts: one associated with spin-up single-electron states and the other associated with spin-down single-electron states.
2. On what basis does one choose the spatial direction (axis) with respect to which spin up and spin down are defined?
3. a) What is the density of states in the spin up band? (in terms of \(D_c\))
b) What is the relationship between \(n_\uparrow\) and \(E_f^{\uparrow}\)?
c) What is the relationship between \(n_\downarrow\) and \(E_f^{\downarrow}\)?
d) Suppose for a fixed value of n, we allow \(n_\uparrow\) and \(n_\downarrow\) to vary. What is the band energy of all n electrons as a function of \(n_\downarrow\) and \(n_\downarrow\)?
e) Express this band energy as a function of \(n_\uparrow - n_\downarrow\).
f) Plot this band energy as a function of \(n_\uparrow - n_\downarrow\). What is the domain of this graph? What configuration(s) of the system have the lowest energy?
4. For most of this quarter we have been using single-electron states (filling them with many electrons), and we have been ignoring the coulomb repulsion between electrons. Sometimes the electron-electron (e-e) doesn't make much difference, but sometimes it makes a lot of difference. A fairly simple model from the e-e interaction if to write: \(V(n_\uparrow, n_\downarrow) = U n_\uparrow n_\downarrow /n\).
a) Plot this as a function of \(n_\uparrow - n_\downarrow\).
b) What is its value at \(n_\uparrow - n_\downarrow = 0 \)?
c) What is its value at \(n_\uparrow - n_\downarrow = n \)?
d) What simple symmetry does this function have?
e) What configuration of the system would be favored by this coulomb interaction?
5. In the Stoner model one combines the coulomb interaction energy associated with electron-electron repulsion with the band energy associated with the filled one-electron states.
a) Do this. For a given value of U, for what values of bandwidth the system will have a ground state with spin alignment (a ferromagnetic ground state).
b) For the case where the system is spin-aligned (magnetic), explain the nature of the origin of the magnetism? What is the driving mechanism that leads to the ferromagnetic state?
6. On can add a entropy related term to this model and explore why magnetism disappears at higher temperature. Generally, \( S= k ln(\Omega)\). For small B, perhaps one can approximate \(ln(\Omega) \) by \(ln(\Omega) \approx \frac{2}{3} n - \frac{1}{2n} (n_\uparrow- n_\downarrow)^2 - \frac{1}{8n^3} (n_\uparrow- n_\downarrow)^4 \).
a) For U=2 eV and B = 1 eV, find the ground state of the system as a function of temperature, T. You can assume the the ground state is the state with the lowest free energy, that is, the lowest value of F= E-TS.
b) Plot the ground state value of \((n_\uparrow- n_\downarrow) \) as a function of kT.
c) extra credit: basically one can think of \(\Omega\) as the number of possible states of the system with a given value of n and \(n_\uparrow\). What is that? That is, what is the value of \(Omega\) in terms of \(n_\uparrow\) and \(n_\downarrow\).
7. another optional special project: Calculate the magnetic susceptibility of the system from problem 6 as a function of temperature. Do this only in the temperature range above where it becomes magnetic, and show that it diverges as you approach the transition temperature from above. What is the form of that divergence? What sort of "power law"?
8. Optional special project: This project I think will show the how electrons with align spins tend to avoid each other. Consider an infinite square well with two electrons in it. Let's try making two electron states from products of the one electron states. Let's call the one electrons states \(\psi_n\) where \(\psi_n(x) = \sqrt{2} sin (n \pi x)\) where x is in nm and the well extends from x = 0 to x= 1 nm.
a) Consider a two-electron state in which the 11th and 12th states are occupied. Is \(\psi_{11}(x_1) \psi_{12}(x_2) \) an appropriate two electron state? Why or why not?
b) Suppose both electrons have the same spin (\(\uparrow \uparrow\) ). What is the appropriate two electron (spatial) state for this case?
c) On the other hand suppose that the 11th and 12th states are occupied, but the spin state is \(\frac{1}{\sqrt{2}} [ \uparrow \downarrow - \downarrow \uparrow ] \), then what is the two electron (spatial) state?
d) How are these (spatial and spin) states different? Why? Discuss here if you like. Don't wait too long. Do it now.
The expectation value of \( |x_1 - x_2| \) can tell us how close the electrons tend to be in these two-electron wave states.
e) for the state from b), calculate the expectation value \( |x_1 - x_2| \).
f) for the state from c), calculate the expectation value \( |x_1 - x_2| \).
g) are they different? by how much? what do you infer from this?
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Solution notes link:
https://drive.google.com/file/d/0B_GIlXrjJVn4V0g5V0NERFVlU3M/view?usp=sharing
Does anyone remember why in class we used this integral to represent the total energy:
ReplyDelete\(\int_{E_c}^{E_f} D(E)*E*dE = E_{total}\)
It seemed to make sense at the time but I can't justify it to myself right now.
Dude, your a boss. I am still having issues with \(D(E)*E\) being the integrand though. I thought D(E) was the density of states function, which you said as well, so multiplying by \(E\) just gives us total states and not total energy. I think I am just gonna sleep this one off and figure it out tomorrow.
ReplyDeleteI think if you re-write \(n_\uparrow^2+n_\downarrow^2\) as \((n_\uparrow+n_\downarrow)^2 + (n_\uparrow-n_\downarrow)^2\) it comes out nicely.
ReplyDeleteOops, there is also a one half term in there to correct.
ReplyDeleteSo I am attempting to work on the extra credit 6c and I am curious as how to start the problem. I am starting with the fact that electrons can be in either two states, up and down. It seems that Omega can therefore be 2n*number of electrons.
ReplyDeleteAnyone have any ideas?
Are you sure it's B < 2U? I got the case of B < 4U. But then again, I got a different e-e function than in class because we were using a different constant in class.
ReplyDeleteCorrect me if I'm wrong. I should probably head to sleep soon...
Actually it is the ground state of the many electron system we are referring to. n is basically the number of electrons in the conduction band (for a one centimeter cubed chunk of material). Normally the electrons which fill the n/2 lowest energy band states. That would force half the electrons to be spin up and half to be spin down. That gives you the lowest band energy.
ReplyDeleteIn this model the ferromagnetic state is when all the electrons are spin up. Or when all the electrons are spin down.
So I been thinking about problem 6c and the solution I got for Omega is
ReplyDelete$$Omega = 2^{n-1} $$
As discussed in office hours on Monday, the number of elements given the number of n electrons and the number of electrons with either spin up or spin down is given by:
$$\dbinom{n}{n_{down}} = \frac{n!}{n_{down}!(n-n_{down}!)} = \frac{n!}{n_{down}!n_{up}!}$$
However, there can be up to n electrons. Hence, Omega is the sum of the combination of n up electrons (direction doesn't matter) in n. In other words:
$$\Omega = \sum_{k=0}^{n}\binom{n}{k}$$
where k is the number of electrons in that spin state.
There is an identity that states the above sum is
$$\Omega = \sum_{k=0}^{n}\binom{n}{k} = 2^{n}$$
However, because of the indistinguishably principle, the number of spin combination of electrons is reduce in half. Therefore, the above expression becomes:
$$\Omega = 2^{n-1}$$
Which in terms of spin states:
$$\Omega = 2^{n_{up}+n_{down}-1}$$
Any thoughts on this? I want to thank my fellow classmates who went to office hours on Monday for really guiding me with the extra credit problem.
Bah. The second expression should be:
Delete$$\binom{n}{n_{down}} = \frac{n!}{n_{down}!(n-n_{down})!} = \frac{n!}{n_{down}!n_{up}!}$$