This looks great. Let's make this a 3-part problem.
part A: Calculate the spontaneous magnetization as a function of temperature for \(e^2/a = 1.5\; eV\) and \(E_{band}\) = 1.4 eV. (This part is for zero applied magnetic field). For this part create a graph of magnetization vs T. Define a \(T_c\) and look at the behavior in the vicinity of that. The part in the square root is a lot more important that the 1/T outside that. What is the critical exponent?
Part B: Calculate the magnetic susceptibility in the normal state, as Arjun has explained.
For both parts, the most important thing is your graph. That is what I will look at first. Do a really nice graph, hand drawn, not too large, with a nice title, labels and scales, and with an excellent caption that explains everything in a succinct and beautiful manner.
Part C. (added Feb 6) Calculate and plot the magnetic susceptibility as a function of T for \(e^2/a = 1.5\; eV\) and \(E_{band}\) = 1.7 eV.
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I decided to go back to my entropy calculation after something Zack mentioned, and I found a sign error in my arithmetic. \( C_0 \) remains the same, but the expression for the polynomial up to order \( O(x^m) \) is:
\( \sum_{k \in \text{evens} }^m 2^k \left[ \frac{1}{(k-1)N^{k-1} } - \frac{(1-N)}{kN^k} \right] \left( N_\uparrow - \frac{N}{2} \right)^k \)
Alternatively written as
\( \sum_{k \in \text{evens} }^m 2^k \left[ \frac{N - (k-1)}{k(k-1)N^k} \right] \left( N_\uparrow - \frac{N}{2} \right)^k \)
The first two terms (to order \( O(x^6) \) ) are:
\( \frac{N-1}{2N^2} 2^2(N_\uparrow -\frac{N}{2})^2 + \frac{N-3}{12N^4} 2^4(N_\uparrow -\frac{N}{2})^4 \)
And this is definitely 100% correct. I checked it against Mathematica and everything. This is a Taylor expansion from
\( 0.5 \ln (N^2 - N_\Delta ^2 ) + N_\Delta \tanh ^{-1} \left( \frac{N_\Delta }{N} \right) \quad \text{where} \quad N_\Delta = N_\uparrow - N_\downarrow \)
which is itself part of the Stirling Approximation for
\( \ln \left( \frac{N!}{N_\uparrow ! N_\downarrow ! } \right) \)
-Aaron
EDIT: Added the revised expression that Aaron derived for the entropy and changed the symbol for the band energy to avoid confusion. -Arjun
Zack--
ReplyDeleteI hope this describes what you envisioned as the midterm question. Please let me know if there are any changes that need to be made. I'd be happy to edit the LaTeX document and re-post it here.
Looks great!
ReplyDeleteA correction in language:
N is the number of occupied states, not states. i.e., it is the number of electrons occupying states in that band.
(same comment for Nup)
Ah, I realized that I used B for both magnetic field and the band energy. Hopefully nobody gets them confused.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteI decided to go back to my entropy calculation after something Zack mentioned, and I found a sign error in my arithmetic. \( C_0 \) remains the same, but the expression for the polynomial up to order \( O(x^m) \) is:
ReplyDelete\( \sum_{k \in \text{evens} }^m 2^k \left[ \frac{1}{(k-1)N^{k-1} } - \frac{(1-N)}{kN^k} \right] \left( N_\uparrow - \frac{N}{2} \right)^k \)
Alternatively written as
\( \sum_{k \in \text{evens} }^m 2^k \left[ \frac{N - (k-1)}{k(k-1)N^k} \right] \left( N_\uparrow - \frac{N}{2} \right)^k \)
The first two terms (to order \( O(x^6) \) ) are:
\( \frac{N-1}{2N^2} 2^2(N_\uparrow -\frac{N}{2})^2 + \frac{N-3}{12N^4} 2^4(N_\uparrow -\frac{N}{2})^4 \)
And this is definitely 100% correct. I checked it against Mathematica and everything. This is a Taylor expansion from
\( 0.5 \ln (N^2 - N_\Delta ^2 ) + N_\Delta \tanh ^{-1} \left( \frac{N_\Delta }{N} \right) \quad \text{where} \quad N_\Delta = N_\uparrow - N_\downarrow \)
which is itself part of the Stirling Approximation for
\( \ln \left( \frac{N!}{N_\uparrow ! N_\downarrow ! } \right) \)
In other words, the entropy term is actually:
ReplyDelete\( C_0 - \sum ( \text{new stuff} ) \)
where "new stuff" is the expression I found. Note that they're being subtracted!!
sorry if this appears twice but i hit publish and my comment disappeared. So if they were proportional in the case B=0, wouldn't that mean M=0 too. And that can't be the case other wise we wouldn't have spontaneous magnetisation.
ReplyDeletethat's exactly what i was thinking!
ReplyDeleteIs anyone else getting very ugly results for \(N_{\uparrow}-N/2\) when minimising F with the added energy due to a magnetic field.
ReplyDeleteI think that the entropy is a maximum at x=0, if \( x= N_{\uparrow} - N/2\). And I think all the terms in the taylor series expansion are negative.
ReplyDeleteSo I think you actually should have something like \(-Ax^2 + Bx^2 + KT(Cx^2+Dx^4\). (where A comes from the e-e interaction and B is the Band-related term.)
Just calculate the susceptibility for the normal state. Not for the spontaneously magnetic state. (see problem statement above.)
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ReplyDeleteJust want to make sure you have not lost sight of the problem statement. It might be best to start with part C.
Deletepart A: Calculate the spontaneous magnetization as a function of temperature for e2/a=1.5eV and Eband = 1.4 eV. (This part is for zero applied magnetic field). Create a graph of magnetization vs T. Define a Tc and look at the behavior in the vicinity of that. The part in the square root is a lot more important that the 1/T outside that. What is the critical exponent?
Part B: Calculate the magnetic susceptibility in the normal state. Plot Chi is as a function of T. Discuss the most interesting aspects.
Part C. (added Feb 6) Calculate and plot the magnetic susceptibility as a function of T for e2/a=1.5eV and Eband = 1.7 eV.
************For all three parts, the most important thing is your graph. That is what I will look at. Do some really nice graphs, of F vs x and of Chi vs T and M vs T. Hand drawn, not too large, with a nice title, labels and scales, and with an excellent caption that explains everything in a succinct and beautiful manner.
I am a little confused about the difference between part b and c?
ReplyDeletedifferent value of Eband. Significantly different!
DeleteAlright so here is what I'm gathering from this discussion. It seems to me that the expression given in the question is wrong. The \(e^2\) is inherently negative so we know there should be a minus sign in front of that term in the Helmholtz energy equation that Arjun gave us.
ReplyDeleteA question I have is whether to use the values for {\e^2/a\) and (\E_band\) that were given in part A for our calculations for part B.
A second question is with regards to our plots of the magnetic susceptibility vs. temperature. I am getting that the magnetic susceptibility is depended on both N and B after some calculation.(This seems fishy to me because I don't think it shouldn't depend of B) but either way if there is a N dependence how are we supposed to find exact values for the magnetic susceptibility without being given a value of N. Should our values on the Temperature axis be in terms of N?
Another question is what the value of \(\mu)\ is.
A question I have is whether to use the values for {\e^2/a\) and (\E_band\) that were given in part A for our calculations for part B.
DeleteYes.
I see what you mean about Chi. Let's redifine Chi by the equation \(m=\Chi B\) where m=M/N. That way m will be intensive and \(\Chi\) will be an intrinsic property that does not depend on N. We can call m the magnetization density.
DeleteWe are also getting a chi dependent on B but agree that it should be independent of the magnetic field. Have we gone wrong?
DeleteThe magnetic susceptibility is defined such that it describes a material's response to an external field. We don't usually think of susceptibilities unless we are discussing external fields.
Delete"We are also getting a chi dependent on B but agree that it should be independent of the magnetic field. Have we gone wrong?"
DeleteYes.
maybe it is a small higher order dependence? not relevant for linear response?
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ReplyDeleteNellie, Moritz and I are arguing about the sign of the energy due to the magnetic field \(2 \mu B X \). The sign of this term should depend on the direction of the magnetic field B and whether X is in terms of Nup or Ndown? Having X in terms of Nup and having a B-field in the Nup direction, would this give us a positive or negative contribution to the free energy? What do you guys think?
ReplyDeleteThis comment has been removed by the author.
DeleteIf \( x \) is in terms of \( N_{\uparrow} \), then we would get that \(U = -2 \mu B (N_{\uparrow} - \frac{N}{2}) \). I think that if the B field is pointing in the upwards direction (parallel to N up), then it would lower the total energy, since the highest energy would be where the spin was antiparallel to the field.
DeleteLOL
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