Solutions for problem set 1
1. (i) If there are no interactions then theN-particle classical Hamiltonian is just the sum ofN copies of the single particle Hamiltonian, and the quantum version is obtained by replacing momenta and locations by the corresponding operators, so: ˆ H= N X i=1 1 2m ˆ ~ pi+e ~A(ˆ~ri, t) 2 −eφ(ˆ~ri, t) = N X i=1 1 2m ˆ ~ p2i + e 2m ˆ ~ piA(ˆ~ ~ri, t) +A(ˆ~ ~ri, t)ˆ~pi −eφ(ˆ~ri, t)
if we can ignore theA~2 term because it is small. The middle term cannot be further simplified (for a general gauge)
because momentum and position operators do not commute.
(ii) Because we only have one-particle operators in the Hamiltonian, in the 2ndQ we can write, for this basis: ˆ
H = X
~k,~k0,σ,σ0
h~kσ|ˆh|~k0σ0ic~†
k,σc~k0,σ0
Using the fact that the Hamiltonian is spin independent, the spin part will just give a δσ,σ0 which will kill one sum
over spins and set both operators to have the sameσ. For the translational part, we have :
h~k|ˆh|~k0i=h~k| 1 2m ˆ ~ p2+ e 2m ˆ~
p ~A(ˆ~r, t) +A(ˆ~ ~r, t)ˆ~p−eφ(ˆ~r, t)|~k0i
where so far I simply used the expression of the one-electron operator ˆh. We can now use the fact that ˆ~p|~ki= ¯h~k|~ki, i.e. these basis states are the eigenstates of the momentum operator – that allows us to simplify things a bit to:
h~k|hˆ|~k0i=¯h 2k2 2m δ~k,~k0+ e¯h 2m( ~k+~k0)h~k|A(~~ r, t)|~k0i −eh~k|φ(~r, t)|~k0i
Now we have to calculate those matrix elements, for instance h~k|A(~~ r, t)|~k0i = R
d~rφ∗~
k(~r)
~
A(~r, t)φ~k0(~r). As discussed
in class, these basis wavefunctions are φ~k(~r) = e
i~k~r
√
V so the integral becomes h
~k|A(~~ r, t)|~k0i= R
d~re−i(~k−~k
0)~r
V A(~~ r, t) =
~
A~k−~k0/V (i.e., the Fourier transform with that momentum), and the electric potential is just the same. In the exam I
was happy if the students left these integrals as integrals (which is what is needed for the next point, anyway). Note that you could have left the~poperators inside the matrix element and replace them with the usual derivatives, and the answer would be precisely the same (as it better be!!!).
To conclude, we have ˆ H= X ~ k,~k0,σ h~k|ˆh|~k0ic†~ k,σc~k0,σ where h~k|ˆh|~k0i=¯h 2 k2 2m δ~k,~k0+ e¯h 2m( ~k+~k0)Z d~re−i( ~ k−~k0)~r V ~ A(~r, t)−e Z d~re −i(~k−~k0)~r V φ(~r, t)
I hope you’ll try the “for fun” part as well, it is actually fairly simple to write the matrix element. Please ask me if you have any questions about it.
(iii) We can now compare this to the general expression given, to read off: ˆ ~j(~r) = e¯h 2m X ~ k,~k0,σ (~k+~k0)e −i(~k−~k0)~r V c † ~k,σc~k0,σ
as the density current operator. Of course, you should also compare the expression for the density operator and see that it is precisely what you get if you second quantize that operator.
Regarding the ”for fun” part: we could also get this operator starting from its classical expression ~j(r) = eP
i ~ pi
mδ(~r−~ri). However, unlike for the density operator, here some care is needed when we quantize this
be-cause momentum and position operators do not commute, so this classical expression needs to be handled with some care to obtain a proper (Hermitian) quantum expression. But if you do that and then 2ndQ, you’ll get exactly the same expression like in the line above. As we should!
Important note: the homework has a typo, I changed the sign for the 3rd term in the Hamiltonian (first equation in part (iii)) by mistake. If you use this wrong definition, then the expression you extract for the number density
operator ˆn(~r) will have a minus compared to the expression we got in class (because the typo propagates). This is how you could see that there was a typo, it’s impossible to have a negative number density, that makes no sense because it means there is a negative number of particles. The lesson is that (a) you should always analyze whether your result makes sense, and (b) if it doesn’t, then find a mistake in your calculations, or a typo in the text, or something to solve the issue. If you were in that quandary here, hopefully it was easy enough to see that that 3rd term had a different sign in the two equations, and to suspect that that might be the problem. Also, ask me if unsure whether you’ve figured things out.
But please, never give as your answer something you know is physically impossible to be a correct answer (like a negative number of particles, or a negative number density). At the very least say ”I know this is wrong but I cannot find the mistake”.
2. AnN-fermion Slater determinant is:
Ψα1,...,αN(x1, ..., xN) = 1 √ N! X P∈SN ξPφα1(xP1)...φαN(xPN)
Since for any permutationP, ˆA=PNi=1Aˆi=PNi=1AˆPi, and since ˆAPi acts only on the particle atxPi and leaves the
rest unchanged, we find: ˆ AΨα1,...,αN(x1, ..., xN) = N X i=1 X β Aβ,αiΨα1,...,β,...,αN(x1, ..., xN)
where in the Slater determinant, the indexαi →β. I skipped a few steps here because writing all the steps is a bit
nasty, but I hope I gave you enough pointers to get to this result.
Now let’s check that we get the same answer in 2nd quantization. First, what is the state corresponding to Ψα1,...,αN(x1, ..., xN)? Clearly, this is the state where states α1, ..., αN are occupied (in this order!), and everything
else is empty, in other words it is the statea†α1...a†αN|0i. Consider then: ˆ Aa†α 1...a † αN|0i= X α,β Aβαa†βaα·a†α1...a † αN|0i
We use the identity a†βaαa†αi = a
†
αia
†
βaα+δα,αia
†
β. You should be able to prove this using the identity [AB, C] =
ABC−CAB =ABC+ACB−ACB−CAB =A{B, C} − {A, C}B and the usual anticommutation operators for a, a†. This means that:
a†βaαa†α1a † α2...a † αN|0i=δαα1a † βa † α2...a † αN|0i+a † α1a † βaαa†α2...a † αN|0i
where in the second term, we now have a†βaα moved past a†α1. We continue to do so past each operator (getting
a new delta term each time). After the last step, we have terms proportional to δααi for each i = 1, N, plus
a†α1a†α2...a†αNa†βaα|0i= 0 becauseaα|0i= 0. So indeed, we find:
ˆ Aa†α 1...a † αN|0i= N X i=1 X β Aβαia † α1...a † β...a † αN|0i
where theβ operator is in theith position. This is precisely the counterpart of the result we had in 1st quantization, so this must be the correct expression for ˆA (it acts in the correct way on Slater determinants). Because any state can be decomposed into Slater determinants, it means that this operator acts correctly on any state, ie gives the counterpart of the answer we would find in 1st quantization.
If you thought this was too easy, repeat for 2-particle operators or for bosons (where the same state may be occupied by multiple particles). Sooner or later you’ll get the hang of working with 2nd quantization notation and you’ll start to appreciate its simplicity.
3. (a) First thing when solving a many-particle problem (and two qualifies as many, anything more than one is “many” in our language) is to see if those particles are interacting or not. In the latter case, we know that we can solve the problem for a single particle (which is always the simplest problem), and then fill up one-particles levels to get the many-particle eigenstates. If there are interactions, however, we have to solve the many-particle Schrodinger equation directly, which can be a complicated task.
If both electrons have spin-up and if the model is the Hubbard model (like we chose), then they do not interact! There is no interaction term of the typec†c†ccin this Hamiltonian that removes two spin-up electrons and adds two other spin-up electrons, potentially in other states.
Aside: at this moment you might think that is wrong, surely the Coulomb repulsion acts between all electrons, no matter their spin, so how can this be? The answer is that when we limited ourselves to the Hubbard model, we truncated (threw out) interaction terms when the electrons are on different sites. Those terms do act on electrons with same spin as well, but we make the approximation that they are very small and can be safely ignored, so they’re not in our model. We only kept on-site interactions, and because Pauli principle forbids two electrons with the same spin to be on the same site, it then follows that within this Hubbard model, two spin-up electrons do not interact. It is an approximation, and sometimes a very good one. The point is that we’ve chosen our model, it contains some approximations (whether those are reasonable or not is important to know, but a separate issue) and our task now is to find the solution for this model.
Because two spin-up electrons do not interact, the easy approach is to find the eigenstates for a single spin-up electron, and then make the two spin-up electron eigenstates and see which is the lowest energy = the ground-state.
In the absence of interactions, we only have hopping and we can follow the solution discussed in class for the general N case, but here N = 3. The general one spin-up particle eigenstate is of the form |φi = P3i=1φic†i↑|0i. Asking thatH|φi=E|φi leads to three coupled equationsEφ1 =−t(φ2+φ3) and cyclic permutations. These can
be solved by brute force (a 3x3 matrix is at the limit of what I’ll ask you to handle by hand). A better way is to use symmetries, we know we have invariance to translations so momentum is a good quantum number, thus we must have φn = exp(ikna)/
√
3, n = 1,2,3, and the PBC fix k = m2π/3a where m =−1,0,1 are the three values that fall inside the BZ (again, in class we discussed the generalN case, just repeat that for N = 3). The eigenenergy is (k) = −2tcos(ka), and for the allowed momenta we find values (0) = −2t and (±2π
3a) = +t. It should feel right
that the latter two are degenerate, because the only difference is whether the particle goes clockwise or anticlockwise around the molecule but with the same|2π/3a|momentum, so surely it must have the same energy in both cases.
So now we’re ready to answer the question. If we have two non-interacting particles, we get their GS by placing them in the lowest one-particle levels allowed. Here, this means that one can be in thek= 0 state with(0) =−2t, but the second must go into one of the upper states (because Pauli doesn’t allow us to put it also on k= 0, this is already taken by a same-spin electron), with (±2π
3a) = t and so the lowest possible total energy for two same-spin
electrons is E = −2t+t = −t, and we can see that it is doubly degenerate. We can also infer that there is one excited state with energy 2t, non-degenerate, obtained by occupying the two±2π/3 states. We could also easily get the two-particle wavefunctions, if we wanted.
Aside: I can now also easily figure out the three spin-up electron spectrum. Pauli forces us to put a particle on each single-particle level, so there is a single eigenstate and its energy is−2t+t+t= 0. Does this make sense? Yes, because the only possible such state isc†1↑c†2↑c†3↑|0i, so it’s non-degenerate. And because Pauli prevents any of these electrons from hopping, its eigenenergy must be 0.
So (i) is solved ... but let’s also see how we could solve it by brute force. Maybe you didn’t realize that spin-up electrons are non-interacting in this model (although I hope that from now on, you’ll remember that that’s the first thing to check, because it makes life much easier). Or, even better, you did solve it that way, and now you want to check that it indeed works.
First, we need to decide how many states we have in the basis, and what to call them. We could use the general rule and label our basis states|n1↑n1↓n2↑n2↓n3↑n3↓iwhere eachn= 0,1 depending on whether there is a particle with that spin at that site. This way, the two spin-up particle states are|1,0,1,0,0,0i, |1,0,0,0,1,0iand |0,0,1,0,1,0i. There’s nothing wrong with this, except it’s cumbersome and obviously going to be really unpleasant when we deal with longer chains. So we’re allowed to use simpler names, so long as we define them clearly.
Let me define the new notation |iji = c†i↑cj†↑|0i, i 6= j being the site indexes, as a short-hand notation for these states with the up-spin electrons at sites i and j (i = j is not allowed by Pauli). Note that |iji = −|jii, which means we only have 3 independent such states, let me pick them to be|12i,|23i,|31i. These are exactly the same states as before, eg |12i =c†1↑c†2↑|0i =|1,0,1,0,0,0i, I just gave them shorter names. Note also that I chose
|31i=c†3↑c†1↑|0i=−|1,0,0,0,1,0i. This is simply for convenience. I’ve solved such problems before and I expect that because of the symmetry of the problem (where everything is invariant to cyclic rotations) this choice may make the calculation simpler than working with +|1,0,0,0,1,0i=|13iinstead. But either choice is valid and will give the same results in the end (check it, if this is not obvious!).
Because the two electrons are never at the same site, the on-site repulsion never acts, so effectively hereH= ˆT. It is then straightforward to check thatH|12i= ˆT|12i= +t(|23i+|31i). To get this you can either act withT and use the anticommutation rules, or use common sense: hopping can move the electron from site 1 of state|12ionly to site 3 (hoping to 2 is not allowed), taking the system to state|32i=−|23i; and similarly for the other electron. So this
explains why the overall sign of the hopping is “wrong”, i.e. +tinstead of−t. The action ofT on the other states is found similarly, or obtained using circular permutations.
The most general eigenstate in this subspace is|Ψi=a|12i+b|23i+c|31i. Asking thatH|Ψi=E|Ψileads to a homogeneous 3x3 equation, which has non-trivial solutions if its determinant vanishes:
−E t t t −E t t t −E = 0→E1,2=−t, E3= +2t
So theSz= +1 triplet ground-state is doubly degenerate and Egs,t=−t(for hopping betweens-orbitals, t >0). So
hooray, this agrees with what we found with the simpler first method. However, I hope you can see that using the first method is simpler, especially if we plan to makeN larger and/or have more particle.
You can now easily check that precisely the same solution works for the other triplet components, i.e. that it makes no difference if we had instead both spins down, or if we had|iji= √1
2
c†i↑c†j↓+c†i↓c†j↑|0i. We still have|iji=−|jii
and the action of H on these states is still precisely the same, so the answer remains the same, as guaranteed by symmetries.
(b) In this case we do have interactions between the spin-up and spin-down electron, so we have to solve the two-particle problem from scratch, we can’t use the single-particle solution to built a many-particle one (this is the problem with interactions!).
Let me now define two-electron states with singlet symmetry |ijis = √12
c†i↑c†j↓−ci†↓c†j↑|0ias the counterparts of the triplet states we had before, BUT note that now |ijis = |jiis, so we don’t get the − sign! This should be
expected. We know that the overall state must be antisymmetric to particle interchange. In the singlet, the spin part is antisymmetric so the spatial part must be symmetric, and indeed it doesn’t matter the order in which we list the sites. For the triplets, it did matter because the spin part was symmetric so the spatial part had to be antisymmetric, hence the sign change when we changed the order.
Now we find: H|12is = −t(|23is+|31is)−t
√
2 (|1i+|2i) where we get new states |ii = c†i↑c†i↓|0i with both electrons at the same site, because these are now allowed. The action ofHon|23is,|31isis obtained similarly.
We also needH|1i=U|1i −t√2 (|12is+|31is), etc.
The eigenstates are now linear combinations of these 6 basis states, leading to a rather scary looking 6×6 matrix. However, we were asked to find the ground-state in the limit U → ∞. In this case, the doubly-occupied states |ii
cannot contribute to the ground-state, because they cost an energy ∼ U which is infinite. So we must still have
|Ψi=a|12is+b|23is+c|31is (for the ground-state) and acting with the Hamiltonian on it (and erasing all doubly
occupied states, as they cannot occur), we get the same solutions as before, but with t → −t because the matrix elements have a different sign than at (a). This means that nowEgs,s=−2t.
Clearly, we could not get this result by populating one-electron states. If there were no interactions U = 0, then we could use that here as well, and we would find the singlet ground-state to haveE =−4t, coming from thek= 0 single particle state with=−2tbeing occupied by a spin-up and a spin-down electron. The next energy state would be at−2t+t=−t, there is no non-interacting two-particle state with energy−2t, like we found! That is because our problem has interactions and those change the eigenstates and makes them different from those of the non-interacting system. In particular, it should feel ”right” that the GS energy went up from−4tto−2tbecause we added a repulsion in the system, so the energy should be higher. If we had added an attraction between the electrons (nevermind how that could be arranged), then we would expect the energy of the interacting system to be lower than that of the non-interacting one.
Another important lesson is the difference between Pauli principle and ”infinite” interactions. It might look as if formally both do the same thing, they don’t allow two particles to be at the same site. But for same spin electrons that is a consequence of them being Fermions with the same spin, it comes from a symmetry of our universe regarding indistinguishable particles, and it’s true whether there are interactions or not. For the singlet case, this is an approximation, it only becomes true asymptotically asU → ∞, it’s not a general ”built-in” feature of the universe (we can distinguish a spin-up electron from a spin-down one!). If we make the repulsion weaker (see below) we’ll get a different answer, and if we remove the repulsionU = 0, we get the−4t energy discussed above. So even though in both cases the consequences seem the same (cannot have two particles at the same site) in the first case it is because of symmetries but the problem is non-interacting. In the second case is because of (very particular) interactions. There’s no reason to expect the same results, and indeed we don’t get the same results, the singlet GS energy is different from the triplet GS energy. This difference vanishes asN→ ∞.
For fun: IfU is finite then all 6 eigenstates in the basis are expected to contribute to the GS, so we need to keep all of them. Solving the 6×6 determinant is still not a particularly pleasant way to spend our time, so let’s be smart
about it. If you look at the symmetry of the U → ∞solution, you will find that|Ψgsi= √13(|12is+|23is+|31is),
i.e. it is a state with total momentumk= 0. (You can check that the other two states have momentumka=±2π/3, which is precisely what periodic boundary conditions for aN= 3 system says we should have. Ha ha, symmetries to the rescue, again!).
If the ground-state at U t has k = 0 momentum, and the ground-state when U = 0 has k = 0 momentum (because we occupy two single electrons states with 0 momentum, so the total momentum is still zero) it is reasonable to assume that by continuity, the momentum will remain 0 at allU. So we look for an eigenstate ofk= 0 momentum, for which the most general possibility is
|ψi= √a
3(|12is+|23is+|31is) + b
√
3(|1i+|2i+|3i)
(it’s best to always use normalized states, hence the √3 factors). Asking that H|ψi = E|ψi now leads to a 2×2 matrix fora, b, and from its determinant we find that:
Egs,s=
1 2
h
U−2t−p(U+ 2t)2+ 32t2i
Let’s check if this is reasonable. For U → ∞, we indeed recover −2t so that’s fine (you can now also understand better what happens with the wavefunction - it turns out that ifU → ∞, thenb/a∼1/U →0 so the probability to find both electrons at the same site vanishes). We can also check the limitU = 0, i.e. non-interacting electrons. The formula above givesEgs,t=−4t, which is again correct.
Of course, you can now go ahead and find all other eigenstates using the fact that we know their momentum, i.e. the general “shape” they can have. Hopefully this gives you an idea (i) why exploiting symmetries is so useful - they limit the possible form of the wavefunctions, making calculations easier; and (ii) why dealing with interacting systems is tough – we have to solve the many-particle Schrodinger equation by brute force, and that can be very tough. Symmetries can help a little, as we showed here by dividing the problem into triplets and singlets, and then using the fact that the momentum is a good quantum number (because of invariance to translations). But there are limits to how much such symmetries help because there’s only a few of them and if we haveN ∼1023 particles, the
general allowed forms of the wavefunctions will still be extremely complicated, and exact calculations tend to become impossible. There are a handful of exceptions, i.e. a very few models with interactions that can be solved exactly, but generically we are forced to use approximations. I’ll tell you more in class about models with exact solution, let me here just say that they tend to be very special in that the number of their symmetries scales with the number of particles, so there are enough constraints on the allowed form of the many-particle eigenstates that we can find them. Sometimes this happens in 1D, so some one-dimensional models (including our good friend the Hubbard model) are exactly solvable in 1D but not in higher dimensions, and not if we generalize them (eg, if we add nearest-neighbor repulsion between electrons, we kill those additional symmetries and we’re back to needing approximations).
