Let us consider tetrahedron with vertices, labeled as 1, 2, 3, 4. Let us have $N$ electrons which can be placed in this vertices. Let the hamiltonian of this system satisfy such condition: $$ \langle i|\hat{H}|j\rangle = \Delta, \: i \neq j, \: 1 \leq i,j \leq 4 .$$ Our task is to find the energy spectrum of this system in cases when 1) $N = 1$; 2) $N = 2$; 3) $N = 3$; 4) $N = 4$.
I have proceeded in case of $N=1$. My approach was to obtain eigenvalues of single-particle hamiltonian directly. My result is that we have two energy levels: ground state with energy $-\Delta$ which is triply degenerated and excited state with energy $3\Delta$.
Now I am stuck with the case of two electrons. The problem is that we should consider electrons as fermions with spin. I have two ideas how to proceed further, but misunderstand how to make it real:

  1. To construct a good basis in the space of states. It should contain antisymmetrized mixtures of single-particle states. It is something like: $$ |\text{state}_1\!\rangle = \frac{1}{\sqrt{2}} (|1 ;2\!\rangle- \:|2;1\!\rangle)$$ And so on. The first number in the brackets corresponds to the position of the first electron and the second number corresponds to the second electron. On this step I am not sure how to account for spin variables. If we add spin variable, then in formula above, it will look like: $$ |\text{state}_1\!\rangle = \frac{1}{\sqrt{2}} (|1,\sigma_1 ;2, \sigma_2\!\rangle-|2,\sigma_2 ; 1,\sigma_1 \!\rangle)$$ The hamiltonian should be rewritten in terms of this basis states. I tried to "extend" initial hamiltonian doing this: $$ \hat{H} = \hat{H_1}+\hat{H_2}$$ $$ \langle i,\sigma_1;k,\sigma_2|\hat{H_1}|j,\sigma_x; k,\sigma_2\rangle \:= \Delta, \: i \neq j, \: 1 \leq i,j \leq 4, \; \sigma_x = \pm\frac{1}{2} . $$ $$ \langle i,\sigma_1;k,\sigma_2|\hat{H_2}|i,\sigma_1; m,\sigma_x\rangle \:= \Delta, \: k \neq m, \: 1 \leq k,m \leq 4, \; \sigma_x = \pm\frac{1}{2} . $$ But this does not include Pauli exclusion principle. It can be inserted in this formula by imposing restrictions on indices like: $$ \langle i,\sigma_1;k,\sigma_2|\hat{H_1}|j,\sigma_x; k,\sigma_2\rangle \:= \begin{cases} \Delta, \: i \neq j, \: 1 \leq i,j \leq 4, \; \sigma_x = \pm\frac{1}{2}, \; \{j,\sigma_x\} \neq \{k, \sigma_2\} \\ 0, \; \text{otherwise} . \end{cases}$$ But this expression now looks too complicated to consider its matrix elements in new basis. Moreover, the Pauli exclusion principle makes this matrix elements dependent from spin. This confuses me about the dimensionality of this matrix. If we forget about spin, then we have $4 \times 3 = 12$ pairs of different indices and therefore 12 basic states. If we then consider the spin indices, then, as I see, we will get $12 \times 4 = 48$ basic states, because each electron in pair has two options of spin. It seems like I am to recollect this $48 \times 48$ matrix, iterating over all the indices and tracking all the restrictions. It is hard to be done for me. If I had somehow the resultant matrix, then I would analyze its eigenvalues (this problem also looks scary in perspective). Is such procedure correct in principle?
  2. Probably the problem may be simplified in secondary quantization representation. The hamiltonian can be considered as combination of annihilation operator in one vertice and creation operator in another: $$ \hat{H} = \Delta \sum_{i \neq j} a^\dagger_i a_j + \Delta \sum_{i \neq j} b^\dagger_i b_j$$ Here $a$ corresponds to the first electron and $b$ corresponds to the second. But I am not sure about this expression because it doesn't include spin and it doesn't forbid electrons to violate Pauli exclusion principle (electrons don't see each other). Moreover, even if I had the expression like this, it would be hard for me to calculate spectrum of this combination. Maybe it can be worked out with commutative relations for fermion operators. Is it correct way to approach the problem?
  3. The last idea that I have is to include "exchange interaction" adding the term like: $$ \widehat{H_{\text{ex}}} = J \sigma_1 \sigma_2 $$ I tried to consider position and spin variables separately, found eigenvalues of $16\times 16$ "positional hamiltonian" ($-\Delta$ and $15\Delta$) and then added the split caused by exchange interaction: $-\frac{3J}{4}, \frac{J}{4}$. Therefore 64 "stationary states" are obtained. It looks not very good for me, because it again doesn't account directly for antisymmetry, I don't know where I can get $J$ from and 64 is also as I think incorrect number, because I expect 48 from what was written above (Pauli principle sorts out redundant states, as I think). But may be something can be done in this manner?

I would be really grateful for any comments and remarks.

  • $\begingroup$ I forgot to mention one more idea. It is to use the information about the symmetry of thetrahedron, to describe it as a group of its movements, which conserve its structure. Maybe this can help to make the calculations with $48 \times 48$ hamiltonian shorter? $\endgroup$
    – goroshek
    Nov 11, 2021 at 0:43
  • 1
    $\begingroup$ Do you mean tetrahedron? $\endgroup$
    – mike stone
    Nov 11, 2021 at 0:45
  • 1
    $\begingroup$ Right now it is a bit hard to understand your setup. What are the states $|j\rangle$? Is $|2\rangle$ the state where one electron is located at vertex 2? Also what is the energy for the diagonal elements? i.e. What is $\langle i|\hat H|i\rangle, i=1,\,\dots,\, 4$? $\endgroup$ Nov 13, 2021 at 10:35
  • 4
    $\begingroup$ Item 3 in the question suggests that you haven't quite decided what the Hamiltonian should be. Are you imagining a physical system and asking what Hamiltonian should be used for that system? What forces are keeping the electrons at/near the vertices of the tetrahedron? $\endgroup$ Nov 13, 2021 at 20:19
  • 2
    $\begingroup$ In the question, you describe how $H$ acts on single particle states. The answer will depend on what exactly you specify as the 'interaction' term (if any). If you are presuming something like the Hubbard model, then you should tell us that. If there is no interaction term then obviously point 3 is not applicable, which is I think why @ChiralAnomaly also asks about this $\endgroup$
    – Sal
    Nov 18, 2021 at 18:55

1 Answer 1


In my mind the problem should be considered as simpler. Since you have non interacting fermions the spectrum is simply given by the total energy after you place electrons in the different states.

The eigenfunction which is not asked above is then simply a Slater determinant taking as entry you different states.

For electrons without spins the possible (energies,degeneracy) couple are then (energies in $\Delta$ scale):

$N=1: (-1,3), (3,1)$

$N=2: (-2,3), (2,3)$

$N=3: (-3,1), (1,3)$

$N=4: (0,1)$

For electrons with spin since one can have spin degeneracy, one get

$N=1: (-1,3), (3,1)$

$N=2: (-2,15), (2,12), (6,1)$

$N=3: (-3,20), (1,30), (5,6)$

$N=4: (-4,15), (0,40), (4,15)$

  • $\begingroup$ Hello! It is preferable to use MathJax (LaTeX) to display formulas. You can find a tutorial at MathJax basic tutorial and quick reference. Please edit your answer accordingly. Thanks! $\endgroup$
    – Jonas
    Nov 18, 2021 at 20:20
  • $\begingroup$ Thank you very much for your answer! $\endgroup$
    – goroshek
    Nov 19, 2021 at 10:33
  • $\begingroup$ Can you please say in general, how have you proceeded? I try to imagine how to calculate total energy by myself, but don't quite understand the procedure, if I only have hoppings energies. $\endgroup$
    – goroshek
    Nov 19, 2021 at 10:49
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    $\begingroup$ In some sense the total energy is just the sum of the different occupied eigneenergies. for N=1 you can occupy either one of the 3 eigen states with energy $-\delta$ or the one of energy $3\delta$ For N=2 you can occupy two state of energy -1 or one of energy -1 and one of energy +3, or 2 of energy +3.... The degeneracy then comes form the possible way of doing this, taking into account that fermion are undistinguishable By the way, I just spotted a mistake since the possible energies the eigenenergies are either -1 or 3 and not -1 or 1: I corrected the text above $\endgroup$ Nov 19, 2021 at 16:15

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