# Fermionic operators

1. Oct 29, 2014

Hi ! I have a doubt about fermionic operators with the anticommutation relations. I know they follow anticommutation, that is,

\begin{eqnarray}
\lbrace c_{i}^{\dagger},c_{j}\rbrace=\delta_{i,j}
\end{eqnarray}

That is for fermionic operators. But, suppose I have two different kind of fermionic operators, i.e, one refers to the left "wire" of a system, and the other to the right "wire".
\begin{eqnarray}
c_{i, L}^{\dagger} c_{j,R}^{\dagger}
\end{eqnarray}
Here, L denotes that this fermionic operator act on the left wire of a system, while R means the right wire.

My question is, do this operators COMMUTE? I mean, because they are acting on different parts of a system (like a subsystem), do they commute? Iam getting confuse about it! The thing is, can I do this?

\begin{eqnarray}
[ c_{i,L}^{\dagger},c_{j,R}]=0
\end{eqnarray}

2. Oct 30, 2014

### DrDu

As long as the systems are completely isolated from each other, it doesn't matter whether you assume commutation or anti-commutation relations. However, whenever the electrons from the left and right subsystem interact with each other, it is important to use anti-commutation relations.

3. Oct 30, 2014

That is the thing, that I have a hamiltonian of the type:
\begin{eqnarray}
H=\sum_{i} e_{L,i}c_{i,L}^{\dagger}c_{i,L} + e_{i,R}c_{i,R}^{\dagger}c_{i,R} - t(c_{L,0}^{\dagger}c_{R,0} + c_{R,0}^{\dagger}c_{L,0} + h.c)
\end{eqnarray}
where h.c means complex conjugate. If I use anticommutation at point x=0 for the two creation and annihilation operators, that is, with the " -t" term, this part is null. Any ideas?

4. Oct 30, 2014

### DrDu

What do you want to calculate?

5. Oct 30, 2014

The idea is to diagonalise this part of the Hamiltonian, so I tried to do a change of basis for new fermionic operators, when it becomes diagonal, and I failed. The thing is that having diagonalised this part, then I want to treat interaction terms using Green´s function. However, because couldn´t diagonalise it first, I just calculate the interaction Green function supposing that the "t" term is my interaction, and having the rest as Ho. However, the calculated Green function diverges when I try to integrate it in time, because I only got two non interacting Green functions, and time dependence cancels out. Because of that, I think the "t" term can´t be treated as an interaction, and there should be one way to diagonalise the hamiltonian of the expression above all together

6. Oct 30, 2014

### DrDu

I can't see any reason, why your hamiltonian should not be diagonalizable. What transformation did you try?

7. Oct 30, 2014

### Jilang

If the particles are in different wires should they not be treated as distinguishable? Is wave function overlap the same as interaction? I don't think so so, but I stand to be corrected.

8. Oct 30, 2014

### DrDu

For i=0 you obviously have some interaction of L and R in the hamiltonian.
You can write this in matrix form $(c_{L0}^\dagger, c_{R0}^\dagger) \begin{pmatrix}e_{L0}&-t \\ -t &e_{R0} \end{pmatrix} (c_{L0}, c_{R0})^T$. If you call the matrix in the center H which is diagonalized by the unitary matrix V, i.e. $H=VV^\dagger H V^\dagger V$, then V obiously also transforms the creation and anihilation operators as $V(c_{L0}, c_{R0})^T$.

9. Oct 30, 2014

Yes, to concrete, the equivalent problem in 1D corresponds with a Dirac delta function at x=0, so obtaining the eigenfunctions is something that can be easily done for one particle. Thus, I will write the hamiltonian in the following way
\begin{eqnarray}
H=\sum_{k1}{e_{k1,L}c_{k1,L}^{\dagger}c_{k1,L}} + \sum_{k2}{e_{k2,R}c_{k2,R}^{\dagger}c_{k2,R}} - t\sum_{k1,k2}T_{k1,k2} (c_{k1,L}^{\dagger}c_{k2,R} + c_{k2,R}^{\dagger}c_{k1,L} )
\end{eqnarray}
Here T_{k1}{k2} is the transmission coefficient calculated with the continuity conditions. What I would like to do now is to find a transformation to diagonalize this hamiltonian, but it seems to be non trivial ...Ideas?

10. Oct 30, 2014

Sorry I forgot to say that I changed to momentum basis. Also, consider the hamiltonian as this because, at the beginning of the post, I think I add more terms in the "t" hopping. That is, the idea is that the wires are only "separated" at x=0, which implies a delta function, but it exists some probability of hopping from the left side to the right side and viceversa. That`s what the second term means.

11. Oct 31, 2014

### DrDu

I understand your hamiltonian. But as in my former post you just have to diagonalize a matrix with the e's on the diagonal and terms -tT on outer diagonal positions. You probably have to do this numerically in the general case, but the unitary matrix which diagonalizes this matrix will also transform from the old to new creation operators.

12. Oct 31, 2014

I will try some transformation between the operators and I will see. Thanks!

13. Nov 2, 2014

Ok, coming back to the topic, I have tried several transformations between the creation and destruction operators and I got nothing. I understand what you say. in this case, the hamiltonian, as it appears above, is just the product of a "row vector" that contains the creation and destruction operator, by a matrix, times a "column vector" that contains also the creation and destruction operators, but it is the transposed conjugate of the row vector. The thing is that, if I do it (it is a 4x4 matrix) I get the eigenvalues andeigenvectors with Mathematica... But the thing is that thesenew creation and destruction operators (new basis) depend on the others, the ones which form the previous basis. But, because these new operators are also fermionic operators, they must satisfy anticommutation relations, and with this transformation it seems that they don´t. Actually, mathematica gives me, for the new creator operator, a linear combination of the previous creation operators... But this can be understood as a Bogoliubov transformation for the new operators.

But Bogoliubov transformation can´t be applied to fermionic operators... So the problem is still there!

14. Nov 3, 2014

### DrDu

Yes, you can view this as a Bogoliubov transformation, although it doesn't mix creation and anihilation operators with each other. I don't quite understand why you say that Bogoliubov's method isn't applicable to fermionic operators. In fact, the classical example of the B transfomation occurs in the theory of superconductivity where it is applied to describe the fermionic excitations in a superconductor.
As you correctly observed, the new operators are linear combinations of the old ones and linear unitary transformations preserve the anti-commutation relations: Let $b_i=V_{ij}c_j$ and $b^\dagger_k=c^\dagger_l V^\dagger_{lk}$ (summation over equal indices is understood), then $\{b^\dagger_k, b_i\}=b^\dagger_k b_i+b_i b^\dagger_k=V_{ij}V^\dagger_{lk}\{c^\dagger_l,c_j\}=V_{ij}V^\dagger_{lk}\delta_{jl}=(VV^\dagger)_{ik}=\delta_{ik}$. In the last step, we use that V is unitary.

15. Nov 3, 2014

Yes but the thing is that I have 4 different operators... Two correspond to the left side (left wire) and the other two correspond to the right side, and I don't know how to combine them to diagonalize the hamiltonian.. As you can see, they are crossed in the second term, and also, there is a double summation

16. Nov 3, 2014

### DrDu

Doesn't look so complicated. E.g. you could use for the left vector: $(\ldots c^\dagger_{kL}c^\dagger_{kR}\ldots)$ and accordingly for the right one $(\ldots c_{kL}c_{kR}\ldots)^T$.
Then the matrix H in the middle has 2x2 blocks on the diagonal (i.e. for $k_1=k_2=k$) $\begin{pmatrix}e_{kL}&-tT_{kk}\\-tT_{kk}&e_{kR}\end{pmatrix}$ and away from the diagonal $\begin{pmatrix}0&-tT_{k_1k_2}\\-tT_{k_1k_2}&0\end{pmatrix}$.

17. Nov 3, 2014