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I am currently going through the Kitaev paper on quantum wires [1].

Given a Hamiltonian of the form $H = 1/2 i \epsilon b' b''$, where $b'$ and $b''$ are majorana modes, we can solve it by introducing two fermionic modes $a = (b'+ i b'') /2$ and $a^\dagger = (b' - i b'') / 2$, yielding $H= \epsilon (a^\dagger a -1/2)$.

I am currently wondering if the choice of $a$ and $a^\dagger$ is arbitrary to the extent that i could also choose $a = (b' - i b'')/2$ and $a^\dagger$ accordingly. This choice also diagonalizes the Hamiltonian, but yields the negative spectrum.

So, which is the "correct" choice? Is there any constraint or line of thought i am missing?

[1] http://arxiv.org/abs/cond-mat/0010440

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Both choices are correct because the spectrum is particle-hole symmetric for Majorana bilinear Hamiltonian, meaning the spectrum is invariant under $\epsilon\to-\epsilon$.

More precisely, for any Majorana bilinear Hamiltonian $$H=\frac{1}{2}\sum_{i,j}\text{i}A_{ij}\chi_i\chi_j,$$ where $\chi_i$ denotes the Majorana fermion operator. The matrix $A$ must be real and antisymmetric ($A^\intercal=-A$). It is a mathematical fact that such matrix can be brought to $2\times2$ blocks $A_n$ under orthogonal transformation $O^\intercal A O=\oplus_n A_n$, where each block is of the form $$A_n=\epsilon_n\left[\begin{matrix}0&-1\\1&0\end{matrix}\right],$$ with real positive $\epsilon_n$. This means all single-particle levels of the Hamiltonian comes in $\pm\epsilon_n$ pairs. So negating the spectrum simply maps the positive levels to negative levels and the whole spectrum remains unchanged, which is a consequence of the particle-hole symmetry. The Hamiltonian can be rewritten as $$H=\sum_{n}\epsilon_n(\psi_n^\dagger\psi_n-1/2).$$ Under the particle-hole transformation: $$\epsilon_n\to-\epsilon_n, \psi_n\leftrightarrow\psi_n^\dagger,$$ the Hamiltonian remains invariant. So we are free to redefine the complex fermion operator $\psi_n$ to its conjugate, the physics will remain the same, as long as we also remember to negate the energy level $\epsilon_n\to-\epsilon_n$. This is a special property of fermions, that one can not discriminate between particles and holes. Which particle should be called electron/positron is merely a matter of choice.

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    $\begingroup$ Hello, I just stumbled over this when looking at Majorana Hamiltonians. I wonder why we can set the diagonal of $A_{ij}$ to zero? I can derive the constraint of being real antisymmetric for all off-diag elements but not for the diagonal. I suppose it it because the Majoranas square to 1 and we can thus neglect the constant off-set? $\endgroup$
    – Marsl
    Dec 17, 2018 at 10:46

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