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I work on the Kitaev toy model for Majorana fermions. He writes that in Majorana basis the Hamiltonian becomes in general

$$ H = \frac{i}{4} \sum_{l,m} A_{l,m}\gamma_{l}\gamma_{m} $$

where $\gamma$ is the creation operator for Majorana fermions and $A$ is a skew-symmetrix matrix. Now he say that the ground state of this Hamiltonian has an even fermionic parity for the case without Majorana fermions at the end of the wire.

To calculate the eigenvalues of the Hamiltonian we can bring the Hamiltonian in block diagonal form with a Transformation $WAW^{T}$ where W is a real orthogonal matrix whose rows are eigenvectors of A. If $W$ has the form $W = e^{D}$ for some skew-symmetrix matrix $D$ or if $\rm{det}(W) = +1$ the parity doesn't change. Otherwise the Transformation $WAW^{T}$ changes the parity of the ground state (here now $\rm{det}(W) = -1$).

My question is now: Why changes the parity of the ground state when $\rm{det}(W) = -1$? Further, what is the reason for the even parity of the ground state (he writes that the ground state is superposition of states with even number of electrons but what happens when when the number of electrons is odd)?

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1 Answer 1

a) The parity of the ground state is the sign of the pfaffian of $A$. With the transformation (12), $A'= WA W^t$, you get

pfaffian ($A'$) = det $W$ pfaffian ($A$)

So, if det $W=-1$, then pfaffian $(A')$ = $-$ pfaffian $(A)$, so sgn(pfaffian $(A')$) $= -$sgn(pfaffian $(A')$), so the parity of the ground state is changed.

b) Every (original) hamiltonian can be reduced to its canonical form $(11)$ with this kind of transformation (12). The ground state for an hamiltonian in its canonical form has an even parity. So you may deduce the parity of the ground state for the original hamiltonian.

[EDIT]

One may write (see formulas (1)(2)(5)):

$$(-ic_{2j-1}c_{2j}) = (1 - 2 a_j^+a_j)$$

The ground state $|0\rangle$, for the hamiltonian in the canonical form
$H_c = \sum_j (a_j^+a_j - \frac{1}{2})$, is the usual ground state for fermionic fields, which is anihilated by each $a_j$ : $a_j|0\rangle = 0$.

The action of the parity operator on the ground state is :

$$P|0\rangle = \Pi_j(-ic_{2j-1}c_{2j})|0\rangle =\Pi_j(1 - 2 a_j^+a_j)|0\rangle = |0\rangle$$

So, the ground state has a even parity.

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