In section 2.A.2 of Quantum Ising Phases and Transition in Transverse Ising Models by Suzuki, et. al. the authors give the following in their derivation of the Bogoliubov transformation for a Hamiltonian

$H = \sum_{ij} c_i^\dagger A_{ij} c_j +\frac{1}{2} \sum_{ij}c_i^\dagger B_{ij}c_j^\dagger$

with $A$ Hermitian, and $B$ antisymmetric, and $c_i$ Fermionic operators.

One makes a linear transformation of the form

$$\eta_q = \sum_i \left( g_{qi}c_i+h_{qi}c_i^\dagger \right)$$

$$ \eta_q^\dagger = \sum_i \left( g_{qi}c_i^\dagger + h_{qi}c_i \right)$$

where $g_{qi}$ and $h_{qi}$ can be chosen to be real. For $\eta_q$ to satisfy fermionic anti-commutation relations we require

$$ \sum_i \left( g_{qi}g_{q'i} + h_{qi}h_{q'i} \right) = \delta_{qq'}$$ $$\sum_i \left( g_{qi}h_{q'i} - g_{q'i}h_{qi} \right) = \delta_{qq'}$$

My questions are the following:

  1. I see why, if $B =0$, $A$ being a Hermitian matrix ensures that $H$ is a Hermitian operator. But, I don't see how this Hamiltonian is Hermitian for nonzero $B$.

  2. I do not see how the second equation $$\sum_i \left( g_{qi}h_{q'i} - g_{q'i}h_{qi} \right) = \delta_{qq'}$$ follows. I can see that if comes from $[ \eta_q, \eta_{q'}]_+ = 0$

But, when I make the computation of $[ \eta_q, \eta_{q'}]_+$ I get:

$$\begin{align} [ \eta_q, \eta_{q'}]+ &= [\sum_i \left( g_{qi}c_i+h_{qi}c_i^\dagger \right), \sum_j \left( g_{q'j}c_j+h_{q'j}c_j^\dagger \right) ]_+ \\ &= \sum_{ij} g_{qi}g_{q'j} [c_i,c_j]_+ + g_{qi}h_{q'j} [c_i,c_j^\dagger]_+ h_{qi}g_{q'j} [c_i^\dagger, c_j]_+ + h_{qi}h_{q'j} [c_i^\dagger,c_j^\dagger]_+ \end{align}$$

Then, using, that $[c_i,c_j]_+ = [c_i^\dagger,c_j^\dagger,]_+ = 0$ and $[c_i,c_j^\dagger]_+ = [c_i^\dagger, c_j]_+ = \delta_{ij}I$, we get

$$\sum_i \left( g_{qi}h_{q'i} + g_{q'i}h_{qi} \right) = \delta_{qq'}$$

which is in conflict with the equation above. I could see the equation above following if we had commutation relations, but we're specifically talking about Fermionic operators. Where am I going wrong? Is the book accidentally doing the Bosonic case?


1 Answer 1


I found the answers I was looking for. Answering my questions in order we have:

  1. The Hamiltonian is not Hermitian as written. I omitted the $h.c.$ at the end, because I didn't know what it meant and assumed it was irrelevant. It turns out it was extremely relevant because it meant "Hermtian conjugate," meaning we add in the Hermitian conjugate of whats written. Doing this makes the Hamiltonian Hermitian.

  2. My computation is correct. The result in the book comes from this paper . The paper gives my result, which is actually the result used in the rest of the derivation.

If you are using this book, note that - at least in this section - there are numerous typos. Below this derivation the authors write $$ [ \eta_q, H]_+ - \omega_q \eta_q =0$$

when they mean $$ [ \eta_q, H] - \omega_q \eta_q =0$$

and later they refer to one equation when they mean another. Be aware.


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