# Is BdG transformation unitarily equivalent?

Consider the fermionic Hamiltonian of the form $$H = \sum_{i,j=1}^n\left( \alpha_{ij} c^\dagger_i c_j + \frac12 \gamma_{ij} c^\dagger_i c^\dagger_j + \frac12 \gamma^*_{ji}c_ic_j \right).$$ In BdG formalism, we use the transformation of the form $$a_i = A_{ij} c_j + B_{ij} c_j^\dagger, \quad i=1,\ldots, n,$$ with the matrix $$A_{ij}$$ and $$B_{ij}$$ is defined suitably to preserve the anticommutation relation and to make the Hamiltonian $$H = \sum_{i=1}^n E_i \left( a^\dagger_i a_i - \frac12\right) + \mathrm{const}$$ in the simple form. This transformation is always possible (Theorem 38 of arXiv:0908.0787).

However, finding the transformed Hamiltonian is not the end of the problem. What we are really interested is the eigenvalues and eigenstates of $$H$$. To find them, a natural idea is the following:

1. Try to find the ground state, characterized by the condition $$a_i |\psi\rangle = 0$$ for each $$i=1,\ldots, n$$.

2. Obtain the excited states by acting $$a^\dagger_i$$ on $$|\psi\rangle$$.

However, this approach gives me two questions:

1. Is the characterization in 1. well-defined? In other words, does there exist $$|\psi\rangle$$ that satisfies $$a_i |\psi\rangle = 0$$? Also, it would be much better if there is a simpler way to obtain the ground state, by not solving $$a_i |\psi\rangle = 0$$.

2. Can all the eigenvectors be obtained by procedure 2.?

It would be best if there is a unitary operator $$U$$ such that $$U a_i U^{-1} = c_i$$. In this case, the answers of 1. and 2. is automatically positive. Does such $$U$$ exist? If so, how can one find $$U$$?

Any help will be appreciated, although it partially answers my question.

Indeed, it is true that one can find a unitary $$U$$ to relate $$a$$ to $$c$$. To find this unitary, it is most convenient to write everything in the Majorana basis: $$c_i=\frac{1}{2}(\chi_i+i\eta_i)$$, where $$\chi_i^2=\eta_i^2=1$$ are Majorana operators. Call them $$\gamma_a, a=1,2,\dots,2n$$ （not to be confused with the $$\gamma_{ij}$$ parameters in your Hamiltonian), then the Hamiltonian can be rewritten as $$H=\frac{i}{2}\Gamma^T A\Gamma$$, where $$A$$ is real skew-symmetric matrix, and $$\Gamma$$ is the column vector $$(\gamma_1,\gamma_2,\dots,\gamma_{2n})^T$$. Then the Bogoliubov transform means that one finds a SO$$(2n)$$ transformation $$V$$ on $$\gamma_a$$'s: $$\Gamma'=V\Gamma$$ ($$\Gamma'$$ is basically equivalent to your $$a$$, written in Majorana basis), such that $$A$$ is "diagonalized" into the following form:
$$VAV^T=\begin{pmatrix} 0 & \theta_1 & & &\\ -\theta_1 & 0 & & &\\ & &\ddots & &\\ & & & 0 & \theta_n\\ & & & -\theta_n & 0 \end{pmatrix}$$ Here $$\theta_i>0$$ (assuming no zero eigenvalues for simplicity). These $$\theta_i$$'s are basically the energy $$E_i$$'s, but let me use a different notation for clarity. The diagonalization can always be done. Then we define a unitary transformation
$$U=\prod_{j=1}^n \left(\cos\frac{\theta_j}{2}+\sin\frac{\theta_j}{2}\Gamma'_{2j-1}\Gamma'_{2j}\right)$$
This $$U$$ satisfies $$U\Gamma U^\dagger = \Gamma'$$. There is a more compact expression for $$U$$:
$$U=\exp \left( \frac{1}{4} \Gamma^T D \Gamma\right)$$ where the matrix $$D$$ satisfies $$e^D=V$$.