# Bogoliubov transformation for fermion (exercise in Piers Coleman)

I am trying to solve the exercise 3.2 in Piers Coleman's Introduction to many body physics. It's about fermionic Bogoliubov transformation with only 2 fermion operators $a_{1}^{\dagger}$, $a_{2}^{\dagger}$, $a_{1}$, $a_{2}$.

The canonical transformation is \begin{align} & c_{1} = u a_{1} + va_{2}^{\dagger} \\ &c_{2}^{\dagger} = -va_{1} + ua_{2}^{\dagger} \end{align}

The starting Hamiltonian is \begin{equation} H = \epsilon(a_{1}^{\dagger}a_{1} - a_{2}a_{2}^{\dagger}) + \Delta(a_{1}^{\dagger}a_{2}^{\dagger} + H.C.) \end{equation}

I have already successfully transformed the above $H$ into \begin{equation} H = \sqrt{\epsilon^{2} + \Delta^{2}} (c_{1}^{\dagger}c_{1} + c_{2}^{\dagger}c_{2} - 1) \end{equation} where $u=cos\theta$ and $v=sin\theta$ with $\theta$ satisfies $tan2\theta = \frac{\Delta}{\epsilon}$.

I am now struggling with finding the ground state of the Hamiltonian in terms of (1) vacuum state of $a_{1}$ and $a_{2}$ (2) $a_{1}^{\dagger}$ and (3) $a_{2}^{\dagger}$.

I know that the ground state energy is $-\sqrt{\epsilon^{2} + \Delta^{2}}$ and the ground state should also satisfies \begin{align} & c_{1}\left| {G} \right\rangle = 0\\ & c_{2}\left| {G} \right\rangle = 0 \end{align}

I have browsed some website and it seems that it is possible that I assume the ground state as \begin{equation} \left| {G} \right\rangle = x \left| {0,0} \right\rangle + y \left| {1,1} \right\rangle \end{equation} where \begin{align} & a_{1} \left| {0,0} \right\rangle = 0\\ & a_{2} \left| {0,0} \right\rangle = 0\\ & a_{1}^{\dagger} a_{2}^{\dagger} \left| {0,0} \right\rangle = \left| {1,1} \right\rangle \end{align}

But I found that I can not get a proper $\{x,y\}$ coefficients to make \begin{equation} \left| {G} \right\rangle = x \left| {0,0} \right\rangle + y \left| {1,1} \right\rangle \end{equation} well-defined.

I think this is a rather easy exercise but I just can not find the ground state haha. I will be extremely grateful for any suggestion! :)

Edit :

It turns out that I find that answer and it is rather easy, as follow.

We want to find a state $\left| {G} \right\rangle$ that satisfies \begin{align} & c_{1} \left| {G} \right\rangle = 0\\ & c_{2} \left| {G} \right\rangle = 0 \end{align}

The key properties are $\{ c_{1},c_{1}\} = 0$ & $\{ c_{2},c_{2}\} = 0$.

We now construct a state $\left| {\psi} \right\rangle = c_{1}c_{2} \left| {0} \right\rangle$ where $a_{1}\left| {0} \right\rangle = a_{2} \left| {0} \right\rangle = 0$, meaning that $\left| {0} \right\rangle$ is the vaccum w.r.t the original fermionic operators $a_{1}$ & $a_{2}$.

Let's see whether this trial $\left| {\psi} \right\rangle$ can satisfy what we require for the ground state.

First, we apply $c_{1}$ on $\left| {\psi} \right\rangle$ and we have $c_{1}c_{1}c_{2} \left| {0} \right\rangle = 0$ according to $\{ c_{1},c_{1}\} = 0$.

Similarly we apply $c_{2}$ on $\left| {\psi} \right\rangle$ and we again get $0$.

This proves that $\left| {\psi} \right\rangle = c_{1}c_{2} \left| {0} \right\rangle$ is the $\left| {G} \right\rangle$ we are searching for.

So, if we write $\left| {G} \right\rangle$ explicitly, it will be \begin{align} c_{1}c_{2} \left| {0} \right\rangle &= (u a_{1} + va_{2}^{\dagger})(-va_{1}^{\dagger} + ua_{2}) \left| {0} \right\rangle \\ & = ( -uva_{1}a_{1}^{\dagger} + u^{2} a_{1}a_{2} -v^{2} a_{2}^{\dagger}a_{1}^{\dagger}+uva_{2}^{\dagger}a_{2} ) \left| {0} \right\rangle\\ & = (-uv - v^{2}a_{2}^{\dagger}a_{1}^{\dagger} )\left| {0} \right\rangle \\ & \propto (u + va_{2}^{\dagger}a_{1}^{\dagger} )\left| {0} \right\rangle \end{align}

By acting the Hamiltonian on the $\left| {G} \right\rangle$ we can easily confirm the ground state energy is exactly $-\sqrt{\epsilon^{2} + \Delta^{2}}$

I think the reason that I can not get the correct coefficient is because I view the $a_{2}^{\dagger}a_{1}^{\dagger} \left| {0} \right\rangle$ as $\left| {1,1} \right\rangle$. However, because of the anticommute properties for fermion, the arrangement of the operator, and the $+$ or $-$ sign become very important. So it will be safer to write $a_{2}^{\dagger}a_{1}^{\dagger} \left| {0} \right\rangle$ rather than $\left| {1,1} \right\rangle$.

## 1 Answer

Take $\psi=x \psi_{00}+w \psi_{01}+z\psi_{10}+y\psi_{11}$ then apply $c_1, c_2$ set equal zero. Then you get $w=z=0$ and one of $x,y$, and remember you need normalization condition to determine the last one.