I have a number operator $a^\dagger a + b^\dagger b$, where $a^\dagger$ and $b^\dagger$ are fermion operators. If a unitary transformation $U$ is performed, the number operator is written in the new fermion operators $(\tilde{a},\tilde{b})^T = U(a,b)^T$, $\tilde{a}^\dagger \tilde{a} + \tilde{b}^\dagger \tilde{b}$. It says that the number of fermion is identical regardless of the fermion basis related by unitary transformations.

However it is not the case for a Bogoliubov transformation. For example, if we transform the fermion operator in the following way $(\tilde{a}^\dagger, \tilde{b}^\dagger, \tilde{a}, \tilde{b})^T = M (a^\dagger, b^\dagger, a, b)^T$ with $$ M = \frac{1}{2}\begin{bmatrix} 1 & -1 & 1 & 1 \\ 1 & 1 & -1 & 1 \\ 1 & 1 & 1 & -1 \\ -1 & 1 & 1 & 1 \\ \end{bmatrix}. $$

$M$ is a unitary matrix or exactly an orthogonal matrix. $MM^\dagger = I$ ensures the anti-commutation relation. The number operator can be written as $\frac{1}{2}(a^\dagger,b^\dagger,a,b)diag(1,1,-1,-1)(a^\dagger,b^\dagger,a,b)^T+1$ or

$$\frac{1}{2}(a^\dagger,b^\dagger,a,b)M^\dagger Mdiag(1,1,-1,-1)M^\dagger M(a^\dagger,b^\dagger,a,b)^T+1 = \\ \frac{1}{2}(\tilde{a}^\dagger,\tilde{b}^\dagger,\tilde{a},\tilde{b})Mdiag(1,1,-1,-1)M^\dagger(\tilde{a}^\dagger, \tilde{b}^\dagger,\tilde{a},\tilde{b})^T+1.$$

Result of $Mdiag(1,1,-1,-1)M^\dagger$ has off diagonal elements $$ \begin{bmatrix} 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ \end{bmatrix}. $$

The number operator after Bogoliubov transformation is not a form of number operator anymore due to the off diagonal elements. That means fermion number is different after the Bogoliubov transformation. It also says that fermionic parity operator $(-1)^N$ is not conserved after Bogoliubov transformation.

It is strange for me. I think that at least the fermionic parity operator should be conserved. So how to understand it physically?


1 Answer 1


There are a couple of inaccuracies in your question.

If you have $N$ fermions with annihilation operators $a_1,…,a_N$ satisfying by definition: $$ \{a_i,a_j^\dagger\}=\delta_{ij} $$ and are transformed to $\tilde a_1,…,\tilde a_N$ by a Bogoliubov transformation: $$ \begin{align} \begin{pmatrix} \tilde a^\dagger \\ \tilde a \end{pmatrix} &= M\begin{pmatrix}a^\dagger \\ a\end{pmatrix} & M &=\begin{pmatrix}U^* & V^* \\ V & U\end{pmatrix} \end{align} $$ then the canonical anti-commutation relations are preserved iff: $$ \begin{align} UV^T+VU^T&=0 & UU^\dagger+VV^\dagger &= I \end{align} $$ or in terms of $M$: $$ M \begin{pmatrix}0 & I\\ I & 0\end{pmatrix} M^T=\begin{pmatrix}0 & I\\ I & 0\end{pmatrix} \tag{*} $$ Note that the criterion is neither $M$ to be unitary nor is it to be orthogonal. However, if you assume $V=0$ then the conditions are equivalent to $U$ being unitary. In your case: $$ \begin{align} U &= \frac{1}{2}\begin{pmatrix} 1 & -1\\ 1 & 1\end{pmatrix} & V &= \frac{1}{2} \begin{pmatrix}1 & 1\\ -1 & 1\end{pmatrix} \end{align} $$ using that $V=U^T$, you can check that the previous identities hold so the CAR’s are preserved.

The issue is that the total number operator corresponds to: $$ N=\begin{pmatrix} a^\dagger & a \end{pmatrix}\begin{pmatrix}0 & I\\ -I & 0\end{pmatrix}\begin{pmatrix} a^\dagger \\ a \end{pmatrix} $$ so if your Bogoliubov transformation preserves the total number operator iff: $$ M^T\begin{pmatrix}0 & I\\ -I & 0\end{pmatrix}M=\begin{pmatrix}0 & I\\ -I & 0\end{pmatrix} \tag{**} $$ or equivalently: $$ \begin{align} V^\dagger U - U^TV^* &= 0 & U^\dagger U-V^T V^* &= I \end{align} $$ Note that $M$ being unitary is neither sufficient nor necessary for preserving the number operator. What is true though is when $V=0$, then the condition is equivalent to $U$ being unitary, which is already imposed to preserve the CAR’s.

You can check by direct computation that this is not the case for your example. In general, the number operator is not preserved by a Bugoliubov (this is actually the reason why they are introduced), so it is hardly surprising.

Hope this helps.

Answer to comments

The group of Bogoliubov transformation preserving the CAR's is isomorphic to $O(2N)$. This can be seen directly by $(*)$ which is a standard alternative definition used in the study of root systems and Cartan subalgebras. One way of recovering the standard definition is by looking at Majorana fermions: $$ \begin{align} \gamma_i &= \frac{a_i+a_i^\dagger}{\sqrt 2} & \gamma_{N+i} &= \frac{a_i-a_i^\dagger}{i\sqrt 2} \end{align} $$ so that you have $\gamma_1,...\gamma_{2N}$ hermitian operators ($\gamma^\dagger = \gamma$) satisfying: $$ \{\gamma_i,\gamma_j\} = \delta_{ij} $$ A Bogoliubov transform is now given by a real matrix $O$: $$ \tilde\gamma = O\gamma $$ and to preserve the CAR's: $$ OO^T = I $$ i.e. $O$ is orthogonal. In particular, the orthogonal group has finite dimensional unitary representations. It can therefore be represented by unitary operators on a finite dimensional Hilbert space representing the finite number of fermionic modes. In fact, given $M$ or $O$ as an exponential, it is not hard to construct the corresponding unitary operator as the fermionic analogue of a squeeze operator.

The Bogoliubov transformations preserving the number operators do satisfy the same identity as the symplectic group $Sp(2N,\mathbb C)$ (see $(**)$), however due to the preservation of CAR's, they are actually a strict subgroup. In fact, they are isomorphic to a subgroup of $O(2N)$ (in particular, this means it is compact), from the previous discussion, so still have finite dimensional unitary representations.

Once again, the use of Majorana fermions brings insight to this. The number operator is now: $$ N \propto \gamma ^T \begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix}\gamma $$ so its preservation is equivalent to: $$ O^T\begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix} O = \begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix} $$ i.e. $O\in Sp(2N,\mathbb R)$. The transformations preserving the CAR's and the total number operator is therefore $Sp(2N,\mathbb R)\cap O(2N)$ in the defining representation, so is isomorphic to $U(N)$. These are therefore exactly given by the transformations where $V=0$, physically, the transformations that do not mix particles and holes.

Note that for boson modes, the results are reversed. The Bogoliubov transformations preserving the CCR's is now isomorphic to the symplectic group $Sp(2N,\mathbb R)$. This time, there are no unitary finite dimensional representations, but this is to be expected since for a representations of the CCR's you need an infinite dimensional Hilbert space. This time, the Bogoliubov transformations preserving the number operator are a subgroup of $O(2N)$ (subgroup since they must additionally preserve the CAR's). Once again, preserving both gives you the transformations forming a group isomorphic to $U(N)$ where the particles and holes are not mixed.

  • $\begingroup$ The fermionic parity operator should be preserved by a Bogoliubov transformation, right? $\endgroup$
    – lsdragon
    Jul 7, 2023 at 7:04
  • $\begingroup$ In fact, the relations that you given to preserve anti-commutation relation is equivalent to have an unitary $M$. You can directly check it by multiplication and using the relations you derived. $\endgroup$
    – lsdragon
    Jul 7, 2023 at 7:15
  • $\begingroup$ For the preservation of CAR’s, while $M$ acts as a unitary operator on the Hilbert space, the matrix $M$ is not unitary in general but rather satisfies $(*)$. You can easily construct counterexamples with two modes. $\endgroup$
    – LPZ
    Jul 7, 2023 at 11:13
  • 1
    $\begingroup$ @lsdragon $M$ must be symplectic and there are no unitary representation of real symplectic matrices in finite dimensional space. $M$ could be unitary in the infinite dimensional Hilbert space of states, but not in the finite dimensional space spanned by your operators. (Basically same ideas as with the Lorentz group.) $\endgroup$ Jul 7, 2023 at 12:49

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