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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.

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.

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.

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: $$ M^T\begin{pmatrix}0 & I\\ -I & 0\end{pmatrix}M=\begin{pmatrix}0 & I\\ -I & 0\end{pmatrix} $$ 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.

The Bogoliubov transformations preserving the number operators do satisfy the same identity as the symplectic group $Sp(2N,\mathbb C)$, 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.

Note that for boson modes, the results are almost 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).

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.

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.

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: $$ M^T\begin{pmatrix}0 & I\\ -I & 0\end{pmatrix}M=\begin{pmatrix}0 & I\\ -I & 0\end{pmatrix} $$ 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.

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: $$ M^T\begin{pmatrix}0 & I\\ -I & 0\end{pmatrix}M=\begin{pmatrix}0 & I\\ -I & 0\end{pmatrix} $$ 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.

The Bogoliubov transformations preserving the number operators do satisfy the same identity as the symplectic group $Sp(2N,\mathbb C)$, 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.

Note that for boson modes, the results are almost 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).