# Can I always find an unitary operator $B$ such that $\{A,B\}= 0$ for a given, unitary operator $A$?

Considering an arbitrary unitary operator $A$, what is the least criteria this operator must satisfy in order that it is possible to find at least another unitary operator $B$ that anti-commutes with it ? Is there a generic way to build this operator $B$ ?

For example, if one consider a displacement operator $\mathcal{D}(\alpha_1) = \exp(\alpha_1\hat{a}^{\dagger}-\alpha_1^*\hat{a})$, then, if $\alpha_1 \neq 0$, it is always possible to find a displacement operator that is anti-commuting with it. If $\mathrm{Im}(\alpha_1\alpha_2^*) = \pm\pi/2$, then we have $\{\mathcal{D}(\alpha_1),\mathcal{D}(\alpha_2)\} = 0$.

• You're already given a counterexample with the identity, so how could you be interested in a proof that it's always possible? What's the exact question here? – ACuriousMind Mar 9 '16 at 18:43
• maybe you should ask the least criteria that any linear operator must satisfy, in order that it is possible to find at least another operator that anti commutes with it. – Sudeepan Datta Mar 9 '16 at 20:11
• To reopen this question (v4), consider to make a precise statement without simple counterexamples, plus consider to harmonize title and main body. – Qmechanic Mar 10 '16 at 11:26

Case 1: Operators over finite dimensional space:

First, assume operators are from a finite dimensional vector space named $V$ to $V$. Let $N=dim(V)$.

We are seeking a unitary operator that anti-commutes with $A$. Let $\{e_i\}$ an orthogonal basis for $V$ and $\sigma$ a permutation from $\{1,...,N\}$ to itself. Define $B$ by $B(e_i)=e_{\sigma(i)}$. It can be shown that $B$ is unitary since its matrix columns are perpendicular. Using this motivation, we find a necessary and sufficient condition for the question. The condition is:

if $\lambda_i$ is an eigenvalue of $A$ with $K_i$ independent eigenvectors, then there is a unitary operator that anti-commutes with $A$ if and only if $-\lambda_i$ is an eigenvalue of $A$ with exactly $K_i$ independent eigenvectors.

Now let's prove it. First consider $A$ meets the condition. Let $\{\lambda_1,...,\lambda_k\}$ positive eigenvalues of $A$. Let's show orthogonal eigenvectors of $\pm\lambda_i$ with $e^{\pm}_{i,j}$ that $j\in\{1,...,K_i\}$. Define $B$ by $B(e^{\pm}_{i,j})=e^{\mp}_{i,j}$. Indeed, we have defined a permutation from $\{1,...,N\}$ and so $B$ is unitary. $B$ anti-commutes with $A$ iff for every $e^{\pm}_{i,j}$, $AB(e^{\pm}_{i,j})+BA(e^{\pm}_{i,j})=0$. It can be seen:

$AB(e^{\pm}_{i,j})=A(e^{\mp}_{i,j})=\mp\lambda_ie^{\mp}_{i,j}$

$BA(e^{\pm}_{i,j})=\pm\lambda_iB(e^{\pm}_{i,j})=\pm\lambda_ie^{\mp}_{i,j}$

Thus, $B$ is unitary and anti-commutes with $A$.

Now consider there is a unitary operator named $B$ that anti-commutes with $A$. $A$ is unitary, so it is normal operator means $A^{\dagger}A=AA^{\dagger}$. Thus it can be represented in a diagonal form with eigenvalues $\lambda_i$ and eigenvectors $\{e_i\}$ that is $i\in\{1,...,N\}$. None of these eigenvalues can be zero since $A$ is invertible. $B$ anti-commutes with $A$, so $AB(e_i)+BA(e_i)=0\Rightarrow (A+\lambda_iI)B(e_i)=0$. $B(e_i)$ is not zero since $B$ is unitary and has inverse. So, $A-(-\lambda_i)I$ is not invertible and so $-\lambda_i$ is eigenvalue of $A$. Now, assume there are $K_i$ independent eigenvectors for $\lambda_i$ and there are $K^{\prime}_i$ independent eigenvectors for $-\lambda_i$. Since $B$ is invertible, so $det(B)\neq0$ and so $B(e_i)$s are independent. Using the above notation, we know $B(e^{+}_{i,j})$ that $j\in\{1,...,K_i\}$ is eigenvector for $-\lambda_i$ and these are independent. So, $K_i\leq K^{\prime}_i$. By the same argument, it can be shown $K^{\prime}_i\leq K_i$. Hence, $K^{\prime}_i=K_i$ and the proof is finished.

Case 2: Operators over a separable Hilbert space:

Now, assume operators are from a separable Hilbert space named $\mathcal{H}$ to $\mathcal{H}$ (over real or complex field). It means they have a denumerable set that the set of all finite compositions of its elements is dense in $\mathcal{H}$. In fact, previous case is a special case of this case since a (finite) basis for $V$ is such a set.

$A$ is unitary, so it is normal. The spectral theorem states there is an orthonormal basis of eigenvectors of $A$. In the other words, the set of all finite compositions of eigenvectors of $A$ is dense in $\mathcal{H}$. $A$ is normal and by definition a normal operator is a continuous operator that commutes with its adjoint. We are seeking another unitary (thus continuous and normal) operator like $B$ that anti-commutes with $A$. Consider $D$ a dense subset of $\mathcal{H}$. It is known that a continuous map from a set $X$ to itself can be uniquely determined by its value over $D$. So it is sufficient to determine $B$ over a dense subset of $\mathcal{H}$. It means it is sufficient $B$ anti-commutes with $A$ over $D$. By this general discussion, we can state a similiar necessary and sufficient condition holds in this case:

if $\lambda_i$ is an eigenvalue of $A$ with $K_i=card(M_{\lambda_i})$ where $M_{\lambda_i}$ is a (linear algebra) basis for eigenspace of $\lambda_i$, then there is a unitary operator that anti-commutes with $A$ if and only if $-\lambda_i$ is an eigenvalue of $A$ which $K_i=card(M_{-\lambda_i})$.

First consider $A$ meets the condition. $A$ by spectral theorem has a orthogonal set like $E$ that all finite compositions of $E$ is dense in $\mathcal{H}$. Then define $B$ in the same manner we did in the previous case by $e_i$s in accordance to $E$. It can be done since it concerns only finite compositions of $E$. It can be verified that $B$ is unitary and anti-commutes with $A$.

Conversely, consider there is a unitary operator anti-commutes with $A$. Again we can show $-\lambda_i$ is eigenvalue of $A$ since $B$ is invertible. $B(e_i)$s are independent (due to finite sums) because $B$ is injective. Thus, one can make a bijection from $M_{\lambda_i}$ to $M_{-\lambda_i}$ and so show they have the same cardinality.

Remark: I stated the latter condition using cardinality. $A$ is injective so does not have zero eigenvalue. As I know, eigenspaces for non-zero eigenvalues of self-adjoint operators, are finite so it is not necessary to use cardinality. But I am not sure the same holds for unitary (or normal) operators. So I used cardinality.

• Thanks for your awnser. Do you know if there is any way to extend this proof to the case of infinite dimension, countable or uncountable ? – Dimitri Mar 16 '16 at 9:40
• @Dimitri I edited my answer for separable Hilbert space. Hope it be helpful. – Kiarash Mar 16 '16 at 22:45
• Is it possible to use the same kind of reasoning when the dimension of the space is uncountable ? – qdr Mar 30 '16 at 14:52