# Are antiunitary operators associative?

Is the antiunitary operator used in quantum mechancal time reversal symmetry (and Wigner's Theorem) associative?

If we have $$\Theta AB$$ where $\Theta$ can be decomposed as $$U K$$ where K performs a complex conjugation and U is a unitary operator , and A and B also operators, and we examine both ways to group the calculations, either $$( \Theta A ) B = A^*B$$ or $$\Theta (AB) \equiv \Theta C = C^* = (AB)^* = A^*B^*$$ These results are different, so $\Theta$ is not associative.

If this is the case, which way is implied when doing quantum mechanical calculations?

• $\Theta A = A^*$ has no meaning. Jan 13 '18 at 15:08
• please explain. Jan 15 '18 at 11:14
• What do you mean for $A^*$? Jan 15 '18 at 15:07
• I guess $A^*$ should be a sort of complex conjugated operator. Please try to define it very precisely. Are you dealig with the Hilbert space $L^2$ or some generic abstract Hilber space? Jan 15 '18 at 18:28
• Edited the definition in above Jan 17 '18 at 12:29

This question raises several issues simultaneously requiring a closer scrutiny before answering the question itself.

1. Complex conjugation.

If $${\cal H}$$ is a complex Hilbert space and $$B=\{u_i\}_{i\in I}$$ is a Hilbert basis therein, an antilinear operation is defined $$K_B : \sum_i c_i u_i \mapsto \sum_i c^*_i u_i\tag{1}$$ called complex conjugation associated to $$B$$.

Antilinearity means $$K_B(au+bv) = a^*K_Bu + b^*K_Bv\quad \forall u,v \in {\cal H} \quad \forall a,b \in \mathbb C.$$

It is evident that this definition depends on the choice of $$B$$. If $$B'= \{u'_j\}_{j\in I}$$ is another Hilbert basis of $${\cal H}$$, it turns out that $$K_B=K_{B'} \quad \Leftrightarrow \quad \langle u_i|u'_j\rangle = \langle u_i|u'_j\rangle^* \quad \forall i,j \in I\:.$$

From the definition, we have that $$K_BK_B =I\tag{2}$$ and $$\langle K_B v|K_B u\rangle = \langle v|u\rangle^* \quad \forall u,v \in {\cal H}\:,\tag{3}$$ so that, in particular, $$K_B$$ is isometric because $$||K_Bu||= ||u||\quad \forall u \in {\cal H}\:.$$

As an example, consider $${\cal H}= L^2(\mathbb R, dx)$$, the Hilbert space of a spinless particle living along the real line. A natural conjugation is defined like this. $$K : L^2(\mathbb R, dx) \ni \psi \mapsto \psi^*\:,$$ where $$\psi^*(x) := (\psi(x))^*\quad \forall x \in \mathbb R\:.$$ It is quite simple to prove that $$K=K_B\:,$$ where $$B=\{f_n\}_{n \in \mathbb N}$$ is the Hilbert basis $$f_n(x) = \frac{H_n(x)e^{-x^2/2}}{\sqrt{\pi^{1/2} 2^n n!}}$$ constructed out of real Hermite polynomials $$H_n$$, defining in particular an orthonormal basis of eigenstates of the Hamiltonian of the harmonic oscillator usually denoted by $$f_n = |n\rangle$$. Notice that no complex phases are permitted.

Another possible conjugation is $$K'$$ obtained with the same recipe as $$K$$, but working in momentum picture, $$K' : \psi \mapsto F^{-1} ((F \psi)^*)\:,$$ where $$F(\psi)(p) := \hat{\psi}(p) := \frac{1}{\sqrt{2\pi}} \int e^{-ipx} \psi(x) dx$$ is the standard Fourier-Plancherel transform of $$\psi$$ (I assume $$\hbar=1$$). It is easy to check that $$K\neq K'\:.$$

1. Notion of complex conjugation of an operator

Given a linear operator $$A : D(A) \to {\cal H}$$, where the domain $$D(A)\subset {\cal H}$$ is a subspace, and a conjugation $$K_B : {\cal H} \to {\cal H}$$ (depending on the basis $$B$$), it is possible to define another linear operator $$A^{*_{K_B}}$$ that we may call the complex conjugated operator of $$A$$ with respect to $$K_B$$.

$$A^{*_{K_B}} := K_B A K_B\tag{*}$$

provided $$K_B (D(A)) \subset D(A)$$. I stress that $$K_B$$ appears twice in the right-hand side of the definition above. This is because we want that $$A^{*_B}$$ is linear as $$A$$ is: A definition like this $$A^{*_{K_B}} := K_B A\quad \mbox{(wrong),}$$ would instead produce an antilinear operator: $$(K_BA)(au) = K_B(aA(u))= a^* K_BAu\:.$$

Also observe that, in the absence of issues with the domains of the operators, (*) implies $$(AB)^{*_{K_B}} = A^{*_{K_B}}B^{*_{K_B}}\tag{4}\:.$$ Consider for instance the momentum operator restricted to the subspace of Schwartz' functions $${\cal S}(\mathbb R)$$. As is well known, $$P\psi = -i \frac{d}{dx} \psi \:,\quad \psi \in {\cal S}(\mathbb R)\:.$$ It is immediately proved that, referring to conjugations $$K$$ and $$K'$$ discussed in 1, $$P^{*_K} = +i \frac{d}{dx} = -P\:,\tag{5}$$ whereas $$P^{*_{K'}} = -i \frac{d}{dx} = P\:.$$

1. Antiunitary operators.

Consider an antiunitary operator $$V : {\cal H} \to {\cal H}$$. By definition it is antilinear and $$\langle V u|V u\rangle = \langle u|v\rangle^* \quad \forall u,v \in {\cal H}\:. \tag{3'}$$ If $$B$$ is a Hilbert basis and $$K_B$$ the associated conjugation, we have from (2), $$V = VK_BK_B$$ so that $$U_B := VK_B$$ is linear and $$V_B$$ is also unitary $$\langle U_B u|U_B v\rangle = \langle u|v\rangle \quad \forall u,v \in {\cal H}\:,$$ from (3) and (3'). We conclude that

Proposition. Given a Hilbert basis $$B$$, an antiunitary operator $$V$$ can be decomposed as $$V= U_B K_B$$ where $$U_B$$ is unitary and $$K_B$$ is the conjugation associated to the said basis.

1. OP's question.

The question have an overall elementary answer without entering into the details of unitary/antiunitary operators. In fact, both unitary or antiunitary operators are functions $$f: {\cal H} \to {\cal H}$$ and the composition of functions (linear or antilinear operators does not matter) is associatve.

OP's mistake relies on the identity $$\Theta A =A^*$$ that has no clear meaning. Let us consider the simplest case of $$\Theta = K_B$$.

(i) Suppose that $$K_B A =A^*$$ is the definition of the right-hand side.

This has nothing to do with the correct definition of conjugated operator with respect to a basis as I defined in 2 because this $$A^*$$ is antilinear, whereas the complex conjugated of an operator with respect to a basis is linear. Also it does not produce natural identities like (5) $$\left(-i\frac{d}{dx}\right)^* = i\frac{d}{dx}$$ established above, where the relevant conjugation is the standard conjugation of wavefunctions (I denoted by $$K$$). Finally, this definition of $$A^*$$ would not satisfy (4) and the contradiction suggested by the author is based on the validity of (4).

(ii) Suppose that $$K_B A =A^*$$ is not the definition of the right-hand side. In this case, author's argument immediately stops since we do not know what the right-hand side is.

With the right definition of conjugated operator (with respect to a Hilbert basis) as in (*) everything goes right also because $$K_B A \neq A^{*_{K_B}}$$.

• Could you, please, elaborate on the part “It is evident that this definition depends on the choice of B”? How do you derive that $\left< u_i, u'_j \right> = \left< u_i, u'_j \right>^*$? And is there a more general family of operators that doesn't depend on the choice of basis? (For example is the definition antiunitary opeators base-dependent?) Thank you!
– m93a
Feb 29 '20 at 9:59

You're confusing complex conjugation of the state with complex conjugation of operators. The right way to evaluate $\Theta A | v \rangle$ is $$\Theta A | v \rangle = \Theta (A |v \rangle) = (A |v \rangle)^*.$$ You've replaced $\Theta A$ with $A^*$, but this is incorrect as $$A^* | v \rangle \neq (A |v \rangle)^*.$$ To see why, consider a two-dimensional complex vector space $(z_1, z_2)$, which is also a four-dimensional real vector space $(x_1, x_2, y_1, y_2)$. Then any linear operator $A$ on the complex vector space must look like $$A' = \begin{pmatrix} A & 0 \\ 0 & A \end{pmatrix}$$ on the real vector space, by linearity. The complex conjugation operator is $$\Theta = \begin{pmatrix} I & 0 \\ 0 & -I \end{pmatrix}$$ because it flips the imaginary part. Then the two quantities we had above are $$\Theta A' = \begin{pmatrix} A & 0 \\ 0 & -A \end{pmatrix}, \quad {A'}^* = \begin{pmatrix} A^* & 0 \\ 0 & A^* \end{pmatrix}$$ and the two are completely different.

• $|v\rangle^*$ does not mean anything for a generic complex vector space, without further information. Jan 18 '18 at 15:22
• @ValterMoretti What I really mean is 'choose some canonical basis and then complex conjugate all the coefficients in this basis'. I know there is no nice mathematical definition in general, but it suffices to choose a specific instance where the conjugate exists to show that it's not the same as $\Theta A$, which is what I think would help the OP most. Jan 18 '18 at 15:25

2. On the other hand, complex conjugation (where it is implicitly assumed that it takes operators in operators) is an operator in a higher sense, so to speak. Understood in this sense, OP's expression $\Theta AB$ is not merely a composition $\Theta \circ A\circ B$ of 3 operators that take vectors in vectors.