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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?

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  • $\begingroup$ $\Theta A = A^*$ has no meaning. $\endgroup$ – Valter Moretti Jan 13 '18 at 15:08
  • $\begingroup$ please explain. $\endgroup$ – psitae Jan 15 '18 at 11:14
  • $\begingroup$ What do you mean for $A^*$? $\endgroup$ – Valter Moretti Jan 15 '18 at 15:07
  • $\begingroup$ 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? $\endgroup$ – Valter Moretti Jan 15 '18 at 18:28
  • $\begingroup$ Edited the definition in above $\endgroup$ – psitae Jan 17 '18 at 12:29
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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 $K_B A \neq A^{*_{K_B}}$.

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  • $\begingroup$ 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! $\endgroup$ – m93a Feb 29 at 9:59
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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.

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  • $\begingroup$ $|v\rangle^*$ does not mean anything for a generic complex vector space, without further information. $\endgroup$ – Valter Moretti Jan 18 '18 at 15:22
  • $\begingroup$ @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. $\endgroup$ – knzhou Jan 18 '18 at 15:25
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Comments to the post (v2):

  1. On one hand, Operators (where it is implicitly assumed that they take vectors in vectors) are manifestly associative under composition, independently of whether they are linear, antilinear, unitary, antiunitary, or neither. Complex conjugation (where it is implicitly assumed that it takes vectors in vector) is an example of such an operator, and it therefore satisfies associativity.

  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.

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  • $\begingroup$ The antiunitary operator does perform a complex conjugation, right? Does that disqualify it as an operator? $\endgroup$ – psitae Jan 11 '18 at 15:06
  • $\begingroup$ Interesting. I'll think about how an antiunitary operator's decomposition contains a complex conjugation element, yet is still an operator. Perhaps it's the subject of another question. $\endgroup$ – psitae Jan 11 '18 at 15:10

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