I was going through time reversal symmetry and came across the term 'conjugate relation'. The text is

Next, in classical mechanics, the initial conditions of a motion $x(t)$ transform under time reversal according to $(x_0, p_0) → (x_0,−p_0)$, so we postulate that the time-reversal operator $\theta$ in quantum mechanics should satisfy the conjugation relations,
$\theta x \theta^\dagger = x$,
$\theta p \theta^\dagger = −p$.

My question is what meant by conjugate relation?

As far as I know conjugate relation is related with the conjugate momenta and position in classical mechanics. How it is related here and how can one convey that time reversal operator cannot be unitary by looking at $\theta x \theta^\dagger = x$, $\theta p \theta^\dagger = −p$.)

  • 1
    $\begingroup$ en.wikipedia.org/wiki/Conjugacy_class $\endgroup$
    – Phoenix87
    Apr 17, 2020 at 16:24
  • $\begingroup$ Which text? Which page? Link? $\endgroup$
    – Qmechanic
    Apr 18, 2020 at 13:47
  • $\begingroup$ @Qmechanic it is a lecture of Robert G. Littlejohn on Time reversal (in pdf). $\endgroup$ Apr 18, 2020 at 14:08

1 Answer 1


the operation $X\mapsto A^{-1}XA$ is usually called conjugation of $X$ by $A$ because this is the language used in group theory. I assume that by $\dagger$ the author means inverse, because antilinear/anti-unitary maps like time reversal do not have an adjoint which is what the $\dagger$ usually denotes.

The "group theory" mention is not important. It's just the origin of the name. When we diagonalize a matrix by $$ X\mapsto A^{-1}XA= {\rm diag}(\lambda_1,\lambda_2\ldots ) $$ the operation is a called a similarity transformation or a conjugation and $X$ and the diagonal matrix are said to be similar or conjugate.

Conjugation is to be distingushed from a diagonalizing matrix by $$ X\mapsto A^TXA= {\rm diag}(\lambda_1,\lambda_2\ldots ) $$ as we do for quadratic forms such as the kinetic energy in a small vibrations problem. In this case $X$ and the diagonal matric are said to be congruent rather than conjugate.

Time reversal has to be antilinear in order to deal with the "$i$" in the time dependent Schrodinger equation. If $\psi(x,t)$ satisfies it and the potential is real, then so does $\psi^*(x,-t)$. So the wavefunction has to be be complex-conjugated and this is an antilinear process: $$ \theta( \alpha |{\bf a}\rangle +\beta |{\bf b}\rangle)= \alpha^* (\theta |{\bf a}\rangle) +\beta^* (\theta|{\bf b}\rangle ) $$

Antilinear/antiuitary maps are tricky and confusing things. An operator $\Omega$ is said to be antiunitary with respect to the usual quantum mechanics (conjugate-symmetric and sesquilinear) inner product $\langle{\phantom -},{\phantom-}\rangle$ if
$$ \langle{\Omega {\bf a}}, {\Omega {\bf b}}\rangle =\langle{{\bf a}},{{\bf b}}\rangle ^*= \langle{\bf b},{\bf a}\rangle. $$ Consider the vector $$ {\bf X}= \Omega( \alpha {\bf a}+\beta {\bf b})- \alpha^* (\Omega {\bf a}) -\beta^* (\Omega{\bf b}). $$ Using the definition of antiunitarity and the antilinearity of $\langle{\phantom -},{\phantom-}\rangle $ in its first slot and linearity in the second, we can expand out $\|{\bf X}\|^2=\langle{{\bf X}},{{\bf X}}\rangle$ and find that it is zero. For a positive definite inner product a vanishing norm implies that ${\bf X}=0$, and so for such a product we have $$ \Omega( \alpha {\bf a}+\beta {\bf b})= \alpha^* (\Omega {\bf a}) +\beta^* (\Omega{\bf b}). $$ Thus an antiunitary operator acting on a positive-definite Hilbert space is necessarily antilinear.

One consequence of the antilinearity is that there is no obvious way to define an adjoint $\Omega^\dagger$. The standard definition of the adjoint (i.e the "Hermitian conjugate") $\Omega^\dagger$ with respect to an inner product is to set $\langle{\Omega^\dagger {\bf a}},{{\bf b}}\rangle = \langle{{\bf a}},{\Omega{\bf b}}\rangle $, but this leads to
$$ \langle{{\bf b}},{{\bf a}}\rangle =\langle{\Omega {\bf a}}, {\Omega {\bf b}}\rangle \stackrel{?}{=} \langle{\Omega^\dagger \Omega {\bf a}}, {{\bf b}}\rangle $$ and a contradiction: the leftmost expression is antilinear in ${\bf b}$ while the rightmost is linear in ${\bf b}$. A similar issue leads to $$ (\langle {\bf a}| \Omega)|{\bf b}\rangle \ne \langle {\bf a}|(\Omega |{\bf b}\rangle) $$ and so makes ``matrix elements'' $\langle{{\bf a}}|{\Omega}|{{\bf b}}\rangle $ ambiguous, and prevents us from defining the usual left action of $\Omega$ on bra vectors $\langle{\bf a}|$. Instead we have $$ \langle{\bf b},{\Omega{\bf a}}\rangle = \langle {\bf a},{\Omega^{-1} {\bf b}}\rangle = \langle\Omega^{-1} {\bf b}, {\bf a}\rangle ^*. $$

  • $\begingroup$ what does it signify actually and with the above expression where does group comes?Also can you simply explain by an example without considering group into picture why time reversal are anti-linear (or it means it is anti-unitary) as i have just started reading about groups $\endgroup$ Apr 18, 2020 at 12:38
  • $\begingroup$ I'll add something about anti-unitary and antilinear to my answer. $\endgroup$
    – mike stone
    Apr 18, 2020 at 12:48
  • $\begingroup$ i have few question $\endgroup$ Apr 20, 2020 at 3:30
  • $\begingroup$ I have few question.How does the definition of diagonalization of matrix changes for a conjugation operator (i mean that the inverse of A in one case and A transpose in another?Are they equivalent or not).2nd is i am not clear with the definition of inner product?Is it the same as multiplying a bra and a ket.3rd is i didnt get the meaning of the line of yours i.e"Using the definition of antiunitarity and the antilinearity of ⟨,⟩ in its first slot and linearity in the second". $\endgroup$ Apr 20, 2020 at 4:53
  • $\begingroup$ Also sir if the adjoint dosent exist then how adjoint can exist? $\endgroup$ Apr 20, 2020 at 6:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.