The time reversal operator $T$ is an antiunitary operator, and I saw $T^\dagger$ in many places
(for example when some guy is doing a "time reversal" $THT^\dagger$),
but I wonder if there is a well-defined adjoint for an antilinear operator?
Suppose we have an antilinear operator $A$ such that $$ A(c_1|\psi_1\rangle+c_2|\psi_2\rangle)=c_1^*A|\psi_1\rangle+c_2^*A|\psi_2\rangle $$ for any two kets $|\psi_1\rangle,|\psi_2\rangle$ and any two complex numbers $c_1^*, c_2^*$.

And below is my reason for questioning the existence of $A^\dagger$:
Let's calculate $\langle \phi|cA^\dagger|\psi\rangle$.
On the one hand, obviously $$ \langle \phi|cA^\dagger|\psi\rangle=c\langle \phi|A^\dagger|\psi\rangle. $$ But on the other hand, $$ \langle \phi|cA^\dagger|\psi\rangle =\langle \psi|Ac^*|\phi\rangle^*=\langle \psi|cA|\phi\rangle^*=c^*\langle \psi|A|\phi\rangle^*=c^*\langle \phi|A^\dagger|\psi\rangle, $$
from which we deduce that $c\langle \phi|A^\dagger|\psi\rangle=c^*\langle \phi|A^\dagger|\psi\rangle$, almost always false, and thus a contradiction!

So where did I go wrong if indeed $A^\dagger$ exists?

  • 3
    $\begingroup$ Your mistake is that you introduced an ambiguous symbol $c A^\dagger$ but you didn't specify clearly whether this operator first multiplies by $c$ and then acts by $A^\dagger$, or vice versa. Instead, you used both interpretations of the symbol on the two "mutually contradicting" lines, so of course you ended up with a contradiction. There isn't any real contradiction here with complex conjugation. If $A$ is antilinear and $A^\dagger$ exists, of course that $A^\dagger$ is also antilinear (much like the Hermitian conjugate of a linear operator is still linear). $\endgroup$ Nov 27, 2012 at 17:20
  • 1
    $\begingroup$ Let me be more specific. If you defined $cA^\dagger$ to be the operator that first acts by $A^\dagger$ on ket and then multiplies by $c$, then its action on bras must still be that it first acts by $A^\dagger$ on the left and then it multiplies by $c$. You used the opposite rule on the second line, and that's why you produced a wrong extra star above $c$ once you interchanged $c$ and $A^\dagger$. $\endgroup$ Nov 27, 2012 at 17:24
  • $\begingroup$ Ac*⁠ = cA” is not a valid identity. See also physics.stackexchange.com/q/43069 $\endgroup$ Oct 22, 2014 at 6:29

1 Answer 1


I) First of all, one should never use the Dirac bra-ket notation (in its ultimate version where an operator acts to the right on kets and to the left on bras) to consider the definition of adjointness, since the notation was designed to make the adjointness property look like a mathematical triviality, which it is not. See also this Phys.SE post.

II) OP's question(v1) about the existence of the adjoint of an antilinear operator is an interesting mathematical question, which is rarely treated in textbooks, because they usually start by assuming that operators are $\mathbb{C}$-linear.

III) Let us next recall the mathematical definition of the adjoint of a linear operator. Let there be a Hilbert space $H$ over a field $\mathbb{F}$, which in principle could be either real or complex numbers, $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=\mathbb{C}$. Of course in quantum mechanics, $\mathbb{F}=\mathbb{C}$. In the complex case, we will use the standard physicist's convention that the inner product/sequilinear form $\langle \cdot | \cdot \rangle$ is conjugated $\mathbb{C}$-linear in the first entry, and $\mathbb{C}$-linear in the second entry.

Recall Riesz' representation theorem: For each continuous $\mathbb{F}$-linear functional $f: H \to \mathbb{F}$ there exists a unique vector $u\in H$ such that $$\tag{1} f(\cdot)~=~\langle u | \cdot \rangle.$$

Let $A:H\to H$ be a continuous$^1$ $\mathbb{F}$-linear operator. Let $v\in H$ be a vector. Consider the continuous $\mathbb{F}$-linear functional

$$\tag{2} f(\cdot)~=~\langle v | A(\cdot) \rangle.$$

The value $A^{\dagger}v\in H$ of the adjoint operator $A^{\dagger}$ at the vector $v\in H$ is by definition the unique vector $u\in H$, guaranteed by Riesz' representation theorem, such that $$\tag{3} f(\cdot)~=~\langle u | \cdot \rangle.$$

In other words, $$\tag{4} \langle A^{\dagger}v | w \rangle~=~\langle u | w \rangle~=~f(w)=\langle v | Aw \rangle. $$

It is straightforward to check that the adjoint operator $A^{\dagger}:H\to H$ defined this way becomes an $\mathbb{F}$-linear operator as well.

IV) Finally, let us return to OP's question and consider the definition of the adjoint of an antilinear operator. The definition will rely on the complex version of Riesz' representation theorem. Let $H$ be given a complex Hilbert space, and let $A:H\to H$ be an antilinear continuous operator. In this case, the above equations (2) and (4) should be replaced with

$$\tag{2'} f(\cdot)~=~\overline{\langle v | A(\cdot) \rangle},$$


$$\tag{4'} \langle A^{\dagger}v | w \rangle~=~\langle u | w \rangle~=~f(w)=\overline{\langle v | Aw \rangle}, $$

respectively. Note that $f$ is a $\mathbb{C}$-linear functional.

It is straightforward to check that the adjoint operator $A^{\dagger}:H\to H$ defined this way becomes an antilinear operator as well.


$^{1}$We will ignore subtleties with discontinuous/unbounded operators, domains, selfadjoint extensions, etc., in this answer.


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