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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/sesquilinear 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 $$\tag{1} f(\cdot)~=~\langle u | \cdot \rangle.$$$$ f(\cdot)~=~\langle u | \cdot \rangle.\tag{1}$$

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

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

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 $$\tag{3} f(\cdot)~=~\langle u | \cdot \rangle.$$$$ f(\cdot)~=~\langle u | \cdot \rangle.\tag{3}$$

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

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's 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

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

and

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

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.

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/sesquilinear 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's 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

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

and

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

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.

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/sesquilinear 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 $$ f(\cdot)~=~\langle u | \cdot \rangle.\tag{1}$$

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

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

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 $$ f(\cdot)~=~\langle u | \cdot \rangle.\tag{3}$$

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

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's 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

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

and

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

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.

Fixed typo in grammar; fixed links http -> https; minor facultative MathJax adjustments.
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I) First of all, one should never use the Dirac bra-ket notationDirac 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 adjointnessadjointness, 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 antilinearantilinear 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 spaceHilbert space $H$ over a fieldfield $\mathbb{F}$, 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 forminner product/sesquilinear 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 theoremRiesz' 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'Riesz's 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)$(2)$ and (4)$(4)$ should be replaced with

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

and

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

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 operatorsdiscontinuous/unbounded operators, domains, selfadjoint extensionsselfadjoint extensions, etc., in this answer.

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

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

and

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

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.

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/sesquilinear 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's 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

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

and

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

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.

replaced http://physics.stackexchange.com/ with https://physics.stackexchange.com/
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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 thisthis 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

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

and

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

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.

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

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

and

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

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.

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

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

and

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

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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