Does every operator have a hermitian adjoint?

Question

If we think of operators as matrices, every matrix can be transposed and its elements can be complex-conjugated. But the identification of the hermitian adjoint with the transpose conjugate comes from inner products. Namely, the hermitian adjoint of $\hat{T}$ is $\hat{T}^{\dagger}$ where $\hat{T}^{\dagger}$ satisfies $$(u,\hat{T}v) = (\hat{T}^{\dagger}u,v).$$ If $T_{ij} = (i,\hat{T}j),$ then $$T_{ij}^{\dagger} = (i,\hat{T}^{\dagger}j) = (j,\hat{T}i)^* = T_{ji}^*.$$ But how do we know that the map $\hat{T}^{\dagger}$ which satisfies the desired relationship exists in the first place? Can we show it without thinking of operators as matrices, just using the inner product? Also, does the relationship $$\big(\hat{T}^{\dagger}\big)^{\dagger} = \hat{T}$$ always hold? Since it seems to be used in the derivation of $T_{ij}^{\dagger}.$

Answer 1

Let $\mathcal{H}$ and $\mathcal{K}$ be Hilbert spaces and $T: \mathcal{H} \to \mathcal{K}$ a continuous linear operator. Let $\mathcal{H}^*$ and $\mathcal{K}^*$ be the topological dual spaces (the bras).

Further let $i_\mathcal{H}, i_\mathcal{K}$ be the Riesz isomorphisms (the maps that sends a ket to a bra i.e. $| \psi \rangle \mapsto \langle \psi |$). Consider the following diagram: $$ \require{AMScd} \begin{CD} \mathcal{H} @. @. \mathcal{H} @>i_\mathcal{H}>> \mathcal{H}^* \\ @VTVV @. @AT^\dagger AA @AAT^*A \\ \mathcal{K} @. @. \mathcal{K} @>i_\mathcal{K}>> \mathcal{K}^*, \end{CD} $$ where $T^*$ is the Banach adjoint of $T$ defined by $T^*( \langle \psi| ) = \langle \psi| T$. Clearly $T^*$ is linear and always exists (and is also continuous). Therefore we can define $T^\dagger$ to be the unique linear and bounded operator that makes the diagram commute ($T^\dagger = i^{-1}_\mathcal{H}T^* i_\mathcal{K} $).

Now the commutativity of the diagram means that for any $\varphi \in \mathcal{H}$ and $\psi \in \mathcal{K}$: $$\langle T^\dagger \psi | \varphi \rangle = T^* (\langle \psi |) |\varphi \rangle = \langle \psi | T \varphi \rangle. $$

So in total $T^\dagger$ always exist for continuous $T$ and is uniquely defined by the property in the preceding equation.

The definition of $T^\dagger$ for operators that are not continuous is more problematic. It can be found in (almost) any functional analysis book.

Answer 2

The OP is concerned with the finite-dimensional setting, as phrased in the question and comments. So let us consider a complex finite-dimensional Hilbert space $H$. The detailed proofs of the theorems below can be found in books on linear algebra/ functional analysis/quantum mechanics.

Theorem (Riesz): For every linear functional $F:H\rightarrow \mathbb C$, there exists a unique $f\in H$ such that $$ F(v) = (f,v) \tag{1} \quad ,$$ for all $v\in H$.

Proof: Let $\{e_n\}_{n=1,2,\ldots, \dim H}$ denote an orthonormal basis and define the vector $$ f:= \sum\limits_{n=1}^{\dim H} F(e_n)^* \, e_n \quad .$$ Expanding any vector $v\in H$ in the said orthonormal basis shows that $(1)$ holds with $f$ defined as above. Uniqueness of $f$ follows from the properties of the inner product.

Now let $A:H\rightarrow H$ denote a linear operator on $H$.

Corollary: For fixed $u \in H$, the map $ H\ni v\mapsto (u,Av)$ is a linear functional and hence for each $u$, there exists a unique $z_u\in H$ such that $$(z_u,v)=(u,Av) \tag{2} \quad ,$$ for all $v\in H$.

This brings us to the following definition.

Definition (Adjoint): The adjoint operator $A^\dagger:H\rightarrow H$ is defined by $$ A^\dagger u:=z_u \tag {3} \quad .$$

Definition $(3)$ makes sense precisely because $z_u$ is unique, i.e. we can associate to every $u\in H$ one and only one $z_u\in H$ and we define $A^\dagger$ to be the function which maps $u$ to $z_u$.

Theorem: The adjoint $A^\dagger$ is linear and it holds that $(A^\dagger)^\dagger=A$.

Proof: Both properties are consequences of the properties of the inner product.

As a trivial consequence of $(2)$ and $(3)$ it holds that $$ (u,Av)=(A^\dagger u,v) \tag{4} \quad ,$$

for all $u,v\in H$. This, in turn, implies that $$(e_m,Ae_n)^*=(e_n,A^\dagger e_m)\quad . \tag 5$$

Equation $(5)$ gives the connection to the matrix representations of both operators, i.e. the matrix of the adjoint is the transpose and complex conjugated matrix of $A$ (in some orthonormal basis).