# Does every operator have a hermitian adjoint?

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 $$T_{ij}^{\dagger}.$$

• Perhaps better to ask this question on Mathematics. It does not seem to be much about physics. Commented Jul 11, 2023 at 10:19
• Check the definition on Wikipedia. Commented Jul 11, 2023 at 10:28
• While this is a math question, I think it is of particular interest for PSE users (too)... Commented Jul 11, 2023 at 17:48

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).

• I read the answer again. The heart of it is in the theorem that says $A^{\dagger}$ is well-defined. Could you at least point me to a proof of that? Stated that way, it's just saying "assume what you want to exist exists" or "there's a theorem that says it does".
– EM_1
Commented Jul 11, 2023 at 18:41
• @EM_1 It is well-defined precisely because $z_u$ is unique. Put differently, using the Riesz representation theorem shows that to each $u$ we can associate one and only one $z_u$ such that $(2)$ holds. This allows us to define a function (operator), which we denote by $A^\dagger$, which maps $u$ to the corresponding $z_u$, which is the content of eq. $(3)$. That $A^\dagger$ is linear follows from the properties of the inner product. Commented Jul 11, 2023 at 18:43
• So for a fixed $u$ there's a corresponding, unique $z_u$, and the operator $A^{\dagger}$ is just the map that associates the first to the second uniquely? Is that a reasonable summary?
– EM_1
Commented Jul 11, 2023 at 18:54
• Yes, indeed - we define the operator $A^\dagger$ to be this map! All the properties you know regarding adjoint operators then follow from this definition. Commented Jul 11, 2023 at 18:54

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.

• The definition for (unbounded) operators defined on a dense subspace, which basically is the case we are interested in QM, is also straightforward, and e.g. given in the corresponding Wikipedia article. Commented Jul 11, 2023 at 11:29
• I'm not familiar with the content of the answer. Is there a way to address the question in linear algebraic terms?
– EM_1
Commented Jul 11, 2023 at 15:59
• @EM_1 Can you be more specific? Which content of the answer are you not familiar with? What do you mean by "linear algebraic terms"? Do you mean the case of $\mathbb{C}^n$ with the usual scalar product? Maybe you could also tell us more about the exact context in which you want to define a hermitian adjoint.
– jd27
Commented Jul 11, 2023 at 17:46
• @EM_1 See e.g. my answer - does that help? Riesz' representation theorem in finite dimensions is more or less trivial. Commented Jul 11, 2023 at 18:25
• @EM_1 You need to use the Riesz isomorphism to show that the Hermitian adjoint exists. The commutative diagram is just a tool to visualize the relevant maps. Saying that it commutes means "that we can follow the arrows in either way". So in the diagram above it means that $i_\mathcal{H} \circ T^\dagger =T^* \circ i_\mathcal{K}$. The only Banach space knowledge required here is the notion of continuity (and the fact that $T^*$ is continuous), which is necessary for infinite dimensional spaces. Without it can only be proven for finite dimensions (see the other answer by Tobias Fünke).
– jd27
Commented Jul 11, 2023 at 18:54