# How to prove $(A|\Psi\rangle)^{\dagger}=\langle\Psi|A$ ?

In the context of quantum mechanics we postulate that every observable operator $$A$$ acting on the corresponding Hilbert space $$\mathcal{H}$$ is self-adjoint (Hermitian), i.e. $$\forall \Psi,\varphi\in\mathcal{H}:\langle\Psi,A\varphi\rangle=\langle A\Psi,\varphi\rangle,$$ which is equivalent to saying $$A=A^{\dagger}$$.

The map $$(\cdot)\mapsto(\cdot)^{\dagger}$$ is transposing and complex conjugating; i.e., in matrix notation: $$(A^{\dagger})_{ij}=A_{ji}^{*}$$ (the star denoting complex conjugation). Finally, we have the bra-ket notation $$(|\Psi\rangle)^{\dagger}:=\langle\Psi|$$.

My question is: How can I prove rigorously, with the above, that $$(A|\Psi\rangle)^{\dagger}=\langle\Psi|A?$$ Is it possible to do so without using the matrix notation (i.e. general statement including continuum states)?

• Not every operator on the Hilbert space is Hermitian. For example, the time-evolution operator is unitary. Sep 18 at 18:50
• @RichardMyers While the time-evolution operator is non-Hermitian, it's worth noting that unitarity and Hermiticity are not mutually exclusive. Sep 19 at 19:40

In general

$$\begin{equation} \langle \varphi| A |\psi \rangle^* = \langle \psi| A^\dagger |\varphi \rangle \end{equation}$$

If $$A$$ is hermitian $$\Rightarrow A = A^\dagger$$

\begin{align} &\Rightarrow \langle \varphi| A |\psi \rangle^* = \langle \psi| A |\varphi \rangle\\ &\iff (|\psi \rangle)^\dagger A^\dagger (\langle \varphi|)^\dagger = \langle \psi| A |\varphi \rangle\\ &\iff (A |\psi \rangle)^\dagger |\varphi \rangle = \langle \psi| A |\varphi \rangle \end{align}

Since this is true for any $$|\varphi \rangle$$

$$\begin{equation} \therefore \quad (A |\psi \rangle)^\dagger = \langle \psi| A \end{equation}$$

I will approach your problem from a pure mathematical perspective:

1. Let me first say that in your post you use the same notation $$\dagger$$ for two distinct mappings which have two different definitions and notations. One mapping acts on the space of linear densely defined operators in a complex separable Hilbert space and the other mapping is from the Hilbert space itself to its (topological) dual with respect to the topology induced by the norm (this mapping assignes a vector to a continuous functional). The first mapping is regularly denoted in physics by the dagger $$\dagger$$, while the second one by the tilde $$\widetilde{,,,}$$

2. Let me redefine your statement to be proved by using proper mathematical notation and dismissing the mathematically complicated Dirac braket notation. You wish to prove that for a self-adjoint $$A:D(A)\subseteq\mathscr{H} \rightarrow \mathscr{H}$$ we have the following equality of functionals:

$$\widetilde{A\Psi} =A^{\times}\widetilde{\Psi}.$$

Denote by $$F_{\Psi}\in\widetilde{\mathscr{H}}$$ a continuous functional on $$\mathscr{H}$$ assigned to an arbitrary vector $$\Psi\in D(A)\subseteq\mathscr{H}$$. We then have:

$$\widetilde{A\Psi} (\varphi) \equiv F_{A\Psi} (\varphi) = \langle A\Psi, \varphi\rangle = \langle \Psi, A^{\dagger}\varphi\rangle = \langle \Psi, A\varphi\rangle, ~ \forall \varphi\in D(A)\tag{1}$$

1. Turning to the right hand side, $$A^{\times}$$ is called the dual operator assigned to a linear operator acting in the Hilbert space ($$A^{\times}:\widetilde{\mathscr{H}}\rightarrow\widetilde{\mathscr{H}}$$). We then use its definition:

$$\left(A^{\times} \widetilde{\Psi}\right) (\varphi) =\left(A^{\times}F_{\Psi}\right)(\varphi)=: F_{\Psi} (A\varphi) = \langle\Psi, A\phi\rangle, ~ \forall \varphi\in D(A) \tag{2}$$

From $$(1)$$ and $$(2)$$ we obtain what we needed to prove.

• These things are indeed much clearer without the Dirac notation.
– Gold
Sep 18 at 23:38

More generally, the adjoint can be defined for an operator $$A:\mathcal H_1\to \mathcal H_2$$ between different Hilbert spaces, and gives an operator $$A^\dagger : \mathcal H_2 \to\mathcal H_1$$. It is then easy to prove that : $$(AB)^\dagger = B^\dagger A^\dagger\tag{1}$$

To apply this definition to $$|\psi\rangle$$, we can see it as an operator $$\mathbb C\to \mathcal H$$ (which sends a complex number $$\lambda$$ to the ket $$\lambda|\psi\rangle$$). Its adjoint is an operator $$|\psi\rangle^\dagger = X: \mathcal H \to \mathbb C$$ such that : $$\forall \lambda \in \mathbb C,\forall |\varphi\rangle \in \mathcal H , (|\varphi\rangle,\lambda|\psi\rangle)_{\mathcal H} = (X|\varphi\rangle,\lambda)_{\mathbb C}$$ ie $$\lambda \langle \varphi|\psi\rangle= (X|\varphi\rangle)^*\lambda$$. This gives $$X|\varphi\rangle= \langle\psi|\varphi\rangle$$ and $$|\psi\rangle^\dagger = \langle \psi|$$.

Using $$(1)$$ with $$A$$ hermitian, we get : $$(A|\psi\rangle)^\dagger = |\psi\rangle^\dagger A ^\dagger = \langle \psi| A$$