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

Question

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)?

Answer 1

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:

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

3. 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.

Answer 2

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}

Answer 3

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 : $$(A|\psi\rangle)^\dagger = |\psi\rangle^\dagger A ^\dagger = \langle \psi| A$$