Quantum Mechanics – How to Prove $(A|\Psi\rangle)^{\dagger}=\langle\Psi|A$?

hilbert-spaceoperatorsquantum mechanics

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)?

Best Answer

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.

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