TL;DR: Yes, it is just a short-cut. The main point is that the complexified map
$$\tag{A} \begin{pmatrix} \phi \\ \phi^{*} \end{pmatrix}
~=~ \begin{pmatrix} 1 & i\\ 1 &-i \end{pmatrix}
\begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} $$
is a bijective map :$\mathbb{C}^2 \to \mathbb{C}^2 $.
Notation in this answer: In this answer, let $\phi,\phi^{*}\in \mathbb{C}$ denote two independent complex fields. Let $\overline{\phi}$ denote the complex conjugate of $\phi$.
I) Let us start with the beginning. Imagine that we consider a field theory of a complex scalar field $\phi$. We are given a Lagrangian density
$$\tag{B} {\cal L}~=~{\cal L}(\phi,\overline{\phi},\partial\phi, \partial\overline{\phi})$$
that is a polynomial in $\phi$, $\overline{\phi}$, and spacetime derivatives thereof. We can always decompose a complex field in real and imaginary parts
$$\tag{C} \phi~\equiv~\phi_1+ i \phi_2 ,$$
where $\phi_1,\phi_2 \in \mathbb{R}$. Hence we can rewrite the Lagrangian density (B) as a theory of two real fields
$$\tag{D}{\cal L}~=~{\cal L}(\phi_1,\phi_2,\partial\phi_1, \partial\phi_2).$$
II) We can continue in at least three ways:
Vary the action wrt. the two independent real variables $\phi_1,\phi_2 \in \mathbb{R}$.
Originally $\phi_1,\phi_2 \in \mathbb{R}$ are of course two real fields. But we can complexify them, vary the action wrt. the two independent complex variables $\phi_1,\phi_2 \in \mathbb{C}$, if we at the end of the calculation impose the two real conditions
$$\tag{E} {\rm Im}(\phi_1)~=~0~=~{\rm Im}(\phi_2). $$
Or equivalently, we can replace the complex conjugate field $\overline{\phi}\to \phi^{*}$ in the Lagrangian density (B) with an independent new complex variable $\phi^{*}$, i.e. treat $\phi$ and $\phi^{*}$ as two independent complex variables, vary the action wrt. the two independent complex variables $\phi,\phi^{*} \in \mathbb{C}$, if we at the end of the calculation impose the complex condition
$$\tag{F} \phi^{*} ~=~ \overline{\phi}. $$
III) The Euler-Lagrange equations that we derive via the two methods (1) and (2) will obviously be exactly the same. The Euler-Lagrange equations that we derive via the two methods (2) and (3) will be just linear combinations of each other with coefficients given by the constant matrix from eq. (A).
IV) We mention for completeness that the complexified theory [i.e. the theory we would get if we do not impose condition (E), or equivalently, condition (F)] is typically not unitary, and therefore ill-defined as a QFT. Recall for starter that we usually demand that the Lagrangian density is real.
References:
- Sidney Coleman, QFT notes; p. 56-57.
Generally speaking the complex conjugate of an operator is not a standard notion of operator theory, though it can be defined after having introduced some general notions.
Definition. A conjugation $C$ in a Hilbert space $\cal H$ is an antilinear map $C : \cal H \to \cal H$ such that is isometric ($||Cx||=||x||$ if $x\in \cal H$) and involutive ($CC=I$).
There are infinitely many such maps, at least one for every Hilbert basis in $\cal H$ (the map which conjugates the components of any vector with respect to that basis).
On $L^2$ spaces there is a standard conjugation $$C : L^2(\mathbb R^n, dx) \ni \psi \mapsto \overline{\psi}\in L^2(\mathbb R^n, dx)\:, $$ where $\overline{\psi}(x) := \overline{\psi(x)}$ for every $x \in \mathbb R^n$ and where $\overline{a+ib}:= a-ib$ for $a,b \in \mathbb R$.
Definition. An operator $H : D(H) \to \cal H$ (where henceforth $D(H)\subset \cal H$) is said to be real with respect to a conjugation $C$ if $$CHx=HCx \quad \forall x \in D(H)$$ (which implies $C(D(H)) \subset D(H)$ and thus $C(D(H)) = D(H)$ in view of $CC=I$, so that the written condition can equivalently be re-phrased $CH=HC$).
The complex conjugate $H_C$ of an operator $H$ with respect to a conjugation $C$, can be defined as $$H_C :=CHC\:.$$ This operator with domain $C(D(H))$ is symmetric, essentially self-adjoint, self-adjoint if $H$ respectively if is symmetric, essentially self-adjoint, self-adjoint. Obviously it coincides with $H$ if and only if $H$ is real with respect to $C$.
Let us come to your issue. Let us start form Schroedinger equation
$$-i \frac{d}{dt} \psi_t = H\psi_t\:.$$
Here we have a vector valued map $$\mathbb R \ni t \mapsto \psi_t \in \cal H\:,$$
such that $\psi_t \in D(H)$ for every $t \in \mathbb R$ and the derivative is computed with respect to the topology of the Hilbert space whose norm is $||\cdot|| = \sqrt{\langle \cdot| \cdot \rangle}$:
$$\frac{d}{dt} \psi_t = \dot{\psi}_t \in \cal H$$
means
$$ \lim_{h\to 0} \left|\left| \frac{1}{h} (\psi_{t+h} -\psi_t) - \dot{\psi}_t \right|\right|=0\:.$$
(See the final remark)
If $C : \cal H \to \cal H$ is a conjugation, as it is isometric and involutive, in view of the definition above of derivative, we have
$$C\frac{d}{dt} \psi_t = \frac{d}{dt} C\psi_t\tag{1}$$
where both sides exist or do not simultaneously.
Summing up, given a conjugation $C$, and the (self-adjoint) Hamiltonian operator $H$, both in the Hilbert space $\cal H$, the complex conjugate of the Schroedinger equation
$$-i \frac{d}{dt} \psi_t = H\psi_t\:.\tag{2}$$
is a related equation satisfied by $C\psi_t$ and just obtained by applying $C$ to both sides of (2) and taking (1) and $CC=I$ into account, obtaining
$$i \frac{d}{dt} C\psi_t = H_C\: C\psi_t\:.$$
If $H$ is real with respect to $C$ (this is the case for a particle without spin described in $L^2(\mathbb R^3)$, assuming the Hamiltonian of the form $P^2/2m + V$ and $C$ is the standard complex conjugation of wavefunctions), the equation reduces to
$$i \frac{d}{dt} C\psi_t = H C\psi_t\:.$$
REMARK. It is worth stressing that $-i \frac{d}{dt}$ is not an operator in the Hilbert space $\cal H$ as, for instance, $H$ is. To compute $H\psi$, it is enough to know the vector $\psi \in D(H)$. To compute
$\frac{d}{dt}\psi_t$ we must know a curve of vectors $$\gamma :\mathbb R \ni t \mapsto \psi_t \in \cal H\:.$$
$\frac{d}{dt}$ computes the derivative of such curve defining another curve
$$\dot{\gamma} :\mathbb R \ni t \mapsto \frac{d}{dt}\psi_t \in \cal H\:.$$
More weakly one may view $\frac{d}{dt}|_{t_0}$ as a map associating vector-valued curves defined in aneighborhood of $t_0$ to vectors $\frac{d}{dt}|_{t_0}\psi_t$. In both cases it does not make sense to apply the derivative to a single vector $\psi$, contrarily to $H\psi$ is well defined.
Best Answer
Whenever you have a function of two variables $a$ and $b$, you can plug in $a = (z + z^*) / 2$ and $b = (z - z^*) / 2i$ to get a function of $z$ and $z^*$. If $a$ and $b$ were real, $z$ and $z^*$ will necessarily be complex conjugates. Nothing fancy is going on. The mapping between $z$ and $z^*$ is no more or less special than the mapping between $b$ and $-b$.
You are just talking about a change of variables but there are two other things which might be confusing you.
It is often desirable that when $a$ and $b$ are traded for $z$ and $z^*$, the resulting function depends on only one of them. This indeed requires the function to use only "a subset of operations" so that it satisfies the Cauchy-Riemann equations. Note that $f(a, b) = a^2 + b^2 \Leftrightarrow f(z, z^*) = z z^*$ (which is the QFT example you referenced) does not.
Sometimes people take a function of $z$ and $z^*$ and promote both of them to independent complex variables. This is an analytic continuation which means it can be unique or not depending on the function.