Do we use both real and imaginary part of the wave function in quantum mechanics as opposed to classical mechanics? In classical mechanics and physics, it's typically said that when we write a wave as $\Psi(x,t)=Ae^{I(kx+\omega t+\delta)}$, what we mean is to take only its real part; that is, $\Psi(x,t)=\Re\left\{Ae^{I(kx+\omega t+\delta)}\right\}$.
Is that the same in quantum mechanics or do we use both real and imaginary parts in QM? And if we use both parts, why is that? Are there two oscillating components in QM?
I found following two posts that are somehow relevant, but none answers the exact question I have!
What does real and imaginary part of wavefunction signify?

Imaginary part of the waveefunction in quantum mechanics [duplicate]

 A: The imaginary part of the wave function includes fundamental information.  This related to the fact that the $x$-dependence of wave function contains information not just about the position of particle, but also its momentum (or velocity).  This differs from the classical case, in which the displacement and velocity information in a wave are described separately.  The usual initial conditions for a classical wave propagation problem involve specifying both $\Psi(x,t=0)$ and $\dot{\Psi}(x,0)$, but since the Schrödinger equation is first-order in time, the only initial condition needed to describe the subsequent time evolution is $\psi(x,0)$.
Consider two exponential plane waves $\psi_{\pm}(x)=e^{\pm ikx}$.  (This is not normalized wave function, but the normalization issue it not key here.)  These have the same real part, $\cos kx$; however, the two states are quite different.  Each plane have is an eigenstate of the momentum operators $p=-i\hbar\,\partial/\partial x$,
$$p\psi_{\pm}=\left(-i\hbar\frac{\partial}{\partial x}\right)e^{\pm ikx}=\pm\hbar k\psi_{\pi},$$
meaning that the plane waves are states of well-defined momentum, but they have opposite momenta $\hbar k$ and $-\hbar k$, so they describe wave trains moving in opposite directions.
This last aspect can also be shown by including the time dependence, $\psi_{\pm}(x,t)=e^{\pm ikx-iEt/\hbar}$, where $E$ is the eigenvalue of the Hamiltonian operators, which is, for a free particle,$$
H=\frac{p^{2}}{2m}=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}},$$
so that the plane waves are both eigenstates of $H$ with eigenvalue $E=\hbar^{2}k^{2}/2m$.  This makes the time-dependent wave functions $$\psi_{\pm}(x,t)=e^{\pm i\left(kx\mp\frac{\hbar k^{2}}{2m}t\right)}$$
move in opposite directions. The waves have phase velocities $\pm \hbar k/2m$ and (if we form a wave packet out of such plane waves) group velocities $\pm \hbar k/m$.
A: Quantum mechanics fundamentally requires complex numbers, whereas generally classical does not.
There are two features of quantum mechanics that lead to the necessity of complex numbers. First, is the fact that quantum mechanics uses probability amplitudes rather than probabilities, and in particular allows for negative amplitudes (which is necessary in order for probabilities to exhibit the phenomena of quantum interference). The second feature of quantum mechanics that necessitates complex numbers is the fact that transformation are continuous. I.e. if you have an allowed transition from one state to another, then you must also have a transition that is part way in-between.
For example consider a qubit with state $\left|\psi\right\rangle=a\left|0\right\rangle+b\left|1\right\rangle$, or
$$\begin{pmatrix}a\\b\end{pmatrix} $$
using a vector representation. Now if $\left|\psi\right\rangle$ is a valid state, then so is $\left|\psi'\right\rangle=a\left|0\right\rangle-b\left|1\right\rangle.$ Therefore, there is a transformation that takes $\left|\psi\right\rangle$ to $\left|\psi'\right\rangle.$ Written in a matrix representation this is simply the transformation
$$\mathbf T=\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}.$$
Now if $\mathbf T$ is an allowed translation, then there is also a valid translation $\mathbf T'$ that is halfway in between. I.e. $\mathbf{T'T'=T}$ or $\mathbf{T'=T^{1/2}}.$ However, because of the $-1$ in $\mathbf T$, there is no way to represent $\mathbf{T^{1/2}}$ with only real numbers. However, you can with complex numbers. This is due to the fact that complex numbers are algebraically complete or closed (i.e. every square root has a solution within that set of numbers), while the real numbers are not (there is no $\sqrt{-1}$ in the set of real numbers).
By contrast classical wave-functions generally describe fundamentally real and measurable quantities, which only make physical sense as real quantities. For instance the electric field amplitude at some point or the height of a water wave. Writing such functions as complex valued is simply a mathematical trick. Technically speaking, writing the wavefunction as complex-valued to do a calculation and then 'taking the real part' of the solution as you mention is known as the method of analytic representation, and is essentially just a mathematical trick that exploits the extra features of complex numbers to simplify mathematical problems that are inherently real-valued.
A: Here is another description, using only a single q-bit, that shows how the imaginary part of $|\psi\rangle$ plays a central role in capturing the state and its dynamics.
A spin-$\frac{1}{2}$ system is described by a two-state Hilbert space, so a general state takes the form
$$|\psi\rangle=\left[\begin{array}{c}
\alpha \\
\beta
\end{array}
\right],$$
with $|\alpha|^{2}+|\beta|^{2}=1$, so that the state is normalized.  With two complex numbers $\alpha$ and $\beta$, subject to one equation $|\alpha|^{2}+|\beta|^{2}=1$, there are three (real) parameters in this $|\psi\rangle$.  However, one of those parameters is superfluous:  the overall phase.  If we define a new spinor $|\psi'\rangle$ equal to $e^{i\chi}|\psi\rangle$, then $|\psi\rangle$ and $|\psi'\rangle$ describe the same physical state.  This means, for instance, that any expectation value $\langle\psi'|{\cal O}|\psi'\rangle$ remains equal to $\langle\psi|{\cal O}|\psi\rangle$, since
$$\langle\psi'|{\cal O}|\psi'\rangle=\langle\psi|e^{-i\chi}{\cal O}e^{i\chi}|\psi'\rangle=\langle\psi|{\cal O}|\psi\rangle,$$
since $e^{i\chi}$ is just a number and so commutes with the operator ${\cal O}$.
So the overall phase of this (or any other wave function) does not contain any physical information.  That leaves two real degrees of freedom that parameterize the physical state $|\psi\rangle$; these may be encapsulated in the single complex number $\alpha/\beta$ which can take any complex value (or be infinite, if $\beta=0$). The quantum-mechanical spin-$\frac{1}{2}$ can describe the same dynamics as a classical vector $\vec{S}$ of fixed length, with the two parameters specifying the state mappable onto the spherical angles $\theta$ and $\phi$ describing the direction of $\vec{S}$.
The connection is made through taking the expectation values of the three Cartesian components of the spin operator,
$$\left\langle\vec{S}\right\rangle=\frac{\hbar}{2}\left[\langle\sigma_{1}\rangle\hat{\imath}+
\langle\sigma_{2}\rangle\hat{\jmath}+\langle\sigma_{3}\rangle\hat{k}\right].$$
The expectation values of the Pauli matrices are
$$\begin{array}{rcl}
\langle\sigma_{1}\rangle=\langle\psi|\sigma_{1}|\psi\rangle & = &
[\alpha^{*}\,\beta^{*}]
\left[\begin{array}{cc}
0 & 1 \\
1 & 0
\end{array}
\right]
\left[\begin{array}{c}
\alpha \\
\beta
\end{array}
\right]=\beta^{*}\alpha+\alpha^{*}\beta=2\,\Re\{\alpha^{*}\beta\}\\
\langle\sigma_{2}\rangle=\langle\psi|\sigma_{2}|\psi\rangle & = &
[\alpha^{*}\,\beta^{*}]
\left[\begin{array}{cc}
0 & -i \\
i & 0
\end{array}
\right]
\left[\begin{array}{c}
\alpha \\
\beta
\end{array}
\right]=i\beta^{*}\alpha-i\alpha^{*}\beta=2\,\Im\{\alpha^{*}\beta\}\\
\langle\sigma_{3}\rangle=\langle\psi|\sigma_{3}|\psi\rangle & = &
[\alpha^{*}\,\beta^{*}]
\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}
\right]
\left[\begin{array}{c}
\alpha \\
\beta
\end{array}
\right]=
|\alpha|^{2}-|\beta|^{2}
\end{array}.$$
Note that the square of the length of the expectation value, $\langle\sigma_{1}\rangle^{2}+\langle\sigma_{2}\rangle^{2}+\langle\sigma_{3}\rangle^{2}$ (which is not the same as the expectation value of the square, $\langle\sigma_{1}^{2}\rangle+\langle\sigma_{2}^{2}\rangle+\langle\sigma_{3}^{2}
\rangle$) is normalized so that
$$\sqrt{\left\langle\vec{S}\right\rangle^{2}}=\frac{\hbar}{2}.$$
However, to have a fully accurate representation of the full space of possible directions evidently requires the use of complex numbers; if $\alpha$ and $\beta$ are both strictly real, then we will inevitably have $\left\langle S_{y}\right\rangle=0$.
It is straightforward to verify that if the direction of the vector $\left\langle\vec{S}\right\rangle$ is described by the polar angle $\theta$ and azimuthal angle $\phi$, meaning
$$\left\langle\vec{S}\right\rangle=\hat{n}=\frac{\hbar}{2}\left(\sin\theta\cos\phi\,\hat{\imath}+\sin\theta\sin\phi\,\hat{\jmath}+
\cos\theta\,\hat{k}\right)=\frac{\hbar}{2}\hat{n}$$
(in terms of a unit vector $\hat{n}$), then $|\psi\rangle$ takes the form
$$|\psi\rangle=\left[\begin{array}{c}
\cos(\theta/2) \\
e^{i\phi}\sin(\theta/2)
\end{array}
\right],$$
(up to a total phase, as always). Moreover, this spinor is also an eigenstate of the projection of the total spin $\vec{S}$ along the $\hat{n}$-direction:
$$\left(\vec{S}\cdot\hat{n}\right)|\psi\rangle=+\frac{\hbar}{2}|\psi\rangle.$$
Finally, when we introduce a time dependence—with, say, the Hamiltonian $H=-\mu B_{0}\,\sigma_{3}$ for a spin carrying magnetic moment $\mu$ in the presence of a magnetic field $\vec{B}=B_{0}\hat{k}$ along the $z$-direction—even an initially real spinor wave function will develop an imaginary part as it evolves in time.  Under the action of
$$H=\mu B_{0}\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}
\right],$$
an initial state
$$|\psi(0)\rangle=\left[\begin{array}{c}
\alpha \\
\beta
\end{array}
\right],$$
will evolve according to
$$|\psi(t)\rangle=\left[\begin{array}{c}
e^{-i\omega t/2}\alpha \\
e^{i\omega t/2}\beta
\end{array}\right].$$
The frequency here is $\omega=2\mu B_{0}/\hbar$. The presence of the imaginary phases is what causes the state to evolve in time. If the initial spinor is strictly real, then the expectation value of the spin will evolve in time according to
$$\left\langle\vec{S}\right\rangle(t)=\frac{\hbar}{2}\left(\sin\theta\cos\omega t\,
\hat{\imath}+\sin\theta\sin\omega t\,\hat{\jmath}+\cos\theta\,{\hat k}\right),$$
which is just the same as the precession of a classical magnetic moment $\vec{\mu}$ in an external magnetic field.
