For the time-dependent quantum mechanics equation, Schrödinger postulate that it takes the form:
-\frac{\hbar}{2m}\nabla \Psi(\mathbf r,t) + V(\mathbf r,t) \Psi(\mathbf r,t) = i\hbar\frac{\partial \Psi(\mathbf r,t)}{\partial t}
Considering an uniform potential (V=0) and the energy of the form:
E= \hbar\omega, \quad k=\sqrt{\frac{2mE}{\hbar^2}}
plain waves of the form:
e^{-i(\omega t \pm kx)} = e^{-i \frac{E}{\hbar} t} e^{\pm i kx}
are solutions, with the minus sign for waves propagating in the forward direction, and the plus for waves propagating in the backward direction.
Schrödinger chose a sign convention for the right hand side, and with this convention a wave \propto e^{ikx} is going in the positive x direction.
Considering a solution of the form \Psi(x,t) = A e^{i(kx - \omega t)}, where A is the amplitude, k is the wave number, and \omega is the angular frequency, it is possible to substitute this form into the Schrödinger equation to see how the choice of sign in the exponential affects the direction of propagation.
i\hbar\frac{\partial}{\partial t} (A e^{i(kx - \omega t)}) = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} (A e^{i(kx - \omega t)})
The time derivative yields -i A \omega e^{i(kx - \omega t)}, and the second spatial derivative yields -A k^2 e^{i(kx - \omega t)}.
Substituting these results into the Schrödinger equation gives:
i\hbar (-I A \omega e^{i(kx - \omega t)}) = -\frac{\hbar^2}{2m}(-A k^2 e^{i(kx - \omega t)})
Simplifying this equation will allow us to understand the relationship between \omega, k, and the direction of the wave propagation.
After simplification:
\frac{A\hbar(-\hbar k^2 + 2m\omega)}{2m} e^{i(kx - \omega t)}
For this equation to hold as an identity (i.e., for all x and t), the coefficient of e^{i(kx - \omega t)} must be zero, which gives the dispersion relation:
-\hbar k^2 + 2m\omega = 0 \Rightarrow \omega = \frac{\hbar k^2}{2m}
This relation shows the energy \hbar \omega of the particle is proportional to the square of its momentum \hbar k, consistent with the free particle energy-momentum relation.
Regarding the sign in the exponential, e^{i(kx - \omega t)} represents a wave moving in the positive x direction because as t increases, for the phase (kx - \omega t) to remain constant (which it must for a point of constant phase on the wave), x must increase if \omega and k are both positive. This is because an increase in t must be compensated by an increase in x to keep the phase constant, indicating motion in the positive x direction.
If the sign before \omega t in the exponential were positive, it would imply e^{i(kx + \omega t)}, which, for the wave’s phase to remain constant as time progresses, would require x to decrease, suggesting a wave moving in the negative x direction.
Therefore, the convention e^{i(kx - \omega t)} for the wave function implies propagation in the positive x direction. This convention is chosen for consistency with the forward movement over time and is rooted in the form of the wave equation and the interpretation of wave function phases.
Before checking this equation further, it is necessary to check that is consistent with the time-independent equation. Considering the time-independent equation:
-\frac{\hbar}{2m}\nabla \psi(\mathbf r) + V(\mathbf r) \psi(\mathbf r) = E \psi(\mathbf r)
A solution \psi(\mathbf r) of this equation, when the energy has a definite E which is an eigenvalue, is not a solution of the time dependent equation, because the right hand side does not have any time dependence.
To solve this apparent contradiction, it is possible to consider a solution of the type:
\Psi(\mathbf r,t) = \psi(\mathbf r) e^{\frac{-iEt}{\hbar}}
Substituting in the time independent equation:
\begin{aligned} & -\frac{\hbar}{2m}\nabla \psi(\mathbf r) e^{\frac{-iEt}{\hbar}} + V(\mathbf r) \psi(\mathbf r) e^{\frac{-iEt}{\hbar}} = E \psi(\mathbf r) e^{\frac{-iEt}{\hbar}} \\ & e^{\frac{-iEt}{\hbar}} \left[ -\frac{\hbar}{2m}\nabla \psi(\mathbf r) + V(\mathbf r) \psi(\mathbf r) \right] = E \psi(\mathbf r) e^{\frac{-iEt}{\hbar}} \end{aligned}
This is possible as e^{\frac{-iEt}{\hbar}} does not depends on \mathbf r; as a consequence the time independent solution is satisfied. Considering now the time dependent equation and substituting:
\begin{aligned} & -\frac{\hbar}{2m}\nabla \Psi(\mathbf r,t) + V(\mathbf r,t) \Psi(\mathbf r,t) = i\hbar\frac{\partial \Psi(\mathbf r,t)}{\partial t} \\ & e^{\frac{-iEt}{\hbar}} \left[ -\frac{\hbar}{2m}\nabla \psi(\mathbf r) + V(\mathbf r) \psi(\mathbf r) \right] = i\hbar\frac{\partial}{\partial t}\left(\psi(\mathbf r) e^{\frac{-iEt}{\hbar}}\right) = i\hbar\frac{-iE}{\hbar}\psi(\mathbf r) e^{\frac{-iEt}{\hbar}} = E\Psi(\mathbf r,t) \end{aligned}
So this wave function also solve the time dependent Schrödinger equation.
So every solution, \psi( \mathbf r), of the time-independent Schrodinger equation with eigenenergy E is also a solution of the time-dependent equation, it by a factor e^{\frac{-iEt}{\hbar}}.
This does not introduce any particular issue on the measurable quantities and they are stable in time even if this oscillating factor is added to the wave function. For example, the probability density:
| \Psi(\mathbf r,x))|^2 = \overline{ \Psi(\mathbf r,x))}\Psi(\mathbf r,x) = \overline {\psi(\mathbf r)} e^{\frac{iEt}{\hbar}} \psi(\mathbf r) e^{\frac{-iEt}{\hbar}} = |\psi(\mathbf r)|^2
So the probability density is indeed not changing with time.
The classical wave equation take the form:
\nabla^2 f(\mathbf r,t) = \frac{k^2}{\omega^2} \frac{\partial f^2(\mathbf r,t)}{\partial t^2}
which has a second time derivative on the left, while Schrödinger equation has a first order derivatives; function of the form:
f(\mathbf r, t) \propto e^{i(kx - \omega t)}
are also a solution of the classical equation.
Furthermore, the Schrödinger equation has a i on the right hand side, so in general the solution has to be a complex function.
As example, for a potential V(\mathbf r)=0, f(x) = e^{i\left(kz - \frac{Et}{\hbar}\right)} is a solution of both, but \sin\left(kz - \frac{Et}{\hbar}\right) is not a solution of the Schrödinger equation, while it is still a solution of the classical version.
In classical wave dynamics, observing a wave’s shape at a given moment doesn’t reveal its direction of motion or confirm if it’s a mix of movements. This snapshot alone is insufficient for predicting future behavior; additional data, like the wave’s time derivative, is required. Conversely, Schrodinger’s time-dependent equation allows for future predictions based solely on the wave’s current state. The wave function encapsulates all possible information about the quantum state, enabling us to foresee its evolution.
With the Schrödinger equation, if the wave function is known at a time t_0, it is possible to calculate the wave function at a future time \delta t as:
\Psi(\mathbf r,t+\delta t) \cong \Psi(\mathbf r,t_0) + \frac{\partial \Psi}{\partial t}\bigg|_{\mathbf r,t_0}\delta t
And therefore, using successive increment \delta t it is possible to predict the future evolution of the wave function; also, integrating back in time it is possible to deduce the past time evolution of the wave function as well.
The idea of superposition is that if there are two solution to a linear equation, then a linear combination of these solution is still a solution; in the time-independent Schrödinger equation, this is not a case that happen often, but in case of the time-dependent, it arise all the time: at a given point in time, any function in space, as far it has a sufficiently well behaved second derivative, it is a solution and therefore, because of linearity, a sum of many solutions it is also a solution.
-\frac{\hbar}{2m}\nabla \Psi(\mathbf r,t) + V(\mathbf r,t) \Psi(\mathbf r,t) = i\hbar\frac{\partial \Psi(\mathbf r,t)}{\partial t}
This equation is linear because it satisfies the two criteria of linearity:
Considering a potential V which is constant in time, each of the energy eigenstates \psi_n(\mathbf r) is a solution of the time-dependent Schrödinger equation (if multiplied by the proper factor e^{-\frac{iE_n t}{\hbar}}, and the set of these eigenfunctions forms a complete set; therefore, the wavefunction at t = 0 can be expanded in the form:
\Psi(\mathbf r,0) = \sum_n a_n \psi_n(\mathbf r)
Since it is known how these function will evolve with time, then, by linear superposition, the solution at time t is:
\Psi(\mathbf r,t) = \sum_n a_n \psi_n(\mathbf r)e^{-\frac{iE_n t}{\hbar}}
This is the solution of the time dependent equation when the potential doesn’t vary with time.
Since a very large number of problems do have potentials that are constant in time, this is a powerful way to look at the time evolution of quantum mechanical systems.