Time-dependent Schrödinger Equation in $\mathbb{R}$
The Schrödinger equation is a fundamental equation in quantum mechanics that describes how the quantum state of a system evolves over time. It determines the behavior of particles at the atomic and subatomic scale using wavefunctions. This equation is essential for understanding the dynamics and probabilities of quantum systems.
$$i\hbar \frac{\partial}{\partial t} \Psi\left(x;t\right) = \left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V \right] \Psi\left(x;t\right)$$
In this case, this equation represents a free particle in quantum mechanics, where the potential $V$ is constant and the solutions are plane waves. $\hbar$ is the reduced Planck constant and $m$ is the mass of the particle.
The probability density of finding the particle at position $x$ at time $t$ is given by the square of the modulus of the wavefunction:
$$P(x;t) = |\Psi(x;t)|^2 = \Psi^*(x;t) \Psi(x;t)$$
where $\Psi^*$ denotes the complex conjugate of $\Psi$. This probability density must be normalized so that the total probability of finding the particle somewhere in space is 1:
$$\int_{-\infty}^{\infty} |\Psi(x;t)|^2 \, dx = 1$$
1. Separation of variables: to solve the Schrödinger equation using separation of variables, assume the wave function can be written as a product of spatial and temporal parts. Substitute this form into the equation, separate the variables, and solve the resulting ordinary differential equations independently.
$$i\hbar \frac{\partial}{\partial t} \Psi\left(x;t\right) = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \Psi\left(x;t\right) + V \Psi\left(x;t\right) \quad \text{and} \quad \boxed{\Psi\left(x;t\right) = \psi\left( x\right) T\left( t\right)}$$
$$i\hbar \psi \frac{dT}{dt} = -\frac{\hbar^2}{2m} T \frac{d^2 \psi}{dx^2} + V \psi T = T \left( -\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} + V \psi \right)$$
$$i\hbar \frac{1}{T} \frac{dT}{dt} = \frac{1}{\psi} \left( -\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} + V \psi \right) = -\frac{\hbar^2}{2m} \frac{1}{\psi} \frac{d^2 \psi}{dx^2} + V$$
$$i\hbar \frac{1}{T} \frac{dT}{dt} = -\frac{\hbar^2}{2m} \frac{1}{\psi} \frac{d^2 \psi}{dx^2} + V = \lambda$$
Thus, we arrive at the following system of ordinary differential equations (ODEs):
$$ \begin{cases} i\hbar \frac{1}{T} \frac{dT}{dt} = \lambda \\ -\frac{\hbar^2}{2m} \frac{1}{\psi} \frac{d^2 \psi}{dx^2} + V = \lambda \end{cases} $$
2. First equation: this is a first-order linear ODE describing the time evolution of a quantum system's phase.
$$i\hbar \frac{1}{T} \frac{dT}{dt} = \lambda \quad \Rightarrow \quad \frac{1}{T}dT = \frac{\lambda}{i\hbar}dt = -\frac{i\lambda}{\hbar}dt$$
$$\int \frac{1}{T} dT = -\frac{i\lambda}{\hbar} \int dt \quad \Rightarrow \quad \ln(T) = -\frac{i\lambda}{\hbar} t + C$$
$$\boxed{T(t) = e^{-\frac{i\lambda}{\hbar} t}}$$
3. Second equation: this is the time-independent Schrödinger equation describing the spatial behavior and energy eigenvalues of a quantum particle in a potential.
$$-\frac{\hbar^2}{2m} \frac{1}{\psi} \frac{d^2 \psi}{dx^2} + V = \lambda \quad \Rightarrow \quad \frac{d^2 \psi}{dx^2} = -\frac{(\lambda - V)2m}{\hbar^2} \psi$$
$$\text{let} \quad k^2 = \frac{(\lambda - V)2m}{\hbar^2} \quad \text{so,} \quad \frac{d^2 \psi}{dx^2} + k^2 \psi = 0$$
$$\text{As} \quad \psi = e^{rx} \quad \Rightarrow \quad \frac{d \psi}{dx} = re^{rx} \quad \Rightarrow \quad \frac{d^2 \psi}{dx^2} = r^2 e^{rx}$$
$$r^2 e^{rx} + k^2 e^{rx} = 0 \quad \Rightarrow \quad e^{rx} \left( r^2 + k^2 \right) = 0$$
$$r^2 + k^2 = 0 \quad \Rightarrow \quad r^2 = -k^2 \quad \Rightarrow \quad r = \pm ik$$
$$\psi(x) = e^{ikx}, \quad e^{-ikx}$$
As we have two solutions the linear combination between them contains all the possible solutions. So the general solution is nothing but:
$$\boxed{\psi(x) = Ae^{ikx} + Be^{-ikx}}$$
4. Final solution: as we assumed that $\Psi(x;t)$ must be the product of the functions $\psi(x)$ and $T(t)$, we can find the final solution.
$$\Psi(x;t) = \psi(x) T(t)$$
$$\Psi(x;t) = \left( Ae^{ikx} + Be^{-ikx} \right) e^{-\frac{i\lambda}{\hbar} t} = Ae^{ikx -\frac{i\lambda}{\hbar} t} + Be^{-ikx -\frac{i\lambda}{\hbar} t} = $$
$$ = Ae^{i(kx -\frac{\lambda}{\hbar} t)} + Be^{-i(kx +\frac{\lambda}{\hbar} t)}$$
Demonstration of $\lambda = E$: from the step 1 we have the following equation,
$$i\hbar \frac{1}{T} \frac{dT}{dt} = \frac{1}{\psi} \left( -\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} + V \psi \right) = \lambda$$
$$\frac{1}{\psi} \left( -\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} + V \psi \right) = \lambda \quad \Rightarrow \quad -\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} + V \psi = \lambda \psi$$
As the Hamiltonian operator $\hat{H}$ is defined as:
$$\hat{H} = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V$$
We have the following:
$$-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} + V \psi = \lambda \psi \quad \Rightarrow \quad \hat{H}\psi = \lambda \psi$$
If we take a look at the time-independent Schrödinger equation $\hat{H}\psi = E \psi$ and considering that $\psi$ does not depend on time we can see that:
$$\hat{H}\psi = \lambda \psi \quad \Leftrightarrow \quad \hat{H}\psi = E \psi$$
Thus $\lambda = E$, and replacing:
$$\Psi(x;t) = Ae^{i(kx -\frac{E}{\hbar} t)} + Be^{-i(kx + \frac{E}{\hbar} t)} \quad \text{if} \quad \omega = \frac{E}{\hbar}:$$
$$\boxed{\Psi(x;t) = Ae^{i(kx -\omega t)} + Be^{-i(kx + \omega t)}}$$
As we are dealing with a free particle, $V = 0$ so:
$$k^2 = \frac{(\lambda - V)2m}{\hbar^2} = \frac{\lambda 2m}{\hbar^2} = \frac{E 2m}{\hbar^2} \quad \Rightarrow \quad E = \frac{\hbar^2 k^2}{2m} \quad \text{and} \quad \omega = \frac{\hbar k^2}{2m}$$
5. Wave packet: The $\Psi(x;t)$ function represents a plane wave that extends infinitely. In quantum mechanics, a single wave does not represent a localized particle. To achieve that, a continuous superposition of many waves (wave packet) with different values of $k$, centered around a $k_0$, is required. The wave packet has the following form:
$$\Psi(x;t) = \int_{-\infty}^{\infty}a(k)e^{i(kx - \omega(k)t)} dk$$
We are only considering $e^{i(kx - \omega(k)t)}$ because if $\omega(k) < 0$ then we have $e^{-i(kx + \omega t)}$. So the function $\omega(k)$ works for both $e^{i(kx - \omega t)}$ and $e^{-i(kx + \omega t)}$.
This wave packet is created from the inverse Fourier transform, where $a(k)$ is the amplitude of the component with wavenumber $k$ (also called the envelope function). $\omega(k) = \frac{\hbar k^2}{2m}$ for free particles is the dispersion relation in non-relativistic quantum mechanics.
The most common form for a quantum wave packet is to use a Gaussian distribution centered at $k_0$:
$$a(k) = \frac{1}{\sqrt{2 \pi \sigma^2}}e^{- \frac{(k - k_0)^2}{2 \sigma^2}}$$
Where $k_0$ is the central wavenubmer (mean value of the momentum), and $\sigma$ is the width of the packet in k-space (inversely proportional to the spatial width).
Substituting $a(k)$ into the integral:
$$\boxed{\Psi(x;t) = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^2}}e^{- \frac{(k - k_0)^2}{2 \sigma^2}} e^{i(kx - \omega(k)t)} dk = \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{-\infty}^{\infty} e^{- \frac{(k - k_0)^2}{2 \sigma^2}} e^{i(kx - \omega(k)t)} dk}$$
This integral cannot always be solved (although it can be approximated numerically).
6. Case with analytical solution: in the particular case where $t=0$, the previous expression becomes the inverse Fourier transform of a Gaussian and results in
$$\Psi(x;0) = \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{-\infty}^{\infty} e^{- \frac{(k - k_0)^2}{2 \sigma^2}} e^{ikx} dk = e^{i k_0 x}e^{-\frac{x^2 \sigma^2}{2}}$$
7. Conclusion: unlike $e^{i(kx - \omega t)}$ the wave packet has a maximum probability of finding the particle within a finite region of space.
The packet propagates with a group velocity:
$$v_g = \frac{d \omega}{dk} = \frac{\hbar k_0}{m}$$
But it also spreads over time, because different k-components travel at different velocities.