Basic Elements of Quantum Mechanics

Let’s derive the Time Dependent Schrödinger equation:

🧠 We know the following:
1. Plane wave description:
  
  $$ \Psi(x,t) = A\,\mathrm{e}^{i(kx-\omega t)} $$
2. De Broglie relation:
  
  $$ \begin{aligned} \lambda&=\frac{h}{p} \\ \text{since: } k=2\pi/\lambda \; \Rightarrow \qquad k&=\frac{2\pi}{h}\,p = \frac1{\hbar}\, p \end{aligned} $$
3. Planck law:
  
  $$ \omega=2\pi\nu=\frac{2\pi}{h}\,E = \frac1{\hbar}\,E $$
4. kinetic energy of a non relativistic particle:
$$ \frac{p^2}{2m}=E $$
💡 Combining the equations:
1. Using equation b, c and d:
  
  $$ \frac{\hbar^2k^2}{2m}=\hbar\omega $$
2. Using equation a:
  
  $$ \frac{\partial^2}{\partial x^2}\Psi=-k^2\Psi \qquad \qquad \frac{\partial}{\partial t}\Psi=-i\omega\Psi $$
3. Finally putting these together we get:
  
  $$ \begin{aligned} \frac{\hbar^2k^2}{2m}\Psi=\hbar\omega\Psi\quad\quad\;& \\ -\frac{\hbar^2}{2m}(\underbrace{-k^2\Psi}{\frac{\partial^2}{\partial x^2}\Psi})=i\hbar(\underbrace{-i\omega\Psi}{\frac{\partial}{\partial t}\Psi})& \\ \boxed{-\frac{\hbar^2}{2m}\,\frac{\partial^2}{\partial x^2}\Psi=i\hbar\,\frac{\partial}{\partial t}\Psi}\quad& \end{aligned} $$
🚀 Adding a potential $V(x,t)$:
1. The kinetic energy of a non relativistic particle becomes
  
  $$ \frac{p^2}{2m}+V(x,t)=E $$
2. By following a similar process to above we find
  
  <aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/51f9300a-5620-44d2-8d62-e07037251025/TDSE.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/51f9300a-5620-44d2-8d62-e07037251025/TDSE.png" width="40px" /> Time Dependent Schrödinger Equation (TDSE):
  
  $$ \left(-\frac{\hbar^2}{2m}\,\frac{\partial^2}{\partial x^2}+V(x,t)\right)\Psi(x,t)=i\hbar\,\frac{\partial}{\partial t}\Psi(x,t) $$
  - $\Psi(x,t)$ is the wavefunction
  - $m$ is mass of the particle in free space
  - $\hbar$ is $h/2\pi \simeq 1.055\times 10^{-34} \;\text{J}\,\text{s}$
  - $V(x,t)$ potential energy of a region of space </aside>
3. We can then convert it to 3 dimensions:
  
  $$ \small{\left(-\frac{\hbar^2}{2m}\underbrace{\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right)}{\nabla^2}+V(\underbrace{x,y,z}{\vec r},t)\right)\Psi(\underbrace{x,y,z}{\vec r},t)=i\hbar\,\frac{\partial}{\partial t}\Psi(\underbrace{x,y,z}{\vec r},t)} $$
4. Which simplifies to:
  
  <aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/538b394e-1a5a-4ddf-bb49-3eca41780856/3D_TDSE.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/538b394e-1a5a-4ddf-bb49-3eca41780856/3D_TDSE.png" width="40px" /> 3 dimensional generalisation of the TDSE:
  
  $$ \left(-\frac{\hbar^2}{2m}\nabla^2+V(\vec{r},t)\right)\Psi(\vec{r},t)=i\hbar\,\frac{\partial}{\partial t}\Psi(\vec{r},t) $$
  
  </aside>

Properties of Wavefunction

Normalization:

$$ \int_{\text{space}}|\Psi(\vec{r},t)|^2\mathrm{d}V=\int_{-\infty}^\infty|\Psi(x,t)|^2\mathrm{d}x=1 $$
$\Psi$ is single values (for each point in the domain it has one unique value in the range)
$\Psi$ is finite everywhere.
$\Psi$ is continuous and smooth except if $V(x)$ has an infinite discontinuity
$\Psi$ is fragile (measuring a property of a particle changes $\Psi$ changing its future value)

Time evolution

🐟 Conditions:
1. Time-independent potential: $V(x,t)=V(x)$
2. TDSE is a linear equation so: $\Psi (x,t)=\psi (x) \, T(t)$ so we find:
  
  $$ \frac{\partial^2\Psi}{\partial x^2}=T\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}\qquad\qquad\frac{\partial\Psi}{\partial t}=\psi\frac{\mathrm{d}T}{\mathrm{d}t} $$
💡 Time Independent Schrödinger Equation:

$$ \begin{aligned} \left(-\frac{\hbar^2}{2m}\,\frac{\partial^2}{\partial x^2}+V(x)\right)\overbrace{\Psi(x,t)}^{\psi(x)\, T(t)}&=i\hbar\,\frac{\partial}{\partial t}\overbrace{\Psi(x,t)}^{\psi(x)\, T(t)} \\ \left(-\frac{\hbar^2}{2m}\,\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}+V(x)\psi(x)\right)T(t) &= i\hbar\psi(x)\,\frac{\mathrm{d}T}{\mathrm{d}t} \\ \frac1{\psi(x)}\left(-\frac{\hbar^2}{2m}\,\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}+V(x)\psi(x)\right) &= \frac{i\hbar}{T(t)}\,\frac{\mathrm{d}T}{\mathrm{d}t} \end{aligned} $$

Both sides depend on different variables so they must be equal to a constant so we can write it:

<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/8e6ca880-3e4b-417b-9b35-de445326ef0e/TISE.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/8e6ca880-3e4b-417b-9b35-de445326ef0e/TISE.png" width="40px" /> Time Independent Schrödinger Equation (TISE):

$$ \left(-\frac{\hbar^2}{2m}\,\frac{\mathrm{d}^2}{\mathrm{d}x^2}+V(x)\right)\psi(x) = E\,\psi(x) $$

It is an eigenfunction equation where $\psi(x)$ is an eigenfunction and $E$ are its eigenvalues In general it has a set of discrete solution $\{ \psi_i (x) \}$ each with their own $E_i$

</aside>
💡 General solution to TDSE:
1. Separation implies:
  
  $$ i\hbar\,\frac{\mathrm{d}T}{\mathrm{d}t}=E\,T(t)\quad\Rightarrow\quad T(t)=A\mathrm{e}^{-iEt/\hbar} $$
2. So general solution is
  
  $$ \Psi(x,t) = \sum_i A_i\,\psi_i(x)\,\mathrm{e}^{-iE_it/\hbar} $$
3. If $A_i\ne 0$
  
  $$ \begin{aligned}|\Psi(x,t)|^2 &= \Psi(x,t)^\,\Psi(x,t) \\&= |A_n|^2\,\psi_n(x)^\,\mathrm{e}^{+iE_nt/\hbar}\,\mathrm{e}^{-iE_nt/\hbar}\,\psi_n(x) \\&= |A_n|^2\,|\psi_n(x)|^2\end{aligned} $$
  
  So the probability distribution doesn’t change with time and the eigenfunctions of the TISE $\psi_i (x)$ are called stationary states