PX275 - G11a - schrödinger equation

PX156 - C1 - the schrodinger equation

in 3D:

i ℏ \frac{\partial ψ}{\partial t} = - \frac{ℏ^{2}}{2 m} \nabla^{2} ψ + V ψ

considering a case without potential, ie: $V = 0$ , and let $D = i ℏ / 2 m :$

\frac{\partial ψ}{\partial t} = D \nabla^{2} ψ

this is identical to the diffusion equation
using separation of variables:

ψ (\vec{r}, t) = R (\vec{r}) T (t) = X (x) Y (y) Z (z) T (t)

\begin{matrix} i ℏ R \frac{d T}{d t} = - \frac{ℏ^{2}}{2 m} T \nabla^{2} R + V R T \\ \frac{i ℏ}{T} \frac{d T}{d t} = - \frac{ℏ^{2}}{2 m R} \nabla^{2} R + V = constant \end{matrix}

rather than deriving solutions for positive, negative and zero constant, considering special, physically-meaningful solutions
suppose the constant of separation is $E$ , the energy associated with $ψ$
the time dependent part:

\begin{matrix} \frac{d T}{d t} = - \frac{i}{ℏ} T E \\ ⟹ T (t) = C \exp (- \frac{i E}{ℏ} t) \end{matrix}

the spatial part, for $E > V$ :

\frac{1}{R} \nabla^{2} R = - \frac{2 m}{ℏ^{2}} (E - V) = - k^{2}

trial solution:

\begin{matrix} R = C \exp (i \vec{k} \cdot \vec{r}) \\ \nabla^{2} R = - k^{2} R \\ ⟹ \frac{ℏ^{2} k^{2}}{2 m} + V = E \\ \frac{p^{2}}{2 m} + V = E \end{matrix}

where, $p = ℏ k$ is the debroglie momentum

the above equation represents the total energy with the first term being the kinetic energy:

T + V = E

particle in a potential well

considering a 1D case:

\begin{matrix} \frac{1}{X} \frac{d^{2} X}{d x^{2}} = - \frac{2 m}{ℏ^{2}} (E - V) = - k^{2} \\ ⟹ X = A \cos k x + B \sin k x \end{matrix}

the boundary conditions de to the well:

\begin{matrix} ψ (x = 0, t) = ψ (x = L, t) = 0 \forall t \\ X (0) = 0 ⟹ A = 0 \\ X (L) = 0 ⟹ k = \frac{n π}{L} \\ ∴ ψ (x, t) = \sum_{n} B_{n} \sin (\frac{n π x}{2}) \exp (- \frac{i E t}{ℏ}) \end{matrix}

now, considering a 3D case:

\begin{matrix} \frac{1}{R} \nabla^{2} R = - \frac{2 m}{ℏ^{2}} (E - V) = - k^{2} \\ \dots \\ \frac{1}{X} \frac{d^{2} X}{d x^{2}} + \frac{1}{Y} \frac{d^{2} Y}{d y^{2}} + \frac{1}{Z} \frac{d^{2} Z}{d z^{2}} = - k^{2} \\ \frac{1}{X} \frac{d^{2} X}{d x^{2}} = - k^{2} - \frac{1}{Y} \frac{d^{2} Y}{d y^{2}} - \frac{1}{Z} \frac{d^{2} Z}{d z^{2}} = - k_{x}^{2} \end{matrix}

the solutions for $X :$

X (x) = A \cos k_{x} x + B \sin k_{x} x

similarly for $Y$ and $Z :$

\begin{matrix} \frac{1}{Y} \frac{d^{2} Y}{d y^{2}} = - k^{2} - \frac{1}{X} \frac{d^{2} X}{d x^{2}} - \frac{1}{Z} \frac{d^{2} Z}{d z^{2}} = - k_{y}^{2} \\ Y (y) = C \cos k_{y} y + D \sin k_{y} y \\ \frac{1}{Z} \frac{d^{2} Z}{d z^{2}} = - k^{2} - \frac{1}{X} \frac{d^{2} X}{d x^{2}} - \frac{1}{Y} \frac{d^{2} Y}{d y^{2}} = - k_{z}^{2} \\ Z (z) = E \cos k_{z} z + F \sin k_{z} z \\ k^{2} = k_{x}^{2} + k_{y}^{2} + k_{z}^{2} = \frac{2 m}{ℏ^{2}} (E - V) \end{matrix}

ψ (\vec{r}, t) = \sum_{n} \sum_{m} \sum_{l} A_{n m l} \sin (\frac{n π x}{L}) \sin (\frac{m π y}{L}) \sin (\frac{l π z}{L}) \exp (\frac{- i E t}{ℏ})

considering a special case, $V = 0$ inside the box:

\begin{matrix} k^{2} = \frac{2 m E}{ℏ^{2}} \\ E = \frac{ℏ^{2}}{2 m} {\frac{π}{L}}^{2} (n^{2} + m^{2} + l^{2}) \end{matrix}

this is the quantization of energy in 3D