Schrödinger equation

Schrödinger’s equation serves the same role in quantum mechanics as Newton’s second law does in classical mechanics. Given initial conditions of the wave function $\Psi$ , solving the Schrödinger equation tells us its behavior for all future time.

i\hbar \frac{\partial\Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + V(x)\Psi.

To solve the SE, let us assume our wavefunction is separable, i.e.

\Psi(x,t) = T(t) \psi(x).

This gives us

\begin{align*} i\hbar \psi \frac{\partial T}{\partial t} &= -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} T(t) + V(x) \psi(x) T(t) \\ \frac{i\hbar}{T} \frac{\partial T}{\partial t} &= -\frac{\hbar^2}{2m} \frac{1}{\psi(x)} \frac{\partial^2 \psi}{\partial x^2} + V(x). \end{align*}

We see that the left and right side of the equation depend on different variables, so the only way for this equation to hold is for it to equal a constant. We will see that this constant is the energy $E$ of the system.

\begin{align*} i\hbar \frac{\partial T}{\partial t} &= ET &\quad E\psi &= -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} + V(x)\psi \\ T &= Ce^{-iEt/\hbar} & E\psi &=: \hat H\psi. \end{align*}

We have defined the Hamiltonian operator $\hat H$ above. We see that what we get is an eigenvalue equation for $\HH$ : the time-independent wavefunction $\psi$ is its eigenvector, and its corresponding energy $E$ is the eigenvalue.

This gives us the time-independent Schrödinger equation:

\begin{align*} -\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + V(x)\psi = E\psi. \end{align*}

The solutions are the eigenfunctions of the Hamiltonian.

In three dimensions

The Hamiltonian is the energy operator, so in 3D it reads

\HH = -\frac{\hbar^2}{2m} \left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right) + V(x,y,z).

We can write this in terms of the Laplacian as

\begin{align*} \left( -\frac{\hbar^2}{2m} \nabla^2 + V(\rr)\right) \psi(\rr) = E \psi(\rr). \end{align*}

Most of the time our potential is spherical, so $V(\rr) = V(r)$ is only a function of the radius.