Deriving the Equation from Operators

For a free particle, we have

$$\bgroup\color{black}{p^2\over 2m}=E\egroup$$

$$\bgroup\color{black}-{\hbar^2\over 2m}{\partial^2\over \partial x^2}\psi=i\hbar{\partial\over \partial t}\psi \egroup$$

Lets try this equation on our states of definite momentum.

$$\bgroup\color{black}-{\hbar^2\over 2m}{\partial^2\over \parti... ...artial t}{1\over\sqrt{2\pi\hbar}}e^{i(p_0x-E_0t)/\hbar} \egroup$$

The constant in front of the wave function can be removed from both sides. Its there for normalization, not part of the solution. We will go ahead and do the differentiation.

$$\begin{eqnarray*} -{\hbar^2\over 2m}{-p_0^2\over\hbar^2}e^{i(p_0x-E_0t)/\hbar} =... ...0^2\over 2m}e^{i(p_0x-E_0t)/\hbar}=E_0e^{i(p_0x-E_0t)/\hbar} \\ \end{eqnarray*}$$

Our wave function will be a solution of the free particle Schrödinger equation provided $E_0={p_0^2\over 2m}$. This is exactly what we wanted. So we have constructed an equation that has the expected wave-functions as solutions. It is a wave equation based on the total energy.

Adding in potential energy, we have the Schrödinger Equation

$${-\hbar^2 \over 2m}{\partial^2\psi(x,t)\over\partial x^2}+V(x)\psi(x,t) =i\hbar{\partial\psi(x,t)\over\partial t}$$

or

$$\bgroup\color{black}H\psi(x,t)=E\psi(x,t)\egroup$$

where

$$\bgroup\color{black} H={p^2\over 2m} + V(x) \egroup$$

is the Hamiltonian operator.

In three dimensions, this becomes.

$$H\psi(\vec{x},t)={-\hbar^2 \over 2m}\nabla^2\... ...(\vec{x})\psi(\vec{x},t) =i\hbar{\partial\psi(\vec{x},t)\over\partial t}$$

We will use it to solve many problems in this course.

So the Schrödinger Equation is, in some sense, simply the statement (in operators) that the kinetic energy plus the potential energy equals the total energy.