Quantum Chemistry/Example 33

Derive the wave function of a free electron.

A free, or unbound, electron is a particle that has no forces acting on it. In other words, it travels in a region of uniform potential. Thus, a free electron travelling in either direction along the x-axis has only kinetic energy, and no potential energy.

$E=\hat{T}+\hat{V}$

since $\hat{V}=V(x)=0$

then, $E=\hat{T}$

The wavefunction describes a particle's behaviour as a wave in terms of its position as a function of time. Since electrons move very quickly, a time-averaged distribution of electron density is observed. Therefore, the wave function of a free electron can be derived using Schrödinger's time-independent wave equation.

$\hat{H}\psi =E\psi$

Recall the definition of the Hamiltonian operator.

$\hat{H}=\hat{T}+\hat{V}$

This definition can be substituted into the time-independent Schrödinger wave equation.

$(\hat{T}+\hat{V})\psi =E\psi$

Recall also the definition of the kinetic energy operator.

$\hat{T}=\frac{-{\hslash }^2}{2m}\frac{d^2}{d{x}^2}$

Now, the definitions of the kinetic energy and potential energy operators can be substituted into the time-independent Schrödinger wave equation.

$(\frac{-{\hslash }^2}{2m}\frac{d^2}{d{x}^2}+0)\psi (x)=E\psi (x)$

$\frac{-{\hslash }^2}{2m}\frac{d^2\psi (x)}{d{x}^2}=E\psi (x)$

This differential equation can be rearranged to isolate for the second derivative on the left-hand side. The result is the Schrödinger equation of a free electron.

$\frac{d^2\psi (x)}{d{x}^2}=-\frac{2mE}{{\hslash }^2}\psi (x)$

Recall the definition of the angular wavenumber.

$k=\sqrt{\frac{2mE}{{\hslash }^2}}$

This definition can be substituted into the Schrödinger equation of a free electron.

$\frac{d^2\psi (x)}{d{x}^2}=-k^2\psi (x)$

The second derivative in the equation above requires solving a second order differential equation for the wavefunction.

$(\frac{d^2}{d{x}^2}+k^2)\psi (x)=0$

This linear second order differential equation can be solved by separating it into two first-order differential equations.

$(\frac{d}{dx}+ik)(\frac{d}{dx}-ik)\psi (x)=0$

The above holds true if either

$(\frac{d}{dx}+ik)\psi (x)=0$

or

$(\frac{d}{dx}-ik)\psi (x)=0$

Now, the first-order differential equations can be solved simultaneously.

$\frac{d{\psi }_{\pm }(x)}{dx}\pm ik{\psi }_{\pm }(x)=0$

$\frac{d{\psi }_{\pm }(x)}{dx}=\pm ik{\psi }_{\pm }(x)$

$\frac{d{\psi }_{\pm }(x)}{{\psi }_{\pm }(x)}=\pm ikdx$

To eliminate the derivative, integrate both sides of the equation.

$ln{\psi }_{\pm }(x)=\pm ikx+C_{\pm }$

To eliminate the natural logarithm, exponentiate both sides of the equation.

${\psi }_{\pm }(x)=e^{\pm ikx+C_{\pm }}$

${\psi }_{\pm }(x)=e^{\pm ikx}{e}^{C_{\pm }}$

Set the $e^{C_{\pm }}$ term equal to a constant, $A_{\pm }$.

${\psi }_{\pm }(x)=A_{\pm }{e}^{\pm ikx}$

This is the wave function of a free electron, where the $\pm$ sign represents the direction in which the wave is propagating, since the free electron does not have defined boundary conditions.