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.

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

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

then, $E={\widehat {T}}$

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.

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

Recall the definition of the Hamiltonian operator.

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

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

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

Recall also the definition of the kinetic energy operator.

${\widehat {T}}={\frac {-\hbar ^{2}}{2m}}{\frac {\ d^{2}}{dx^{2}}}$

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

$\left({\frac {-\hbar ^{2}}{2m}}{\frac {\ d^{2}}{dx^{2}}}+0\right)\psi (x)=E\psi (x)$

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

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.

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

Recall the definition of the angular wavenumber.

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

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

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

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

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

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

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

The above holds true if either

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

or

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

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.