# Quantum Chemistry/Example 33

## 1. Background Information[edit | edit source]

#### 1.1 Wave Function[edit | edit source]

Wave is a verb meaning "to move back and forth". Oscillations happen when a build-up of an opposing force, such water waves, springs, or pendulums, counteracts displacements in either direction. The reason it's called a "wave" function is because it satisfies a wave equation, known as the Schrödinger equation. This equation describes how the wave function evolves over time, much like how classical wave equations describe the propagation of waves in classical physics. So, the term "wave function" reflects both the wave-like behavior of quantum entities and the mathematical form of the function, which satisfies a wave equation. The Schrödinger equation is also 2nd order; for free electron, it has the same oscillatory “wave” solution and is given by;

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

#### 1.2 Free Electron[edit | edit source]

An electron is said to be free if it can move along the x-axis in any direction without any force acting on it (i.e., 𝒱 𝑥 = 0, 𝑥 = −∞, ∞ ). It is not bound by an external force, or equivalently not in a region where its potential energy varies. In quantum mechanics, it indicates that the particle is in a region of uniform potential, which is often set to zero in the area of interest because any point in space can have an arbitrary potential set to zero.

The wave function of a free electron is typically represented as a plane wave, which is a solution to the Schrödinger equation. The Schrödinger equation is a fundamental equation in quantum mechanics that describes how the quantum state of a physical system changes with time. The wave function of a free electron can be written as:

${\displaystyle \psi (x,t)=A*e^{(}i(kx-wt))}$

where:

-  is the wave function, which depends on position x and time t,

-  is the amplitude of the wave,

- k is the wave number, which is related to the momentum of the electron,

-  is the angular frequency, which is related to the energy of the electron,

-  is the imaginary unit.

## Example

#### 2.1 Derive the wave function of a free electron[edit | edit source]

Time-independent Schrödinger wave equation is given by;

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

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

Next define the Hamiltonian,

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

Substitute the potential energy operator

${\displaystyle {\widehat {V}}=V(x)=0}$ (free electron)

and the kinetic energy operator

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

${\displaystyle ({\hat {\mathrm {T} }}+{\widehat {V}})\psi =E*\psi }$

To obtain the Schrödinger equation for a free particle, rearrange the differential equation; isolate 2nd derivative on LHS

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

and make the substitution

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

Since  is the kinetic energy,

${\displaystyle \mathrm {E} ={\frac {p^{2}}{2m}}}$

p and the wave vector k are related, p = ${\displaystyle \hbar }$k

we also could recognize that ${\displaystyle {\frac {2mE}{\hbar ^{2}}}}$ is just ${\displaystyle k^{2}}$as shown in this equation

${\displaystyle {\frac {2mE}{\hbar ^{2}}}=({\frac {2m}{\hbar ^{2}}})({\frac {p^{2}}{2m}})}$

${\displaystyle =({\frac {2m}{\hbar ^{2}}})({\frac {\hbar ^{2}k^{2}}{2m}})}$

${\displaystyle =k^{2}}$

after rearranging and substitution of result from equation, the result is

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

This linear second-order differential equation can be solved by separating into two factors, and each is set equal to 0.${\displaystyle ({\frac {d^{2}}{dx^{2}}}+k^{2})}$${\displaystyle \psi (x)}$${\displaystyle =}$ ${\displaystyle ({\frac {d}{dx}}+ik)}$ ${\displaystyle ({\frac {d}{dx}}-ik)}$${\displaystyle \psi (x)}$${\displaystyle =}$${\displaystyle 0}$Equation is true if either

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

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

By concurrently rearranging and denoting the two equations and their solutions with a + and a - sign, the result is

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

which translates to

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

and finally

${\displaystyle {\psi _{\pm }(x)}}$${\displaystyle =}$ ${\displaystyle A_{\pm }}$${\displaystyle e^{i}}$${\displaystyle ^{k}}$${\displaystyle ^{x}}$

Any function that satisfies the requirements of wave functions and is a solution to this differential equation is a valid wave function for this system.

## Reference[edit | edit source]

1. Libretexts. (2022, April 21). 5.1: The Free Particle. Chemistry LibreTexts. https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book_Quantum_States_of_Atoms_and_Molecules_(Zielinksi_et_al)/_Translational_States/_The_Free_Particle
2. Anderson, J.M. Introduction to Quantum Chemistry, 1969, W.A. Benjamin, Inc, pg.81-91.