Time-independent Schrodinger Equation

The time-independent variant for Schrodinger’s equation is \(Hψ=Eψ\) where \(ψ\) is a wave function, \(E\) is the energy of the wave, and \(H\) is the Hamiltonian operator. \(\psi\) is an eigenfunction with eigenvalue \(E\). Below we will derive the value of \(E\).

Derivation of constant \(E\)

To derive \(E\), the energy (or amplitude) of the wave, we first manipulate the classical equation for kinetic energy to get it in terms of momentum \(p\):

\[E = \frac{1}{2}mv^2 = \frac{(mv)^2}{2m} = \frac{p^2}{2m}\]

Despite the above energy equation being specific to classical mechanics, we can use the particle momentum \(p\), \(p=\frac{h}{λ}\) ,in place of classical momentum \(p\) to get quantized \(E\).

\[E=\frac{p^2}{2m}=\frac{(\frac{h}{λ})^2}{2m}\]

Where \(h\) is Planck’s constant and \(λ\) is the particle’s wavelength. \(λ=\frac{2π}{k}\) (where \(k\) is the spatial frequency of the wave) is then substituted into the equation to achieve

\[E=\frac{h(\frac{k}{2π})^2}{2m}\]

To simplify the appearance of the final equation, we substitute \(ℏ=\frac{h}{2π}\). Thus the constant, \(E\), is equal to

\[E=\frac{ℏ^2 k^2}{2m}\]

Derivation of Hamiltonian Operator \(H\)

From the (proven) relationship \(∇^2 ψ=-k^2ψ\), and the constant \(E=\frac{ℏ^2 k^2}{2m}\), we can derive the expression for the Hamiltonian operator, \(H\). Starting with \(Hψ=Eψ\), the known constant \(\frac{ℏ^2 k^2}{2m}\) for \(E\) is plugged in, as shown below:

\[Hψ=\frac{ℏ^2 k^2}{2m} ψ\]

By slightly manipulating the equation, we obtain the following equivalent expression:

\[Hψ=-\frac{ℏ^2}{2m} (-k^2ψ)\]

The relationship \(∇^2 ψ=-k^2 ψ\) can then be clearly used to obtain the final equation:

\[Hψ=-\frac{ℏ^2}{2m} (∇^2 ψ)\]

Thus, the final expression for the Hamiltonian operator \(H\) for \(Hψ=Eψ\) is equal to

\[H=\frac{-ℏ^2}{2m} ∇^2\]

---

Page Author: Natalie Woods