Q19Engineering Physics

Question

Schrodinger time-independent wave equation.

Answer 1

Erwin Schrödinger originally formulated his wave equation to be time-dependent, describing the continuous, dynamic evolution of a quantum state over time. However, in many fundamental physical problems (such as analyzing the allowed electron orbitals in a hydrogen atom or a particle confined in a rigid potential well), the forces acting on the particle are strictly static. The potential energy field $V (x, y, z)$ depends exclusively on the physical coordinates and is entirely independent of time. Under these specific conditions, the quantum system can settle into what are known as stationary states, where measurable probabilities remain completely constant over time. To analyze these stationary states, we must derive the Time-Independent version of the Schrödinger equation.

1. The Classical Wave Equation Approach

We begin our derivation by drawing an analogy to classical wave mechanics. Consider the standard three-dimensional differential wave equation that describes a classical, non-dispersive wave (like a sound wave or a light wave) traveling with a constant phase velocity $v$ :

\nabla^{2} Ψ = \frac{1}{v ^{2}} \frac{\partial ^{2} Ψ}{\partial t ^{2}}

Where $Ψ (x, y, z, t)$ represents the overall wave amplitude, and the Laplacian operator $\nabla^{2} = \frac{\partial ^{2}}{\partial x ^{2}} + \frac{\partial ^{2}}{\partial y ^{2}} + \frac{\partial ^{2}}{\partial z ^{2}}$ represents the spatial variation.

Because we are exclusively seeking standing, stationary wave solutions, we employ the powerful mathematical technique of Separation of Variables. We assume the total wave function can be perfectly factored into two completely independent functions: one depending purely on space ( $ψ (x, y, z)$ ) and one oscillating purely in time:

Ψ (x, y, z, t) = ψ (x, y, z) \cdot e^{- iω t}

Here, $ω$ represents the angular frequency of the wave.

2. Differentiating the Separated Function

We must substitute this separated function back into the classical wave equation. First, we take the second partial derivative with respect to time ( $t$ ):

\frac{\partial Ψ}{\partial t} = - iω \cdot ψ (x, y, z) e^{- iω t}

\frac{\partial ^{2} Ψ}{\partial t ^{2}} = (- iω) (- iω) ψ (x, y, z) e^{- iω t} = - ω^{2} Ψ

Next, we take the spatial Laplacian. Since the time exponential $e^{- iω t}$ is treated as a constant with respect to space, we have:

\nabla^{2} Ψ = (\nabla^{2} ψ) \cdot e^{- iω t}

Now, we substitute these calculated derivatives back into the original classical wave equation ( $\nabla^{2} Ψ = \frac{1}{v ^{2}} \frac{\partial ^{2} Ψ}{\partial t ^{2}}$ ):

(\nabla^{2} ψ) \cdot e^{- iω t} = \frac{1}{v ^{2}} (- ω^{2} ψ \cdot e^{- iω t})

The time-dependent exponential term $e^{- iω t}$ beautifully cancels out from both sides entirely, leaving us with an equation purely dependent on spatial coordinates—the pure amplitude equation:

\nabla^{2} ψ + \frac{ω ^{2}}{v ^{2}} ψ = 0

3. Introducing Quantum Mechanics (de Broglie and Energy)

We must now convert this classical amplitude equation into a quantum mechanical equation. We utilize standard wave relations: angular frequency $ω = 2 π ν$ , and velocity $v = ν λ$ .

\frac{ω ^{2}}{v ^{2}} = \frac{( 2 π ν ) ^{2}}{( ν λ ) ^{2}} = \frac{4 π ^{2}}{λ ^{2}}

Substituting this back into the amplitude equation gives:

\nabla^{2} ψ + \frac{4 π ^{2}}{λ ^{2}} ψ = 0

Now, we inject the two absolute pillars of quantum theory into the derivation:

De Broglie's Hypothesis: The wavelength of a matter wave is strictly tied to its momentum ( $p$ ): $λ = \frac{h}{p} ⟹ \frac{1}{λ ^{2}} = \frac{p ^{2}}{h ^{2}}$
Classical Conservation of Energy: The total strictly conserved energy ( $E$ ) of a particle is the sum of its Kinetic Energy ( $K$ ) and its Potential Energy ( $V$ ). $E = K + V ⟹ K = E - V$ Since classical kinetic energy is $K = \frac{1}{2} m v^{2} = \frac{p ^{2}}{2 m}$ , we can express the square of the momentum entirely in terms of energy: $\frac{p ^{2}}{2 m} = E - V ⟹ p^{2} = 2 m (E - V)$

Combine these two fundamental concepts by substituting $p^{2}$ into the de Broglie relation:

\frac{1}{λ ^{2}} = \frac{2 m ( E - V )}{h ^{2}}

4. The Final Equation

Finally, we substitute this quantum expression for $\frac{1}{λ ^{2}}$ back into our purely spatial amplitude equation:

\nabla^{2} ψ + 4 π^{2} [\frac{2 m ( E - V )}{h ^{2}}] ψ = 0

To clean up the notation, physicists universally use the "reduced Planck's constant" (or Dirac constant), defined as $ℏ = \frac{h}{2 π}$ . Therefore, $ℏ^{2} = \frac{h ^{2}}{4 π ^{2}}$ . Substituting $ℏ^{2}$ into the denominator of the equation yields the final, elegant form of the Time-Independent Schrödinger Equation:

\nabla^{2} ψ + \frac{2 m}{ℏ ^{2}} (E - V) ψ = 0

This second-order differential eigenvalue equation is arguably the most important formula in all of quantum chemistry and solid-state physics. Solving this exact equation for specific boundary conditions (the potential $V$ ) yields the absolute allowed, quantized energy levels ( $E$ ) and the specific spatial shapes of the atomic orbitals ( $ψ$ ) for any quantum system.

Q19Engineering Physics

Question

Answer

1. The Classical Wave Equation Approach

2. Differentiating the Separated Function

3. Introducing Quantum Mechanics (de Broglie and Energy)

4. The Final Equation