Solution for Simple Harmonic Oscillator

Claimed by Lim, Xuen Zhen (Spring 2022)

Introduction

One of the many situations which can be analyzed with the Schrodinger’s Equation is the one-dimensional simple harmonic oscillator. This situation draws an analogy between a quantum-mechanical particle’s oscillating movement and the movement of a common classical oscillator: an object attached to a spring. Like its classical spring counterpart described under Hooke's Law, a quantum harmonic oscillator has the force function [math]\displaystyle{ F = -k x }[/math] and the associated potential function [math]\displaystyle{ U = \frac{1}{2} k x^2 }[/math], with [math]\displaystyle{ k }[/math] being the force constant (spring constant in classical case). Despite the simplicity of a harmonic oscillator's smooth parabolic potential, it acts as an important foundation to solving more complicated quantum systems due to it being one of the few quantum-mechanical systems with an exact, known analytical solution.

Mathematical Setup

We may use the time-independent Schrodinger's equation to represent the state of a quantum particle in the harmonic potential by substituting the potential [math]\displaystyle{ U }[/math] with [math]\displaystyle{ \frac{1}{2} k x^2 }[/math].
[math]\displaystyle{ \frac{-\hbar^2}{2m} \frac{d^2 \Psi}{d x^2} + \frac{1}{2} k x^2 \Psi = E \Psi }[/math]
The solution to this equation are the wave function [math]\displaystyle{ \Psi }[/math] and the energy function [math]\displaystyle{ E }[/math] that satisfies the above conditions.

Deriving the Solution

For quantum harmonic oscillator, there are no boundaries between different regions, so one condition for the wave function is it must approach 0 as [math]\displaystyle{ x→+∞ }[/math] and [math]\displaystyle{ x→-∞ }[/math]. A simple general wave function that satisfies this requirement is [math]\displaystyle{ \Psi (x) = A e^{-ax^2} }[/math]. We begin the derivation with finding the second order differential of the general wave equation.

[math]\displaystyle{ \frac{d \Psi}{d x} = -2 a x (A e^{-ax^2}) = -2 a x \Psi }[/math]

[math]\displaystyle{ \frac{d^2 \Psi}{d x^2} = -2 a (A e^{-ax^2})-2 a x (-2 a x)(A e^{-ax^2}) = (-2 a+4a^2x^2)A e^{-ax^2} = (-2 a+4a^2x^2)\Psi }[/math]

Substituting the differential equation into the time-independent Schrodinger equation produces

[math]\displaystyle{ \frac{-\hbar^2}{2m} (-2 a+4a^2x^2)\Psi + \frac{1}{2} k x^2 \Psi = E \Psi }[/math]
[math]\displaystyle{ \frac{\hbar^2 a}{m}\Psi-\frac{2 a^2 \hbar^2 x^2}{m}\Psi + \frac{1}{2} k x^2 \Psi = E \Psi }[/math]
[math]\displaystyle{ \frac{\hbar^2 a}{m}-\frac{2 a^2 \hbar^2}{m}x^2 + \frac{1}{2} k x^2 = E }[/math]

One common misconception to be aware of is that this is not an equation to be solved for [math]\displaystyle{ x }[/math]. [math]\displaystyle{ x }[/math] is posing as a free variable here, and the solution is one that makes the Schrodinger equation true for any value of [math]\displaystyle{ x }[/math]. To allow this equation to be consistent for any [math]\displaystyle{ x }[/math], the coefficients to [math]\displaystyle{ x^2 }[/math] must cancel out, leaving the remaining constants to be equal to each other.

[math]\displaystyle{ \frac{\hbar^2 a}{m} }[/math][math]\displaystyle{ - }[/math][math]\displaystyle{ \frac{2 a^2 \hbar^2}{m}x^2 + \frac{1}{2} k x^2 }[/math][math]\displaystyle{ = }[/math][math]\displaystyle{ E }[/math]

[math]\displaystyle{ -\frac{2 a^2 \hbar^2}{m} + \frac{1}{2} k = 0 }[/math] and [math]\displaystyle{ \frac{\hbar^2 a}{m} = E }[/math]

Remember, [math]\displaystyle{ k }[/math], [math]\displaystyle{ \hbar }[/math], and [math]\displaystyle{ m }[/math] are variables which we already know the values, [math]\displaystyle{ a }[/math] and [math]\displaystyle{ E }[/math] are the only unknowns we are trying to find. We will first solve for [math]\displaystyle{ a }[/math] in the equation on the left to help find the wave function.

[math]\displaystyle{ -\frac{2 a^2 \hbar^2}{m} + \frac{1}{2} k = 0 }[/math]
[math]\displaystyle{ \frac{2 a^2 \hbar^2}{m} = \frac{1}{2} k }[/math]
[math]\displaystyle{ a^2 = \frac{k m}{4 \hbar^2} }[/math]
[math]\displaystyle{ a = \frac{\sqrt{k m}}{2 \hbar} }[/math]

Substituting [math]\displaystyle{ a }[/math], the wave function now looks like

[math]\displaystyle{ \Psi (x) = A e^{-(\sqrt{k m}/2\hbar) x^2} }[/math]

We know the total probability of finding the particle from the entire space ([math]\displaystyle{ -∞ }[/math] to [math]\displaystyle{ +∞ }[/math]) is 1. Using this condition, the normalization constant [math]\displaystyle{ A }[/math] can be calculated. (Note that since there are no complex component in this equation, the complex conjugate is simply the equation itself, and |\Psi|^2 is the square of the wave function.)

[math]\displaystyle{ 1 = \int_{-∞}^{+∞} |\Psi|^2 dx = \int_{-∞}^{+∞} A^2 e^{-(\sqrt{k m}/\hbar) x^2} dx }[/math]

Using the Gaussian Integral identity [math]\displaystyle{ 1 = \int_{-∞}^{+∞} e^{-ax^2} dx = \sqrt{\frac{\pi}{a}} }[/math], the integral is turned into

[math]\displaystyle{ 1 = A^2 \sqrt{\frac{\hbar \pi}{\sqrt{k m}}} }[/math]
[math]\displaystyle{ A^2 = \sqrt{\frac{\sqrt{k m}}{\hbar \pi}} = \sqrt{\frac{\sqrt{(m w_{0}^2)m}}{\hbar \pi}} = \sqrt{\frac{m w_{0}}{\hbar \pi}} }[/math]
[math]\displaystyle{ A = (\frac{m w_{0}}{\hbar \pi})^{1/4} }[/math]

The complete ground state wave function is

[math]\displaystyle{ \Psi (x) = (\frac{m w_{0}}{\hbar \pi})^{1/4} e^{-(\sqrt{k m}/2\hbar) x^2} }[/math]

Note that the [math]\displaystyle{ A }[/math] and [math]\displaystyle{ \Psi (x) }[/math] found here is only valid for ground state of harmonic oscillator. The general solution to the wave function is of the form [math]\displaystyle{ \Psi_{n} (x) = A f_{n} (x) e^{-ax^2} }[/math], with [math]\displaystyle{ f_{n} (x) }[/math] being a polynomial in which the highest power of [math]\displaystyle{ x }[/math] is [math]\displaystyle{ x^n }[/math].

One interesting trait of this wave function is, just like the finite potential well, the probability density can pentrate into the forbidden region beyond the classical turning points (the boundary of the region at which the potential energy becomes higher than the energy).

Finding energy function

With [math]\displaystyle{ a }[/math] derived above, we can easily substitute it into the second equation and find the energy [math]\displaystyle{ E }[/math].

[math]\displaystyle{ E = (\frac{h^2}{m}) (\frac{\sqrt{k m}}{2 \hbar}) }[/math]
[math]\displaystyle{ E = \frac{1}{2} \hbar \sqrt{\frac{k}{m}} }[/math]

We can write the energy in terms of the classical frequency [math]\displaystyle{ ω }[/math]₀=[math]\displaystyle{ \sqrt{\frac{k}{m}} }[/math].

[math]\displaystyle{ E = \frac{1}{2} \hbar ω }[/math]₀

By quantizing the energy, we then create the energy function for a simple harmonic oscillator.

[math]\displaystyle{ E }[/math]_n [math]\displaystyle{ = (\frac{1}{2}+n) \hbar ω }[/math]₀   [math]\displaystyle{ n=0,1,2,... }[/math]

Note that, contrary to one-dimensional energy function of a potential well, the energy function here are uniformly spaced with the interval [math]\displaystyle{ ∆E=\hbar w_{0} }[/math].

Applications

While a one-dimensional quantum harmonic oscillator with smooth potential is virtually inexistant in the nature, there exists some systems that behave akin to one, such as a vibrating diatomic molecule like HCl. A complex potential with arbitrary smooth curve can also usually be approximated as a harmonic potential near its stable equilibrium point, also known as the minimum.