Classical and quantum harmonic oscillators

When you think of the classical harmonic oscillator, think of a mass connected to a spring oscillating at a natural frequency which is independent of the initial position or velocity of the mass. The natural frequency will depend only on the stiffness of the spring and the size of the mass.

When you think of the quantum harmonic oscillator, think of quasi-factorising the Hamiltonian operator in Schrödinger’s equation to get creation and annihilation operators, re-expressing the Hamiltonian in terms of these ladder operators, and operating on the system with these ladder operators to increase and decrease the energy of the system by multiples of discrete packets of energy (‘quanta’).

Recall that the classical harmonic oscillator model, say a mass $m$ bouncing up and down on a spring with spring constant $k$ aligned with the $y$ -axis, involves a restoring force $-ky$ on the mass whenever it is away from equilibrium at $y = 0$ . Newton’s second law then gives the differential equation

$m \frac{d^2 y}{dt^2} = -ky$

Defining $\omega = \sqrt{\frac{k}{m}}$ (this will be the natural frequency of the oscillations), we can write this as

$\frac{d^2 y}{dt^2} = -\omega^2 y$

which has the general solution

$y = Ae^{i \omega t} + Be^{-i \omega t}$

$= \alpha \sin(\omega t) + \beta \cos(\omega t)$

We can obtain particular solutions from this general solution by specifying initial conditions. For example, starting off with the mass $m$ at 10cm below the equilibrium position and then releasing it gives the initial conditions

$y(0) = -10$

$y^{\prime}(0) = 0$

Applying these to the general solution we get the equations

$y(0) = \beta = -10$

$y^{\prime}(0) = \alpha \omega = 0 \implies \alpha = 0$

Therefore the particular solution for this situation is

$y(t) = -10 \cos(\omega t)$

The work done by the spring force on the mass in opposing its motion from, say, the equilibrium position to a height $y$ above the equilibrium point is given by

$W(y) = \int_0^y F(v) dv = \int_0^y (-kv) dv = -\frac{ky^2}{2}$

This work can be viewed as transferring energy from the kinetic energy of the mass to the elastic potential energy of the spring (or more strictly speaking, the mass-spring system). The potential energy of the spring for this displacement from equilibrium is thus

$V(y) = -W(y) = \frac{ky^2}{2} = \frac{1}{2} m \omega^2 y^2$

(Also note that, as usual, the original spring force is recoverable as the negative of the first derivative of the potential energy).

If the maximum displacement of the spring before returning to equilibrium is $y_{max}$ , so that it momentarily stops there (maximum potential energy, zero kinetic energy), then the maximum speed $v_{max}$ of the mass which occurs when it is back at the equilibrium point (zero potential energy, maximum kinetic energy) can be calculated using the conservation of total energy equation $K + V = E$ as

$\frac{ky_{max}^2}{2} = \frac{1}{2} m v_{max}^2$

$\implies v_{max} = \sqrt{\frac{k}{m}} y_{max} = \omega y_{max}$

This maximum speed is higher the higher the spring constant $k$ (i.e., the stiffer the spring) and the lower the mass $m$ of the particle.

In the case of the quantum harmonic oscillator, we use the time-independent Schrödinger equation rather than Newton’s second law to get the relevant differential equation. Starting from the total energy equation $K + V = E$ , written as

$\frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2 = E$

where $p = m \dot{x}$ is linear momentum, we simply replace $p$ by the quantum mechanical momentum operator $\hat{p} = -i \hbar \frac{d}{dx}$ and $x$ by the quantum mechanical position operator $\hat{x} = x$ to get the time-independent Schrödinger equation

$\big(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + \frac{1}{2}m \omega^2 \hat{x}^2\big)\psi = E\psi$

The bracketed expression on the left-hand side is the Hamiltonian operator $\hat{H}$ (corresponding to the total energy of the system, kinetic plus potential) acting on the wave function $\psi$ :

$\hat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + \frac{1}{2}m \omega^2 \hat{x}^2$

$= \frac{\hat{p}^2}{2m} + \frac{1}{2}m \omega^2 \hat{x}^2$

$= \frac{1}{2} m \omega^2 \bigg(\hat{x}^2 + \big(\frac{\hat{p}}{m \omega}\big)^2\bigg)$

Inside the brackets we now have an operator sum of squares similar to the algebraic sum of squares $(a^2 + b^2)$ which can be factorized as $(a - ib)(a + ib)$ . We would therefore like to factorize $\hat{H}$ by setting $a = \hat{x}$ , $b = \frac{\hat{p}}{m \omega}$ , and writing

$\frac{1}{2} m \omega^2 \big(\hat{x} - i \frac{\hat{p}}{m \omega}\big)\big(\hat{x} + i \frac{\hat{p}}{m \omega}\big)$

However, multiplying out this expression does not give us $\hat{H}$ , because the operators $\hat{x}$ and $\hat{p}$ do not commute. Instead we get

$\frac{1}{2} m \omega^2 \big(\hat{x} - i \frac{\hat{p}}{m \omega}\big)\big(\hat{x} + i \frac{\hat{p}}{m \omega}\big)$

$= \hat{H} + \frac{i \omega}{2}[\hat{x}, \hat{p}]$

$= \hat{H} - \frac{\hbar \omega}{2}$

where

$[\hat{x}, \hat{p}] \equiv \hat{x}\hat{p} - \hat{p} \hat{x} = i \hbar$

is the quantum mechanical commutator of $\hat{x}$ and $\hat{p}$ . Therefore

$\hat{H} = \frac{1}{2} m \omega^2 \big(\hat{x} - i \frac{\hat{p}}{m \omega}\big)\big(\hat{x} + i \frac{\hat{p}}{m \omega}\big) + \frac{\hbar \omega}{2}$

$= \hbar \omega \bigg( \frac{m \omega}{2 \hbar} \big(\hat{x} - i \frac{\hat{p}}{m \omega}\big)\big(\hat{x} + i \frac{\hat{p}}{m \omega}\big) + \frac{1}{2} \bigg)$

$= \hbar \omega \big(\hat{a}^{\dag} \hat{a} + \frac{1}{2}\big)$

where $\hat{a}^{\dag}$ and $\hat{a}$ are creation and annihilation operators respectively (also known as raising and lowering operators, and collectively as ladder operators) defined as

$\hat{a}^{\dag} = \sqrt{\frac{m \omega}{2 \hbar}}\big(\hat{x} - i \frac{\hat{p}}{m \omega}\big)$

$\hat{a} = \sqrt{\frac{m \omega}{2 \hbar}}\big(\hat{x} + i \frac{\hat{p}}{m \omega}\big)$

If we reversed the order of $\hat{a}^{\dag}$ and $\hat{a}$ we would find that

$\hat{H} = \hbar \omega \big(\hat{a} \hat{a}^{\dag} - \frac{1}{2} \big)$

We can therefore write the Schrödinger equation $\hat{H} \psi = E \psi$ as

$\hbar \omega \big(\hat{a}^{\dag} \hat{a} + \frac{1}{2}\big) \psi = E \psi$

However, the real usefulness of ladder operators becomes apparent when we apply the Hamiltonian written in this form to $\hat{a}^{\dag} \psi$ rather than to $\psi$ . We get

$\hat{H} (\hat{a}^{\dag} \psi)$

$= \hbar \omega \big(\hat{a}^{\dag} \hat{a} + \frac{1}{2}\big) (\hat{a}^{\dag} \psi)$

$= \hbar \omega \hat{a}^{\dag} \big(\hat{a} \hat{a}^{\dag} + \frac{1}{2}\big) \psi$

$= \hbar \omega \hat{a}^{\dag} \big(\hat{a} \hat{a}^{\dag} - \frac{1}{2} + 1\big) \psi$

$= \hat{a}^{\dag} \big(\hat{H} + \hbar \omega\big) \psi$

$= \hat{a}^{\dag} \big(E + \hbar \omega\big) \psi$

$= \big(E + \hbar \omega\big) (\hat{a}^{\dag} \psi)$

Therefore if $\psi$ is a solution to the quantum harmonic oscillator problem, $\hat{a}^{\dag} \psi$ is also a solution, i.e., we can apply the creation operator $\hat{a}^{\dag}$ to the solution $\psi$ and get another solution $\hat{a}^{\dag} \psi$ with an energy eigenvalue $E + \hbar \omega$ instead of $E$ .

Using the same algebra, we find that if $\psi$ is a solution to the quantum harmonic oscillator problem with energy eigenvalue $E$ , then $\hat{a} E$ is another solution with energy eigenvalue $E - \hbar \omega$ .

We therefore call the ladder operator $\hat{a}^{\dag}$ a raising operator because applying it to a quantum state $\psi$ results in a new quantum state $\hat{a}^{\dag} \psi$ whose energy is higher by a quantum of energy $\hbar \omega$ . The term creation operator arises because these quanta of energy actually behave like particles, so the addition of this extra quantum of energy can also be viewed as the creation of a new particle. Similarly, $\hat{a}$ is a lowering operator because applying it to a quantum state $\psi$ results in a new quantum state $\hat{a} \psi$ whose energy is lower by a quantum of energy $\hbar \omega$ . It is also known as an annihilation operator because this process is like removing a particle.

Now, the Schrödinger equation $\hat{H} \psi = E \psi$ for the quantum harmonic oscillator has a solution set consisting of eigenfunctions $\psi_n(x)$ (expressed in terms of Hermite polynomials), each with a corresponding energy eigenvalue

$E_n = \hbar \omega \big(n + \frac{1}{2}\big)$

for $n = 0, 1, 2, \ldots$ . The energy of a quantum harmonic oscillator is therefore indeed quantized in steps of $\hbar \omega$ . The lowest possible energy, namely the zero point energy corresponding to $n = 0$ , is $E_0 = \frac{1}{2}\hbar \omega$ . The corresponding eigenfunction is $\psi_0$ . Since the energy level cannot fall below $\frac{1}{2}\hbar \omega$ , we specify $\hat{a}\psi_0 = 0$ , so trying to apply the lowering operator to $\psi_0$ just gives the zero function. We could actually solve for the explicit form of $\psi_0$ by solving the condition $\hat{a}\psi_0 = 0$ as a simple first-order differential equation:

$\hat{a}\psi_0 = \sqrt{\frac{m \omega}{2 \hbar}}\big(\hat{x} + \frac{i\hat{p}}{m \omega}\big)\psi_0$

$= \sqrt{\frac{m \omega}{2 \hbar}}\big(\hat{x} + \frac{\hbar}{m \omega}\frac{d}{dx}\big)\psi_0 = 0$

$\implies \frac{\hbar}{m \omega} \frac{d \psi_0}{dx} = - x \psi_0$

$\iff \frac{1}{\psi_0} d\psi_0 = -\frac{m \omega}{\hbar} x dx$

$\implies ln(\psi_0) = -\frac{m \omega}{2\hbar}x^2 + c$

$\iff \psi_0 = Ae^{-\frac{m \omega}{2\hbar}x^2}$

Normalizing, we get

$A = \big(\frac{m \omega}{\pi \hbar}\big)^{1/4}$

so the explicit form of the zero point energy eigenfunction is

$\psi_0(x) = \big(\frac{m \omega}{\pi \hbar}\big)^{1/4}e^{-\frac{m \omega}{2\hbar}x^2}$

We can now get all the higher energy eigenfunctions by repeatedly applying the raising operator $\hat{a}^{\dag}$ to this explicit form for $\psi_0$ (with some adjustments for normalization).

Classical and quantum harmonic oscillators

Published by Dr Christian P. H. Salas

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