Part 1 · Oscillators and waves

1.2 Complex numbers and the harmonic oscillator

Source: Postorino, week 1 slides, pp. 9–12

Objectives

Picture. A complex number is an arrow in the plane: length and angle. Multiplying multiplies lengths and adds angles; $e^{i φ}$ is the point at angle $φ$ on the unit circle.
Recipe. Go between $a + ib$ and $A e^{i φ}$ , and compute $∣ c ∣ = c^{*} c$ .
Theorem. Near any stable equilibrium, any potential is a parabola: small displacements oscillate harmonically with $ω_{0} = U^{''} (x_{1}) / m$ .
Vocabulary. Restoring force, natural frequency, stable and unstable equilibrium, harmonic approximation.

The harmonic oscillator is the first model of the course and the one that keeps coming back: a bound electron in light (§3.1 Dielectrics: the Lorentz model), a mode of the radiation in a hot cavity (§5.1 Line spectra and blackbody radiation), an atom in a solid. Complex numbers are the tool that solves it, and every oscillator and wave after it, in a few lines.

Complex numbers

Definition 1.2.1Complex numberslides w1, p. 9

A complex number is $c = a + ib, a, b \in R, c \in C,$ with $i = - 1$ the imaginary unit (unità immaginaria): $i^{2} = - 1$ . $a = Re c$ is the real part, $b = Im c$ the imaginary part.

The picture. $c$ is a point, or an arrow from the origin, in the complex plane: $a$ along the real axis (Re), $b$ along the imaginary axis (Im). The arrow has a length $A$ and makes an angle $φ$ with the real axis.

Theorem 1.2.2Euler’s relationslides w1, p. 9

$e^{i φ} = cos φ + i sin φ .$ It is the point at angle $φ$ on the unit circle: real part $cos φ$ , imaginary part $sin φ$ .

Proofbeyond the lectures

Compare the power series. $e^{x} = \sum_{n} x^{n} / n!$ ; put $x = i φ$ and use $i^{2} = - 1$ , $i^{3} = - i$ , $i^{4} = 1$ : $e^{i φ} = (1 - \frac{φ ^{2}}{2 !} + \frac{φ ^{4}}{4 !} - \dots) + i (φ - \frac{φ ^{3}}{3 !} + \frac{φ ^{5}}{5 !} - \dots) = cos φ + i sin φ .$ The even powers are real and make the cosine series; the odd powers carry one $i$ and make the sine series.

So every complex number has a polar form, length times direction:

Proposition 1.2.3Polar formslides w1, p. 9

$c = a + ib = A e^{i φ} = A [cos φ + i sin φ], A = a^{2} + b^{2}, tan φ = \frac{b}{a} = \frac{imag. part}{real part} .$

Proof

The arrow of length $A$ at angle $φ$ has components $a = A cos φ$ , $b = A sin φ$ . Then $a + ib = A (cos φ + i sin φ) = A e^{i φ}$ by Euler’s relation. Pythagoras gives $A$ , and $b / a = A sin φ / A cos φ = tan φ$ .

WarningThe quadrantbeyond the lectures

$tan φ = b / a$ fixes $φ$ only up to $π$ : $1 + i$ and $- 1 - i$ have the same ratio. The signs of $a$ and $b$ say which quadrant $c$ is in. $- 1 - i$ has $φ = - 3 π /4$ , not $π /4$ .

Definition 1.2.4Complex conjugate and modulusslides w1, p. 9

The complex conjugate $c^{*}$ is the mirror image of $c$ across the real axis: same length, opposite angle. $c^{*} = a - ib = A e^{- i φ} .$ The modulus is the length of the arrow: $c c^{*} = (a + ib) (a - ib) = a^{2} - (ib)^{2} = a^{2} + b^{2} = A^{2}, A = ∣ c ∣ = c^{*} c .$

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Figure 1.2.1 Drag

c

. Its real part

a

and imaginary part

b

are its shadows on the axes; its length

A

and angle

φ

are its polar form. The mirror image is

c^{*}

, and

c c^{*} = A^{2}

is always real. The handle snaps to the unit circle and to multiples of

45°

Two facts used all the time below beyond the lectures:

Proposition 1.2.5Multiplying multiplies lengths and adds angles

$A_{1} e^{i φ_{1}} \cdot A_{2} e^{i φ_{2}} = A_{1} A_{2} e^{i (φ_{1} + φ_{2})}, \frac{A _{1} e ^{i φ_{1}}}{A _{2} e ^{i φ_{2}}} = \frac{A _{1}}{A _{2}} e^{i (φ_{1} - φ_{2})} .$ So $∣ c_{1} / c_{2} ∣ = ∣ c_{1} ∣/∣ c_{2} ∣$ and $ar g (c_{1} / c_{2}) = ar g c_{1} - ar g c_{2}$ .

Proofbeyond the lectures

The exponential turns sums of exponents into products: $e^{i φ_{1}} e^{i φ_{2}} = e^{i (φ_{1} + φ_{2})}$ . Multiply the real lengths separately.

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Figure 1.2.2 Drag

c_{1}

and

c_{2}

. Their product

c_{1} c_{2}

has length

∣ c_{1} ∣∣ c_{2} ∣

and angle

φ_{1} + φ_{2}

. Put

c_{2}

on the unit circle and multiplying by it is a pure rotation; put it at

i

and it is a quarter turn.

Key observation 1.2.6A time derivative of

e^{iω t}

multiplies it by

iω

beyond the lectures

$\frac{d}{d t} e^{iω t} = iω e^{iω t}, \frac{d ^{2}}{d t ^{2}} e^{iω t} = (iω)^{2} e^{iω t} = - ω^{2} e^{iω t} .$ Derivatives become multiplications. This is what turns linear differential equations into algebra (§1.3 Linear equations and the complex substitution). Multiplying by $i = e^{iπ /2}$ is a quarter turn: the velocity of a point going round a circle is a quarter turn ahead of its position.

The harmonic oscillator

A mass $m$ on a spring feels the elastic force (Hooke’s law) $F = - k x,$ with $k$ the spring constant (costante elastica) and $x$ the displacement from rest. The minus sign means the force always points back towards $x = 0$ : it is a restoring force.

Theorem 1.2.7The harmonic oscillatorslides w1, p. 10

Newton’s law $F = m d^{2} x / d t^{2}$ with $F = - k x$ gives $m \frac{d ^{2} x}{d t ^{2}} = - k x ⟹ \frac{d ^{2} x}{d t ^{2}} + ω_{0}^{2} x = 0, ω_{0} = \frac{k}{m},$ whose solution is $x (t) = x_{0} cos (ω_{0} t + φ),$ with the amplitude $x_{0}$ and the phase $φ$ fixed by the initial conditions.

Proofbeyond the lectures

Differentiate twice: $\overset{x}{˙} = - ω_{0} x_{0} sin (ω_{0} t + φ)$ , $\overset{x}{¨} = - ω_{0}^{2} x_{0} cos (ω_{0} t + φ) = - ω_{0}^{2} x$ . ✓ A second-order equation needs two constants, one for the initial position and one for the initial velocity; $x_{0}$ and $φ$ are those two.

$ω_{0}$ is the oscillator’s natural (angular) frequency: stiffer springs and lighter masses oscillate faster. It does not depend on the amplitude.

Key observation 1.2.8An oscillation is the real part of a rotationslides w1, p. 10

$x (t) = Re {x_{0} e^{i (ω_{0} t + φ)}} :$ the shadow on the real axis of a point going round a circle of radius $x_{0}$ at angular speed $ω_{0}$ . This is the way into a general technique for linear differential equations (§1.3 Linear equations and the complex substitution).

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Figure 1.2.3 The arrow

x_{0} e^{i (ω_{0} t + φ)}

turns at

ω_{0}

; its shadow on the real axis is

x (t)

, drawn against time on the right. Drag the arrow to set the amplitude

x_{0}

and phase

φ

t = 0

, then press Play.

The harmonic approximation

A particle moving along $x$ in a potential energy $U (x)$ feels the force $F_{x} = - \frac{\partial U}{\partial x} .$ The force points downhill. It is zero where the slope vanishes, at the extremal points of $U$ :

at a maximum: unstable equilibrium. Push the particle a little and the force pushes it further away.
at a minimum: stable equilibrium. Push it a little and the force pulls it back.
on a flat stretch: neutral equilibrium. Nothing pulls either way.

Theorem 1.2.9Harmonic approximationslides w1, p. 11–12

Near a minimum $x_{1}$ of $U$ , with $k = U^{''} (x_{1}) > 0$ , $U (x) = U (x_{1}) + \frac{1}{2} k (x - x_{1})^{2} + \dots, F_{x} = - k (x - x_{1}) .$ Near any stable equilibrium, any potential looks like a parabola, and the force is the elastic restoring force of a spring centred at $x_{1}$ . Small displacements oscillate harmonically with $ω_{0} = k / m$ .

Proof

Expand $U$ in a Taylor series around $x_{1}$ : $U (x) = U (x_{1}) + \frac{\partial U}{\partial x}_{x_{1}} (x - x_{1}) + \frac{1}{2} \frac{\partial ^{2} U}{\partial x ^{2}}_{x_{1}} (x - x_{1})^{2} + \dots$ The handwritten notes on the slide make two points:

the first-derivative term is crossed out: at a minimum the slope is zero;
the second derivative is written “COSÌ K”: call it $k$ . At a minimum the curvature is positive, so $k > 0$ .

What is left is the parabola. Then $F_{x} = - \partial U / \partial x = - k (x - x_{1})$ . $U (x_{1})$ is a constant: it shifts the energy but exerts no force.

The terms dropped, cubic and higher, matter only for larger displacements. The parabola is symmetric; a real potential usually is not.

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Figure 1.2.4 A bond potential

U (x)

and its harmonic approximation, the

parabola

with the same curvature at the minimum. Drag the energy

E

up. At small

E

the particle swings between turning points that are the same for both curves. At larger

E

the real well is steep on the left and gentle on the right, so the swing becomes lopsided and the

mean position

moves out: thermal expansion.

The spring–ball model of a solid. Atoms are balls (nucleus plus inner electrons) joined by springs (the chemical bonds) in a 3D lattice. Each bond sits at the minimum of its interatomic potential, so by the harmonic approximation it behaves as a spring.

This simple model is the basis for describing a solid:

it predicts the elastic, thermodynamic and transport properties of solids fairly accurately;
it fails for thermal expansion.

RemarkWhy it fails for thermal expansionbeyond the lectures

In a perfectly parabolic well an atom with more energy swings equally far each way, so its average position stays at $x_{1}$ . Heating makes the vibrations larger but leaves the bond length unchanged: no expansion. Real interatomic potentials are asymmetric, with a steep wall at short distance and a gentle slope at long distance, so the average position moves out as the energy grows (Figure 1.2.4). Thermal expansion comes from the anharmonic (cubic) terms the approximation throws away.

Exercises

Exercise 1.2.1beyond the lectures

Write $c = - 1 + i 3$ in polar form, and compute $c^{*}$ , $∣ c ∣$ , $c^{2}$ and $1/ c$ .

Solution

$A = 1 + 3 = 2$ . The point is in the second quadrant, with $tan φ = - 3$ , so $φ = 2 π /3$ : $c = 2 e^{i 2 π /3}$ . $c^{*} = - 1 - i 3 = 2 e^{- i 2 π /3}$ , $∣ c ∣ = 2$ . $c^{2} = 4 e^{i 4 π /3} = - 2 - 2 i 3$ . $1/ c = \frac{1}{2} e^{- i 2 π /3} = c^{*} /∣ c ∣^{2} = (- 1 - i 3) /4$ .

Exercise 1.2.2beyond the lectures

Show that $cos φ = \frac{e ^{i φ} + e ^{- i φ}}{2}$ and $sin φ = \frac{e ^{i φ} - e ^{- i φ}}{2 i}$ .

Solution

Add and subtract Euler’s relation for $φ$ and $- φ$ : $e^{\pm i φ} = cos φ \pm i sin φ$ . The sum is $2 cos φ$ , the difference $2 i sin φ$ .

Exercise 1.2.3beyond the lectures

A particle moves in $U (x) = U_{0} [(a / x)^{12} - 2 (a / x)^{6}]$ , a Lennard-Jones-type bond. Find the equilibrium, the spring constant $k$ , and $ω_{0}$ for a mass $m$ .

Solution

$U^{'} (x) = U_{0} [- 12 a^{12} / x^{13} + 12 a^{6} / x^{7}] = 0$ gives $x_{1} = a$ , with $U (a) = - U_{0}$ . $U^{''} (x) = U_{0} [156 a^{12} / x^{14} - 84 a^{6} / x^{8}]$ , so $k = U^{''} (a) = 72 U_{0} / a^{2} > 0$ : stable. $ω_{0} = 72 U_{0} / (m a^{2})$ . The well is steeper for $x < a$ than for $x > a$ , so the bond also expands when heated.

Exercise 1.2.4beyond the lectures

A mass of 0.2 kg on a spring with $k = 20$ N/m is pulled 5 cm and released from rest. Find $ω_{0}$ , the period, $x (t)$ , and the maximum speed.

Solution

$ω_{0} = 20/0.2 = 10$ rad/s, $T = 2 π / ω_{0} = 0.63$ s. Released from rest at $x = 0.05$ m: $x_{0} = 0.05$ m, $φ = 0$ , so $x (t) = 0.05 cos (10 t)$ m. $v_{max} = ω_{0} x_{0} = 0.5$ m/s.

Exercise 1.2.5beyond the lectures

For $U (x) = - U_{0} cos (x / a)$ (a pendulum-like potential), find the stable and unstable equilibria and the frequency of small oscillations about the stable one.

Solution

$U^{'} = (U_{0} / a) sin (x / a) = 0$ at $x = nπa$ . $U^{''} = (U_{0} / a^{2}) cos (x / a)$ is $+ U_{0} / a^{2}$ at even $n$ (minima, stable) and $- U_{0} / a^{2}$ at odd $n$ (maxima, unstable). Small oscillations: $ω_{0} = U_{0} / (m a^{2})$ .

Exercise 1.2.6beyond the lectures

Where is $i^{i}$ on the real line? (Write $i$ in polar form first.)

Solution

$i = e^{iπ /2}$ , so $i^{i} = e^{i \cdot iπ /2} = e^{- π /2} = 0.208$ : a real number. (Taking $i = e^{i (π /2 + 2 πm)}$ gives other values too, $e^{- π /2 - 2 πm}$ ; complex powers are many-valued.)