How the Harmonic Oscillator Is Solved, Why It Cannot Come to Rest, and Second Quantization

— Following the derivations in the lecture notes “Harmonic Oscillator” (1)(2)(3) and “Second Quantization” (1)–(5) (for second- and third-year undergraduates)

A point mass attached to a spring with spring constant $k = m ω^{2}$ . In classical mechanics the energy can take any value continuously from 0, and the lowest state is “at rest.” Solving the Schrödinger equation gives $E_{n} = (n + \frac{1}{2}) ℏ ω$ , so the energy is discrete, and what is more, the lower limit is lifted by $ℏ ω / 2$ . We derive, step by step, where these two features come from.

The conclusion first “Discrete energies” come from the boundary condition. If we require that the wave function converge to 0 as

x \to \pm \infty

, the series must terminate somewhere, and this becomes the condition

ε = 2 n + 1

. The lifted lower limit is the uncertainty principle itself;

ℏ ω / 2

Δ x Δ p \geq ℏ / 2

translated into energy.

From §7 on we turn to second quantization. This is a route to the same answer that never solves a differential equation. The creation and annihilation operators that appear there are not tools for the harmonic oscillator alone. They become the common language for treating systems with many particles.

0. Setup

This is the starting point of lecture notes (1). The force exerted by the spring is $F = - k x$ , and its potential is

(1)

V (x) = \frac{1}{2} k x^{2} = \frac{1}{2} m ω^{2} x^{2}, ω = \sqrt{\frac{k}{m}}

The time-independent Schrödinger equation is

(2)

- \frac{ℏ^{2}}{2 m} \frac{d^{2} ψ}{d x^{2}} + \frac{1}{2} m ω^{2} x^{2} ψ = E ψ

This form appears everywhere For any potential, if we Taylor-expand around a point of stable equilibrium, the first-order term is 0 (because it is an equilibrium point), and the second-order term is

\frac{1}{2} V^{″} x^{2}

. In other words, near its bottom, every potential is a harmonic oscillator. Molecular vibrations, atoms in a crystal, and the individual modes of the electromagnetic field (the ones counted in Planck’s law) all obey this equation.

1. Intuition First (the 30-Second Version)

Why is the energy discrete? The reason is “unless it goes to 0 at infinity, it cannot be a probability.”

Eq. (2) is a second-order differential equation of the form $ψ ″ = (\dots) ψ$ , so a solution exists whatever value we choose for E.
But far from the origin, for most values of E the solution behaves like $e^{+ X^{2} / 2}$ and diverges. Since the integral becomes infinite, no probability interpretation is possible.
Only for discrete values of E does the diverging component cancel out cleanly, and $e^{- X^{2} / 2}$ is all that remains.

Quantization comes not from the equation but from the boundary condition — this is the most worthwhile point to learn from the harmonic oscillator. In the figure of §2.4, you can actually vary ε and watch the solution diverge.

2. Solving the Equation — Lecture Notes (1)

2.1 Making the equation dimensionless

Eq. (2) has $ℏ, m, ω$ scattered all over it, which makes it hard to read, so we re-measure lengths and energies in their “natural units.”

(3)

X = \sqrt{\frac{m ω}{ℏ}} x, ε = \frac{2 E}{ℏ ω} ⟹ \frac{d^{2} ψ}{d X^{2}} + (ε - X^{2}) ψ = 0

All the constants have disappeared. From here on, no physical constants appear at all. $\sqrt{ℏ / m ω}$ is the “1 meter” of this oscillator, and $ℏ ω / 2$ is its “1 joule.”

2.2 Looking at the form far away first

For large $| X |$ , we have $ε ≪ X^{2}$ , so Eq. (3) becomes $ψ ″ ≃ X^{2} ψ$ . The function that satisfies this is $ψ ≃ e^{\pm X^{2} / 2}$ . The + sign diverges, so we discard it (boundary condition $\lim_{x \to \pm \infty} ψ = 0$ ). Now we look for the solution in the form

(4)

ψ (X) = f (X) e^{- X^{2} / 2}

We guess this form and substitute it. It looks like a clever trick, but all we are doing is “pulling out first the part of the answer that we already know.” Substituting it into Eq. (3), the factor $e^{- X^{2} / 2}$ multiplies every term, and setting the contents of the bracket to 0 gives

(5)

\frac{d^{2} f}{d X^{2}} - 2 X \frac{d f}{d X} + (ε - 1) f = 0

This is the Hermite differential equation.

2.3 Expanding in a series

Writing $f = \sum_{k} a_{k} X^{k}$ and substituting into Eq. (5), then matching the coefficients of $X^{k}$ , we find a relation that links neighboring coefficients.

(6)

a_{k + 2} = \frac{2 k - ε + 1}{(k + 2) (k + 1)} a_{k}

2.4 Here is where quantization happens

Look at Eq. (6). For large $k$ we have

\frac{a_{k + 2}}{a_{k}} \to \frac{2}{k}

But the coefficient ratio of $e^{X^{2}} = \sum_{j} X^{2 j} / j!$ is likewise, if we put $k = 2 j$ , exactly $2 / k$ . In other words,

⚠ If the series does not terminate, it always diverges If the series goes on forever, then

f \sim e^{X^{2}}

, and from Eq. (4),

ψ \sim e^{X^{2}} e^{- X^{2} / 2} = e^{+ X^{2} / 2}

. The divergent solution that we had so carefully discarded comes back to life. The only way to avoid this is for some coefficient to become 0 so that the series ends as a polynomial.

The series stops wherever the numerator of Eq. (6) becomes 0. Suppose it terminates at $k = n$ , then

(7)

2 n - ε + 1 = 0 ⟹ ε = 2 n + 1 ⟹ E = \frac{ℏ ω}{2} ε = (n + \frac{1}{2}) ℏ ω

The energy has become discrete. This is what stands behind “ $ε = 2 n + 1$ (from the series solution)” in the notes. In the figure below, vary ε continuously and see what happens.

Left: ψ obtained by numerically integrating Eq. (3) from $X = 0$ . If ε deviates even slightly from the discrete values, the solution immediately diverges (the vertical axis is logarithmic). Right: the far-field value $ψ (5)$ seen as a function of ε. Only the points where the curve crosses 0 are allowed — choosing an even function gives $ε = 1, 5, 9, \dots$ , choosing an odd function gives $ε = 3, 7, 11, \dots$ , and together they give $ε = 2 n + 1$ .

ε = 1.000 (E = 0.500 ħω) Parity of the solution

3. The Form of the Solution — Lecture Notes (2)

The polynomial obtained when the series is cut off at order $n$ is the Hermite polynomial $H_{n} (X)$ . Fixing the constant with the normalization condition $\int {| ψ |}^{2} d x = 1$ gives $C_{n} = {(\frac{m ω}{π ℏ})}^{1 / 4} / \sqrt{2^{n} n!}$ , as in the notes.

(8)

ψ_{n} (x) = {(\frac{m ω}{π ℏ})}^{1 / 4} \frac{1}{\sqrt{2^{n} n!}} H_{n} (X) e^{- X^{2} / 2}

n	$ψ_{n} (x)$	$E_{n}$	Shape
0	${(\frac{m ω}{π ℏ})}^{1 / 4} e^{- m ω x^{2} / 2 ℏ}$	$\frac{1}{2} ℏ ω$	A single hump with no nodes (Gaussian function)
1	${(\frac{m ω}{π ℏ})}^{1 / 4} \sqrt{\frac{2 m ω}{ℏ}} x e^{- m ω x^{2} / 2 ℏ}$	$\frac{3}{2} ℏ ω$	One node (zero at the center)
2	${(\frac{m ω}{π ℏ})}^{1 / 4} \frac{1}{\sqrt{2}} (\frac{2 m ω}{ℏ} x^{2} - 1) e^{- m ω x^{2} / 2 ℏ}$	$\frac{5}{2} ℏ ω$	Two nodes

These are the same as (i), (ii), (iii) of lecture notes (2). The number of nodes is exactly n, and the more nodes there are, the finer the wave, that is, the larger the kinetic energy. This is why the energy goes up.

4. What the Zero-Point Energy Really Is — Lecture Notes (3)

Even for $n = 0$ , $E_{0} = ℏ ω / 2 \neq 0$ . Why it cannot be made 0 can be understood without solving the Schrödinger equation. This is the argument of lecture notes (3).

In the classical energy $E = p^{2} / 2 m + \frac{1}{2} m ω^{2} x^{2}$ , replace x and p by their “fluctuations.” For a state centered at the origin ( $⟨ x ⟩ = ⟨ p ⟩ = 0$ ), this is just the mean energy $⟨ E ⟩$ . Using the inequality between the arithmetic and geometric means,

(9)

⟨ E ⟩ = \frac{Δ p^{2}}{2 m} + \frac{1}{2} m ω^{2} Δ x^{2} \geq 2 \sqrt{\frac{Δ p^{2}}{2 m} \cdot \frac{1}{2} m ω^{2} Δ x^{2}} = ω Δ x Δ p

The thing to notice is that $m$ drops out cleanly, leaving only $ω Δ x Δ p$ . Inserting the uncertainty principle $Δ x Δ p \geq ℏ / 2$ here gives

(10)

⟨ E ⟩ \geq \frac{ℏ ω}{2}

The zero-point energy is produced by the “fluctuations in position and momentum” This is what lies behind the closing line of the notes, “The zero-point energy was an ‘energy fluctuation’ after all.” (What fluctuates is position and momentum; the energy of the ground state itself has

Δ E = 0

— every measurement gives

ℏ ω / 2

). To come to rest, we would have to make $Δ p \to 0$ , and then $Δ x \to \infty$ , so the potential energy of the spring now becomes infinite. Conversely, if we try to pin it down at a single point, the kinetic energy diverges. In this tug of war, the best compromise is

ℏ ω / 2

, and the

E_{0}

obtained in §2 agrees with it exactly.

This agreement is no accident. Equality in Eq. (9) holds when the two terms are equal, that is, when $Δ x = \sqrt{ℏ / 2 m ω}$ . This is exactly the width of $ψ_{0}$ in Eq. (8). In fact, since $σ_{x} σ_{p} = (n + \frac{1}{2}) ℏ$ , equality in the uncertainty principle holds only when $n = 0$ . The ground state is “a state compressed as far as the uncertainty principle allows.”

5. Absolute Zero and the Third Law of Thermodynamics

In thermal equilibrium at temperature T, the probability of being in level n is given by the Boltzmann distribution $P_{n} \propto e^{- E_{n} / k_{B} T}$ . Summing the geometric series (exactly the same calculation as on the page on Planck’s law) gives

(11)

⟨ E ⟩ = \frac{ℏ ω}{e^{ℏ ω / k_{B} T} - 1} + \frac{ℏ ω}{2} = \frac{ℏ ω}{2} \coth \frac{ℏ ω}{2 k_{B} T}

Relation to Planck’s law The first term is exactly the expression

⟨ E ⟩ = h ν / (e^{h ν / k T} - 1)

that Planck used for cavity radiation. Planck’s law contains no zero-point-energy term. This is because the zero-point term does not depend on temperature, so it does not appear in the energy exchanged as thermal radiation (that is, in the spectrum).

h ν / 2

differs from mode to mode and diverges when summed over all modes, so it cannot be pulled out as a constant common to all modes. Incidentally, in 1912 Planck himself added

h ν / 2

in his second theory.

5.1 Absolute zero does not mean “coming to a stop”

Take the limit $T \to 0$ . In $\coth$ , the argument $ℏ ω / 2 k_{B} T$ diverges to $\infty$ , and $\coth$ approaches 1, so

(12)

\lim_{T \to 0} ⟨ E ⟩ = \frac{ℏ ω}{2} \neq 0

Even when every particle has dropped into the ground state, the jiggling remains. Saying “absolute zero = a state in which all motion has stopped” is natural in the language of classical mechanics, but it is not correct in quantum mechanics. The correct statement is “a state in which everything is in the ground state,” and there, zero-point motion with energy $ℏ ω / 2$ always remains.

Why liquid helium does not solidify Helium remains a liquid at ordinary pressure even if it is cooled to absolute zero. Because the atoms are light (m is small) and the interatomic forces are weak (ω is small), the amplitude of the zero-point motion is so large, compared with the interatomic distance, that it cannot be neglected, and the atoms cannot settle into a crystal lattice. This is one of the few examples in which the zero-point energy is directly visible.

5.2 Why it cannot be reached

The third law of thermodynamics can be stated in two ways.

Nernst’s theorem: as $T \to 0$ , the entropy $S \to 0$ . This is because the ground state is uniquely determined. In fact, $S / k_{B} = u / (e^{u} - 1) - \ln (1 - e^{- u})$ (where $u = ℏ ω / k_{B} T$ ) tends to 0 exponentially as $T \to 0$ .
Unattainability: $T = 0$ cannot be reached by any finite number of operations.

The two are connected. Cooling is the operation of “throwing entropy out of the system,” but as T falls the remaining S becomes smaller, so the amount that can be thrown away in a single operation also decreases. Because the decrease is exponential, no number of stages will ever reach 0. However, how much can be thrown away in one step is determined by the details of the cooling operation (the coupling to the heat bath and the speed of relaxation). This is an explanation that shows the line of reasoning; it is not a theorem that follows from the ground state being unique alone.

⚠ Mind the direction of causality Saying “absolute zero cannot be reached because there is zero-point energy” involves a bit of a leap. More precisely, both start from the same fact: “the ground state is uniquely determined, and there is nothing below it,” but they say different things.

The zero-point energy tells us what kind of state absolute zero is — not a state at rest, but the ground state that keeps vibrating with the energy $ℏ ω / 2$ .
The third law says that the procedure for getting there never ends in a finite number of steps.

“It does not stop even when we cool it” and “it cannot be cooled all the way in the first place” are different claims. The former is a property of the ground state itself, whereas the latter is a property of the cooling operation, and the two are linked through

S \to 0

(Nernst’s formulation).

6. The Road Back to Classical Physics

Then why can’t we see quantum behavior in a spring on a desk? Lining up the numbers makes it obvious at a glance.

System	$ℏ ω$	$ℏ ω / k_{B}$	$σ_{x}$ (n=0)	$⟨ n ⟩$ at 300 K
H₂ molecule stretching mode	0.546 eV	6332 K	8.72 pm	7×10⁻¹⁰
CO molecule stretching mode	0.269 eV	3122 K	3.37 pm	3×10⁻⁵
Si optical phonon (Γ point)	0.0632 eV	734 K	3.43 pm	0.095
Si acoustic phonon (softest mode)	0.00152 eV	17.6 K	22.1 pm	16.5
Desk spring	6.6×10⁻¹⁵ eV	7.6×10⁻¹¹ K	7.3×10⁻¹⁸ m	3.9×10¹²

Two things to take away (1) At room temperature, the desk spring is at $n \approx 4 \times 10^{12}$ . The spacing of one step is only 10⁻¹² of the total, so the staircase looks continuous. The width of the zero-point motion, 7×10⁻¹⁸ m, is even smaller than the radius of a proton and cannot be measured.
(2) Conversely, at room temperature 99.997 % of CO molecules are already in the ground state. In other words, for this vibration, room temperature is, for all practical purposes, “absolute zero.” Even so, it is not at rest: it keeps up a zero-point motion of 3.37 pm (3 % of the bond length).

⚠ A crystal is not “a single harmonic oscillator” So far we have discussed a single oscillator, but a crystal containing N atoms has

3 N

vibrational modes (phonons), and

ω

depends on the wavenumber q and on the branch. The two Si rows in the table above are taken from the Si phonons calculated from first principles by the Mochizuki Research Group, and they are the two extremes. Within the same crystal, the room-temperature

⟨ n ⟩

ranges from 0.095 to 16.5, varying by more than two orders of magnitude. Averaged over the Brillouin zone and all branches, we get

⟨ n ⟩ = 0.59

. The modes are in a range where n = 0, 1, 2, … can be counted, so the quantum nature shows up directly in measurements — the evidence is that the constant-volume molar heat capacity of Si at room temperature is 19.9 J/(mol·K), only 80 % of the Dulong–Petit value

3 R = 24.9

Another route is the correspondence principle. As we increase n, ${| ψ_{n} |}^{2}$ oscillates rapidly, but its overall shape approaches the classical probability density $P_{cl} (x) = 1 / π \sqrt{A^{2} - x^{2}}$ . In tab ① of the simulator, you can check this by raising n to 20.

7. Another Route — Second Quantization (Lecture Notes “Second Quantization” (1)(2))

In §2 we expanded Eq. (3) in a series and, from the requirement that “the series must terminate,” obtained $ε = 2 n + 1$ . This is the right route, but chasing the coefficients of the Hermite polynomials is laborious. Here we take a route that arrives at the same answer without ever solving a differential equation. Moreover, this tool is not limited to the story of a single harmonic oscillator. It becomes the standard language for treating systems with many particles.

7.1 Why do we need such a tool?

The Introduction of the notes When treating the wave function of a many-particle state, it is often more convenient, without writing down the wave function each time, to describe the number of particles and how it changes. For this purpose, the creation operator and annihilation operator of the particle are often used.

When dealing with fermions (such as electrons), the particle number itself seldom changes, but when we consider excited states of electrons, we have to see them as “one electron disappears and one hole is created.” For bosons (phonons, photons, and so on) the particle number changes all the time, so the creation and annihilation operators play the leading role in the description.

A note on notation In the notes, the creation operator is written

{\hat{a}}^{+}

. Here, following the usual textbook convention, we write

{\hat{a}}^{†}

(dagger). They are exactly the same thing. The dagger is the mark of the “Hermitian conjugate,” a notation chosen so that it will not be mistaken for a +.

7.2 Factoring the Hamiltonian

The starting point is the same Schrödinger equation as Eq. (2). With the position operator $\hat{q}$ and the momentum operator $\hat{p} = \frac{ℏ}{i} \frac{d}{d x}$ , it reads

(13)

\hat{H} = \frac{{\hat{p}}^{2}}{2 m} + \frac{1}{2} m ω^{2} {\hat{q}}^{2} = \frac{1}{2} m ω^{2} [{\hat{q}}^{2} + \frac{{\hat{p}}^{2}}{m^{2} ω^{2}}]

This is the starting point. A sum of two squares such as $A^{2}$ + $B^{2}$ can, for ordinary numbers, be rewritten as $(A + i B) (A - i B)$ by factoring. Let us try the same thing here, and we get

\frac{1}{2} m ω^{2} (\hat{q} + i \frac{\hat{p}}{m ω}) (\hat{q} - i \frac{\hat{p}}{m ω}) = ℏ ω \cdot \underset{annihilation operator}{\underset{⏟}{\sqrt{\frac{m ω}{2 ℏ}} (\hat{q} + i \frac{\hat{p}}{m ω})}} \cdot \underset{creation operator}{\underset{⏟}{\sqrt{\frac{m ω}{2 ℏ}} (\hat{q} - i \frac{\hat{p}}{m ω})}}

The two blocks underlined in the notes have names.

(14)

\hat{a} = \sqrt{\frac{m ω}{2 ℏ}} (\hat{q} + i \frac{\hat{p}}{m ω}) = \sqrt{\frac{m ω}{2 ℏ}} (x + \frac{ℏ}{m ω} \frac{d}{d x}) (annihilation operator)

{\hat{a}}^{†} = \sqrt{\frac{m ω}{2 ℏ}} (\hat{q} - i \frac{\hat{p}}{m ω}) = \sqrt{\frac{m ω}{2 ℏ}} (x - \frac{ℏ}{m ω} \frac{d}{d x}) (creation operator)

Inverting these relations, we can rewrite position and momentum in terms of $\hat{a}$ , ${\hat{a}}^{†}$ .

(15)

\hat{q} = \sqrt{\frac{ℏ}{2 m ω}} (\hat{a} + {\hat{a}}^{†}) \hat{p} = - i \sqrt{\frac{ℏ m ω}{2}} (\hat{a} - {\hat{a}}^{†})

However, this factorization is not exact

\hat{q}

and

\hat{p}

are operators that give a different result when the order is swapped (

\hat{q} \hat{p} - \hat{p} \hat{q} = i ℏ

). If we imitate the factoring of ordinary numbers, a discrepancy appears by exactly that much. And that discrepancy is none other than the zero-point energy $ℏ ω / 2$ that we met in §4. We settle the matter in §8 below.

8. Commutation Relations — Where Bosons and Fermions Part Ways (Lecture Notes (3))

8.1 $[\hat{a}, {\hat{a}}^{†}] = 1$

Multiplying out Eq. (14) straightforwardly gives the following two lines from the notes’ Evidence.

\hat{a} {\hat{a}}^{†} = \frac{m ω}{2 ℏ} ({\hat{q}}^{2} + \frac{{\hat{p}}^{2}}{m^{2} ω^{2}}) + \frac{m ω}{2 ℏ} \cdot i (\frac{\hat{p} \hat{q}}{m ω} - \frac{\hat{q} \hat{p}}{m ω})

{\hat{a}}^{†} \hat{a} = \frac{m ω}{2 ℏ} ({\hat{q}}^{2} + \frac{{\hat{p}}^{2}}{m^{2} ω^{2}}) + \frac{m ω}{2 ℏ} \cdot i (\frac{\hat{q} \hat{p}}{m ω} - \frac{\hat{p} \hat{q}}{m ω})

The first terms are identical, so they cancel when we subtract. Only the commutator remains, and substituting $[\hat{p}, \hat{q}] = \hat{p} \hat{q} - \hat{q} \hat{p} = \frac{ℏ}{i}$ gives

(16)

[\hat{a}, {\hat{a}}^{†}] = \hat{a} {\hat{a}}^{†} - {\hat{a}}^{†} \hat{a} = \frac{1}{2 ℏ} {i (\frac{ℏ}{i}) - i (- \frac{ℏ}{i})} = \frac{2 ℏ}{2 ℏ} = 1

8.2 Writing the Hamiltonian in a form with no ambiguity of ordering

This is the line marked “Attention” in the notes. This time, if we add, the commutator part cancels.

\hat{a} {\hat{a}}^{†} + {\hat{a}}^{†} \hat{a} = \frac{m ω}{ℏ} ({\hat{q}}^{2} + \frac{{\hat{p}}^{2}}{m^{2} ω^{2}}) = \frac{2}{ℏ ω} (\frac{{\hat{p}}^{2}}{2 m} + \frac{1}{2} m ω^{2} {\hat{q}}^{2})

The bracket on the right is none other than the $\hat{H}$ of Eq. (13). Therefore

(17)

\hat{H} = \frac{ℏ ω}{2} (\hat{a} {\hat{a}}^{†} + {\hat{a}}^{†} \hat{a})

$\hat{a} {\hat{a}}^{†}$ and ${\hat{a}}^{†} \hat{a}$ are treated on an equal footing, so there is no ambiguity about which one to write first. This is what the notes mean when they write that “the Schrödinger equation of the harmonic oscillator can be expressed as $\frac{ℏ ω}{2} (\hat{a} {\hat{a}}^{†} + {\hat{a}}^{†} \hat{a}) ψ_{n} = E_{n} ψ_{n}$ .”

The discrepancy in the factorization = the zero-point energy The naive factorization in §7.2 was

ℏ ω \hat{a} {\hat{a}}^{†}

. By Eq. (16) we have

\hat{a} {\hat{a}}^{†} = {\hat{a}}^{†} \hat{a} + 1

, so Eq. (17) becomes

\hat{H} = ℏ ω ({\hat{a}}^{†} \hat{a} + \frac{1}{2})

. In other words, the naive factorization was too large by $\frac{1}{2} ℏ ω$ .

This

\frac{1}{2} ℏ ω

is the same quantity as the one obtained from the uncertainty principle in §4. “ $\hat{q}$ and $\hat{p}$ do not commute,” “the uncertainty principle,” and “the zero-point energy” are three ways of saying one thing — here we can see it in the equations.

8.3 For fermions, it is addition, not subtraction

In the left margin of the notes it says, “This commutation relation holds only for bosons. For fermions it is ${\hat{a}, {\hat{a}}^{†}} = 1$ .” This is the passage in question. If we write the fermion operators as $\hat{c}$ , ${\hat{c}}^{†}$ , the rules are

(18)

{\hat{c}, {\hat{c}}^{†}} = \hat{c} {\hat{c}}^{†} + {\hat{c}}^{†} \hat{c} = 1 \hat{c} \hat{c} = {\hat{c}}^{†} {\hat{c}}^{†} = 0

Why does the sign change? When two particles of the same kind are exchanged, the many-body wave function keeps its sign for bosons and changes sign for fermions. Eqs. (16) and (18) are exactly what this “exchange sign” turns into when carried over to the commutation relations of the operators.

Pauli’s exclusion principle comes out in two lines

{\hat{c}}^{†} {\hat{c}}^{†} = 0

means, directly, that “a second particle cannot be put into the same state.” Squaring the number operator

\hat{N} = {\hat{c}}^{†} \hat{c}

and using Eq. (18) just once gives

{\hat{N}}^{2} = {\hat{c}}^{†} \hat{c} {\hat{c}}^{†} \hat{c} = {\hat{c}}^{†} (1 - {\hat{c}}^{†} \hat{c}) \hat{c} = {\hat{c}}^{†} \hat{c} - {\hat{c}}^{†} {\hat{c}}^{†} \hat{c} \hat{c} = \hat{N}

In other words,

\hat{N} (\hat{N} - 1) = 0

. The occupation number can only be 0 or 1. This contrasts with the boson’s

\hat{N}

, which can grow as 0, 1, 2, … without limit (in tab ⑤ of the simulator, you can play with exactly this difference).

Even for the same

N

, a different way of packing gives a different shape Let us put

N

particles into a single harmonic oscillator. If the particles are bosons, all of them can be placed in

n = 0

, so the ground-state energy is

N ℏ ω / 2

, the density stays in the shape of

{| ψ_{0} |}^{2}

and is sharply peaked at the center. Fermions can occupy a state only one at a time, so they have to be stacked up from

n = 0

N - 1

, and the ground-state energy is

(N^{2} / 2) ℏ ω

, the density has to mobilize even the many-noded

ψ_{n}

, and so it becomes broad and flat. If we write the number density of particles as

n (x)

, the total number

\int n (x) d x = N

is the same for both, yet the shape and the ground-state energy are completely different. Matter has volume and cannot be squeezed together because of this latter property.
However, spin has been omitted here. Tab ⑤ of the simulator also displays “Fermions — at most one per level,” but this is a way of counting that leaves out spin. Real electrons have spin ↑↓, so up to two can enter a single level, and if we recount the states as (level, spin) pairs, the argument above holds as it stands.

9. The Number Operator $\hat{N} = {\hat{a}}^{†} \hat{a}$ (Lecture Notes (4))

Let us write ${\hat{a}}^{†} \hat{a}$ out again carefully. We simply substitute $[\hat{q}, \hat{p}] = i ℏ$ into the second line of §8.1.

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\hat{N} = {\hat{a}}^{†} \hat{a} = \frac{1}{ℏ ω} (\frac{{\hat{p}}^{2}}{2 m} + \frac{1}{2} m ω^{2} {\hat{q}}^{2}) - \frac{1}{2} ⟺ \hat{H} = ℏ ω (\hat{N} + \frac{1}{2})

Substituting the result of §2, $E_{n} = (n + \frac{1}{2}) ℏ ω$ , we obtain, as in the notes,

(20)

\hat{N} ψ_{n} = \frac{1}{ℏ ω} (n + \frac{1}{2}) ℏ ω ψ_{n} - \frac{1}{2} ψ_{n} = n ψ_{n}

The meaning of

n

changes Up to §2,

n

was the number that labeled “which level it is.” Here it is the eigenvalue of $\hat{N}$ , that is, the number that counts “how many quanta are in that oscillator.”

E_{n} = (n + \frac{1}{2}) ℏ ω

can be read as “

ℏ ω

per quantum, times

n

quanta, plus the zero-point energy.”

Each kind of system gives these quanta its own name. For the lattice vibrations of a crystal they are phonons, and for the electromagnetic field they are photons. The heart of second quantization is that “moving up a level” and “gaining one particle” are exactly the same mathematics.

In the margin of the notes: “Swap

{\hat{a}}^{†}

and

\hat{a}

and you get +½, right?” Exactly. From Eq. (16) we have

\hat{a} {\hat{a}}^{†} = \hat{N} + 1

, so

\hat{a} {\hat{a}}^{†} = \frac{\hat{H}}{ℏ ω} + \frac{1}{2}

{\hat{a}}^{†} \hat{a}

is smaller than

\frac{\hat{H}}{ℏ ω}

\frac{1}{2}

\hat{a} {\hat{a}}^{†}

is larger than it by

\frac{1}{2}

, so the difference, 1, is exactly Eq. (16). Eq. (17) takes the mean of the two because it passes through the middle of this ±½.

10. Climbing Up and Down the Ladder (Lecture Notes (5))

The names are “creation” and “annihilation,” but what actually happens when we apply them to a wave function? The notes check this using the exact wave function for $n = 1$ . In terms of the dimensionless variable $ξ = \sqrt{\frac{m ω}{ℏ}} x$ , Eq. (14) becomes

\hat{a} = \frac{1}{\sqrt{2}} (ξ + \frac{d}{d ξ}) {\hat{a}}^{†} = \frac{1}{\sqrt{2}} (ξ - \frac{d}{d ξ})

10.1 The calculation in the notes

We apply the creation operator to $φ_{1} (x) = {(\frac{m ω}{π ℏ})}^{1 / 4} \sqrt{2} ξ e^{- ξ^{2} / 2}$ .

{\hat{a}}^{†} φ_{1} = {(\frac{m ω}{π ℏ})}^{1 / 4} [ξ^{2} e^{- ξ^{2} / 2} - (e^{- ξ^{2} / 2} - ξ^{2} e^{- ξ^{2} / 2})] = {(\frac{m ω}{π ℏ})}^{1 / 4} (2 ξ^{2} - 1) e^{- ξ^{2} / 2}

= \sqrt{2} φ_{2} (x)

This is the notes’ line “and so we obtain $= \sqrt{2} φ_{2}$ .” Conversely, applying the annihilation operator changes the sign of the middle term, and $ξ^{2}$ cancels out cleanly,

\hat{a} φ_{1} (x) = {(\frac{m ω}{π ℏ})}^{1 / 4} e^{- ξ^{2} / 2} = φ_{0} (x)

This is the notes’ line “and so we obtain $= φ_{0} (x)$ !” What to notice is that the shape of the wave function itself changes into that of the neighboring level.

10.2 Generalizing to arbitrary $n$

This is the notes’ Theorem.

(21)

{\hat{a}}^{†} φ_{n} = \sqrt{n + 1} φ_{n + 1} \hat{a} φ_{n} = \sqrt{n} φ_{n - 1}

In the figure below, you can check this by actually applying the operators numerically, as differential operators. The curve computed for ${\hat{a}}^{†} φ_{n}$ and the curve for $\sqrt{n + 1} φ_{n + 1}$ coincide exactly.

Applying the ladder operators to a wave function. Gray is the original $φ_{n}$ , and the thick blue curve is ${\hat{a}}^{†} φ_{n}$ or $\hat{a} φ_{n}$ (derivatives by numerical differences), and the red dashed curve is the right-hand side of Eq. (21). If the two curves coincide, Eq. (21) holds.

n = 1 Operator to apply

10.3 The question in the margin of the notes: “Isn’t there a smarter way?”

There is. The notes’ Evidence rewrites the expression partway using ${\hat{a}}^{†} \hat{a} = \hat{a} {\hat{a}}^{†} - 1$ , but if we first collect it into the form of a commutator, it takes just one line.

(22)

[\hat{N}, {\hat{a}}^{†}] = {\hat{a}}^{†} \hat{a} {\hat{a}}^{†} - {\hat{a}}^{†} {\hat{a}}^{†} \hat{a} = {\hat{a}}^{†} [\hat{a}, {\hat{a}}^{†}] = {\hat{a}}^{†} [\hat{N}, \hat{a}] = - \hat{a}

Just apply this to $\hat{N} ψ_{n} = n ψ_{n}$ .

\hat{N} ({\hat{a}}^{†} ψ_{n}) = ({\hat{a}}^{†} \hat{N} + {\hat{a}}^{†}) ψ_{n} = (n + 1) ({\hat{a}}^{†} ψ_{n})

In other words, ${\hat{a}}^{†} ψ_{n}$ is an eigenfunction with eigenvalue $n + 1$ . All that remains is to measure its length (normalization), which fixes the coefficient.

(23)

{| {\hat{a}}^{†} ψ_{n} |}^{2} = ⟨ ψ_{n} | \hat{a} {\hat{a}}^{†} | ψ_{n} ⟩ = ⟨ ψ_{n} | \hat{N} + 1 | ψ_{n} ⟩ = n + 1

Its length is $\sqrt{n + 1}$ and its direction is $ψ_{n + 1}$ , so the first of the two relations in Eq. (21) follows. In the same way, ${| \hat{a} ψ_{n} |}^{2} = ⟨ \hat{N} ⟩ = n$ gives the second one. We have never used the explicit form of the wave function. All we used were Eqs. (16) and (19).

10.4 Quantization appears once more

The equation below Eq. (23), ${| \hat{a} ψ_{n} |}^{2} = n$ , has the square of a length on its left-hand side, so it cannot be negative. That is, $n \geq 0$ . But each time we apply $\hat{a}$ , $n$ decreases by 1. If it does not stop somewhere, we eventually reach $n < 0$ . The only way for it to stop is

\hat{a} ψ_{0} = 0

Counting from there gives $n = 0, 1, 2, \dots$ . The very same conclusion that we derived in §2 from “the series must terminate” has now come out without ever solving a differential equation.

Bonus: the ground state follows from a first-order differential equation Writing

\hat{a} ψ_{0} = 0

out using Eq. (14) gives

(ξ + \frac{d}{d ξ}) ψ_{0} = 0

, that is,

\frac{d ψ_{0}}{ψ_{0}} = - ξ d ξ

. Integrating gives

ψ_{0} \propto e^{- ξ^{2} / 2}

. This is the same as the

ψ_{0}

in Eq. (8), which we obtained in §2 by solving the second-order differential equation with a series. From here, applying Eq. (21) repeatedly gives every

ψ_{n}

ψ_{n} = \frac{1}{\sqrt{n!}} {({\hat{a}}^{†})}^{n} ψ_{0}

11. How This Helps in the Many-Body Problem

So far this has been about a single oscillator. Second quantization shows its worth when there are many particles.

If we try to write out in full the wave function of $N$ particles, $Ψ (x_{1}, \dots, x_{N})$ , the Slater determinant for fermions contains $N!$ terms. For $N = 10$ there are 3.6 million terms; for $N = 20$ there are 2×10¹⁸ terms. That is far too many to write down on paper. (The value of the determinant itself can be found using LU decomposition in $O (N^{3})$ operations. What really gets out of hand is the combinatorial explosion that arises when we superpose this determinant in many different ways to build a many-body state.)

But if we write down only “how many particles are in which level,”

| n_{0}, n_{1}, n_{2}, \dots ⟩ {\hat{c}}_{k}^{†} changes n_{k} \to n_{k} + 1

That is all it takes. The notation does not change however many particles there are. Moreover, the fermion rule that “the sign changes when two particles are exchanged” is absorbed wholesale into the anticommutation relation of Eq. (18). There is no longer any need to write a Slater determinant.

The Hamiltonian, interactions included, can be written in this form for any system.

(24)

\hat{H} = \sum_{k} ε_{k} {\hat{c}}_{k}^{†} {\hat{c}}_{k} + \frac{1}{2} \sum_{k l m n} V_{k l m n} {\hat{c}}_{k}^{†} {\hat{c}}_{l}^{†} {\hat{c}}_{m} {\hat{c}}_{n}

The first term is the one-body energy and the second term is the two-body interaction (the Coulomb repulsion, for electrons). The Hartree–Fock method, perturbation theory, and Feynman diagrams all start from this equation.

Processes that change the particle number translate directly into equations

Absorption of light: annihilate one photon and promote an electron to a higher level — ${\hat{c}}_{l}^{†} {\hat{c}}_{k} {\hat{a}}_{λ}$
Electron excitation: annihilate the electron in level $k$ , and create one in level $l$ — ${\hat{c}}_{l}^{†} {\hat{c}}_{k}$ . The vacated level $k$ is a hole. This is what the Introduction of the notes means by “annihilation of an electron and creation of a hole.”
Lattice vibrations: if we decompose the vibrations of a crystal into normal modes, each mode is an independent harmonic oscillator. For the mode with wavenumber $q$ and branch $ν$ , applying ${\hat{a}}_{q ν}^{†}$ amounts to “creating one phonon.”

Specific heat, thermal conductivity, and the electron–lattice interaction (superconductivity belongs here too) all turn out to be calculations that count the creation and annihilation of these phonons. In tab ② of the simulator we used the quantity

⟨ n ⟩

, which is exactly “the average number of phonons in that mode.”

12. Numerical Checks

What was checked	Agreement
Recurrence-relation version of Eq. (8) vs. the explicit forms in lecture notes (2) for n = 0, 1, 2	Relative difference < 4×10⁻¹⁶
Normalization $\int {\| ψ_{n} \|}^{2} d X = 1$ (n ≤ 30)	Relative difference < 2×10⁻¹³
Orthogonality $⟨ m \| n ⟩ = 0$	< 4×10⁻¹⁵
Whether Eq. (3) is satisfied (numerical second derivative)	Residual < 9×10⁻⁸
$⟨ X^{2} ⟩ = n + 1 / 2$ , $σ_{x} σ_{p} = (n + 1 / 2) ℏ$	Relative difference < 2×10⁻¹³
Eqs. (9)(10): minimum from the arithmetic–geometric mean inequality = $ℏ ω / 2$	Relative difference < 2×10⁻¹⁶
Coefficient ratio of Eq. (6) → 2/k (for k = 6, 10, 18: 0.57, 0.71, 0.82, approaching 1)	Same behavior as e^X²
Numerical integration of Eq. (3): decays only for $ε = 1, 3, 5, 7, 9$	At eigenvalues, 10⁻⁵; elsewhere, 10⁺³
Eqs. (11)(12): ⟨E⟩ at $T \to 0$ is $ℏ ω / 2$ , and at $T \to \infty$ is $k_{B} T$	Relative difference < 1×10⁻¹²
Second quantization (lecture notes (2)–(5))	Agreement
(21) ${\hat{a}}^{†} φ_{n} = \sqrt{n + 1} φ_{n + 1}$ , $\hat{a} φ_{n} = \sqrt{n} φ_{n - 1}$ (n ≤ 12, operators applied by numerical differences)	Max. difference < 3×10⁻¹¹
Coefficient $\sqrt{n + 1}$ measured by an overlap integral	Relative difference < 8×10⁻¹²
$\hat{a} φ_{0} = 0$ (bottom of the ladder, §10.4)	< 2×10⁻¹³
(16) $[\hat{a}, {\hat{a}}^{†}] = 1$ , (19)(20) eigenvalues of $\hat{N}$ are n (12×12 matrix representation)	Max. difference < 2×10⁻¹⁰
(17) diagonal of $\hat{H} = \frac{ℏ ω}{2} (\hat{a} {\hat{a}}^{†} + {\hat{a}}^{†} \hat{a})$ is $n + 1 / 2$	Max. difference < 1×10⁻¹⁰
Discrepancy of the naive factorization $ℏ ω \hat{a} {\hat{a}}^{†} - \hat{H} = ℏ ω / 2$ (§8.2)	Max. difference < 2×10⁻¹⁰
(22)(23) $[\hat{N}, {\hat{a}}^{†}] = {\hat{a}}^{†}$ , ${\| {\hat{a}}^{†} ψ_{n} \|}^{2} = n + 1$	Max. difference < 2×10⁻¹⁰
$ψ_{n} = \frac{1}{\sqrt{n!}} {({\hat{a}}^{†})}^{n} ψ_{0}$ (n ≤ 10)	Max. difference < 2×10⁻¹¹
(18) ${\hat{c}, {\hat{c}}^{†}} = 1$ , ${\hat{c}}^{†} {\hat{c}}^{†} = 0$ , ${\hat{N}}^{2} = \hat{N}$	Exactly 0
Ground-state energy for the same N: bosons $N ℏ ω / 2$ / fermions $(N^{2} / 2) ℏ ω$ , $\int n (x) d x = N$ (tab ⑤)	Relative difference < 4×10⁻¹⁴

13. Summary

Near its bottom, every potential is $\frac{1}{2} m ω^{2} x^{2}$ . That is why the harmonic oscillator appears everywhere (1)(2).
Making the equation dimensionless removes all the constants, leaving the single equation $ψ ″ + (ε - X^{2}) ψ = 0$ (3).
To discard the divergent solution far away, we put $ψ = f e^{- X^{2} / 2}$ (4)(5).
If the series does not terminate, the divergence returns. The condition for termination is $ε = 2 n + 1$ , that is, the quantization of the energy (6)(7). Quantization comes not from the equation but from the boundary condition.
Even at the lowest level, $ℏ ω / 2$ remains. This quantity follows from the arithmetic–geometric mean inequality plus the uncertainty principle alone, and $n = 0$ realizes the equality of the uncertainty principle (9)(10).
Absolute zero is not a “state at rest” but “a state in which everything is in the ground state.” It cannot be reached in a finite number of operations — because the entropy that can be discarded in one operation keeps decreasing, and how fast it decreases also depends on the details of the cooling operation (11)(12).
Rewriting the Hamiltonian in terms of $\hat{a}, {\hat{a}}^{†}$ by factorization gives the same answer without solving the differential equation. The remainder due to the inexactness of the factorization, $ℏ ω / 2$ , is exactly the zero-point energy (13)–(17).
$\hat{N} = {\hat{a}}^{†} \hat{a}$ has eigenvalue n, which is “how many quanta are present.” ${\hat{a}}^{†}$ creates one, and $\hat{a}$ removes one. The fact that the ladder stops at $\hat{a} ψ_{0} = 0$ is itself quantization (19)–(23).
Bosons obey the commutation relation $[\hat{a}, {\hat{a}}^{†}] = 1$ ; fermions obey the anticommutation relations ${\hat{c}, {\hat{c}}^{†}} = 1$ and ${{\hat{c}}^{†}, {\hat{c}}^{†}} = 0$ . From the latter we get ${\hat{c}}^{†} {\hat{c}}^{†} = 0$ , that is, Pauli’s exclusion principle comes out in two lines (16)(18).
When there are many particles, we write not the wave function but “how many particles are in which level.” This is second quantization, and it becomes the common language of many-body quantum theory (24).

Mochizuki Research Group, Tokyo University of Science / Based on lecture notes (in Japanese) 5. Wave Mechanics 3. Harmonic Oscillator (1) “The One-Dimensional Harmonic Oscillator,” (2) “Solutions of the One-Dimensional Harmonic Oscillator and Their Graphs,” (3) “Zero-Point Vibrational Energy and the Uncertainty Principle,” and the lecture notes “Second Quantization”: (1) “What Is Second Quantization?”, (2) “Creation and Annihilation Operators,” (3) “Commutation Relations of the Creation and Annihilation Operators,” (4) “The Number Operator,” (5) “Wave Functions of the Harmonic Oscillator and the Creation and Annihilation Operators.” The thermodynamic part of §5 and the introduction to many-body quantum theory in §11 go beyond the notes and have been added here. Numerical verification uses the recurrence relation for Hermite functions, Simpson’s rule, fourth-order Runge–Kutta, and matrix representations in which the ladder operators are applied with fourth-order central differences. The equations are typeset with the browser’s built-in MathML (there are no external libraries or network communication at all).