Two operators that differ only in a sign do the whole job. Their commutator is a number, and every result in the chapter follows from that one fact.

§6.3 took $E_n = (n+\frac12)\hbar\omega$ on credit and §6.4 spent it twice. This section pays the debt.

Phillips marks it “for mathematically inclined readers” and says it “may be omitted without significant loss of continuity”. Ignore that. The result is worth one page; the method is worth the chapter. It finds the entire spectrum without ever solving a differential equation — build one state, then climb — and chapter 8 constructs the whole theory of angular momentum by repeating the argument almost word for word.

Physics you need first — nothing new, but one habit: non-dimensionalize before you solve

No new physics here at all. What the section needs is a technique.

Equation (6.10) contains three dimensional constants — $\hbar$ , $\omega$ and $m$ . From them you can build exactly one energy, $\hbar\omega$ , and exactly one length, $\sqrt{\hbar/m\omega}$ . Measure $E$ and $x$ in those units and every constant disappears from the equation, leaving pure numbers.

This is not cosmetic. An equation with three parameters looks like a three-parameter family of problems; stripped of units it is revealed as one problem, solved once and for all. Any solution can be scaled back to any mass and any frequency afterwards. Engineers do this constantly — non-dimensional groups, per-unit systems, normalized frequency $\omega/\omega_c$ — and it is the same move for the same reason: fewer symbols, and the structure becomes visible.

Carried over: §6.2 for Eq. (6.10) itself and the boundary condition $\psi \to 0$ at infinity, which does all the selecting here.

Cleaning up the equation

Measure energy in units of $\hbar\omega$ and length in units of $\sqrt{\hbar/m\omega}$ :

E = \epsilon\,\hbar\omega \qquad\text{and}\qquad x = q\sqrt{\frac{\hbar}{m\omega}}\tag{6.28}

Then Eq. (6.10) becomes

\left[-\frac{\mathrm d^2}{\mathrm dq^2} + q^2\right]\psi(q) = 2\epsilon\,\psi(q)\tag{6.29}

The one identity everything rests on

For any function $f(q)$ , expanding and using $\mathrm d[qf]/\mathrm dq = f + q\,\mathrm df/\mathrm dq$ :

\left[q + \frac{\mathrm d}{\mathrm dq}\right]\left[q - \frac{\mathrm d}{\mathrm dq}\right]f = \left[q^2 - \frac{\mathrm d^2}{\mathrm dq^2} + 1\right]f

The bracket on the right is Eq. (6.29)‘s operator plus one. So the eigenvalue equation can be written

\left[q + \frac{\mathrm d}{\mathrm dq}\right]\left[q - \frac{\mathrm d}{\mathrm dq}\right]\psi(q) = (2\epsilon + 1)\psi(q)\tag{6.30}

and repeating the calculation with the two factors swapped gives

\left[q - \frac{\mathrm d}{\mathrm dq}\right]\left[q + \frac{\mathrm d}{\mathrm dq}\right]\psi(q) = (2\epsilon - 1)\psi(q)\tag{6.31}

The entire chapter is in the difference between those two lines

Equations (6.30) and (6.31) contain the same two operators in opposite orders, and the results differ by exactly 2. That is the whole content of the section, and everything else is bookkeeping.

Order matters because $q$ and $\mathrm d/\mathrm dq$ do not commute: applying $\mathrm d/\mathrm dq$ to $qf$ produces an extra $f$ that applying it to $f$ first does not. Everything else follows — that constant 2 is why the rungs are evenly spaced, why the spacing is $\hbar\omega$ and not something that drifts with $n$ , and why an operator can move you exactly one rung at a time.

Notice what has not happened: nobody has solved anything. Two ways of writing the same equation, differing by a constant, are about to produce an entire infinite spectrum.

The ground state, from a first-order equation

Both forms are satisfied trivially if the relevant bracket annihilates $\psi$ .

Try Eq. (6.30). It holds if $2\epsilon + 1 = 0$ and $[q - \mathrm d/\mathrm dq]\psi = 0$ , i.e. $\epsilon = -\frac12$ and

\psi(q) = Ae^{+q^2/2}

which must be discarded — it diverges as $q \to \pm\infty$ and cannot be normalized. Eq. (6.11) throws it away.

Try Eq. (6.31). It holds if $2\epsilon - 1 = 0$ and $[q + \mathrm d/\mathrm dq]\psi = 0$ , i.e. $\epsilon = +\frac12$ and

\psi(q) = Ae^{-q^2/2}

which is perfectly well behaved. So

\epsilon_0 = \tfrac12\qquad\text{and}\qquad \psi_0(q) = A_0e^{-q^2/2}\tag{6.32}

giving, back in ordinary units,

E_0 = \tfrac12\hbar\omega\tag{6.33}

\psi_0(x) = N_0e^{-x^2/2a^2},\qquad a = \sqrt{\frac{\hbar}{m\omega}}\tag{6.34}

A second-order problem solved by a first-order equation

Equation (6.29) is second order, and second-order equations are hard. But $[q + \mathrm d/\mathrm dq]\psi = 0$ is first order — separable, one line to integrate — and its solution is the ground state.

That is the trick, and it is worth naming: factorizing an operator turns one hard equation into two easy ones. It is the same move as factorizing a quadratic rather than using the formula, or writing a second-order ODE as a pair of first-order ones, or an LU decomposition — solve a triangular system twice instead of a full one once.

The discarded root is not a technicality either. $e^{+q^2/2}$ is a perfectly good solution of the differential equation; it is physics that rejects it, in the form of Eq. (6.11). This is the third time the same thing has selected a spectrum — the infinite well’s walls, the finite well’s smooth join, and now normalizability — and it is always the boundary condition, never the equation.

Climbing

Now the step that generates everything. Write Eq. (6.30) for the $n$ -th state and hit both sides with $[q - \mathrm d/\mathrm dq]$ :

step 1 of 4

Four lines, and the only tool is comparing two equations that look alike. This argument is the reason §6.6 is worth reading even though its result was already quoted in §6.3.

1Start from Eq. (6.30) applied to the n-th state, whose eigenvalue is ε_n. Nothing is assumed about ψ_n except that it satisfies the equation.
$[q + \frac{d}{d q}] [q - \frac{d}{d q}] ψ_{n} = (2 ϵ_{n} + 1) ψ_{n}$
2Apply [q − d/dq] to BOTH sides. This is legitimate for the same reason multiplying both sides of any equation by the same thing is, and it is the only creative step in the argument.
$[q - \frac{d}{d q}] [q + \frac{d}{d q}] ([q - \frac{d}{d q}] ψ_{n}) = (2 ϵ_{n} + 1) ([q - \frac{d}{d q}] ψ_{n})$
Read the brackets carefully: the object in parentheses, [q − d/dq]ψₙ, now appears on both sides as a single unit.
3Compare with Eq. (6.31), which says exactly this for a state of eigenvalue ε. The two have identical structure, so whatever [q − d/dq]ψₙ is, it is an eigenfunction — and its eigenvalue can be read off.
$[q - \frac{d}{d q}] [q + \frac{d}{d q}] ψ = (2 ϵ - 1) ψ$
4Match the two constants: 2ε − 1 = 2ε_n + 1. So the new state has ε = ε_n + 1 — one full quantum higher, every time, regardless of which n you started from.
$2 ϵ - 1 = 2 ϵ_{n} + 1 ⟹ ϵ = ϵ_{n} + 1$
That "regardless of n" is where the even spacing comes from — and it traces straight back to the constant 2 separating Eqs. (6.30) and (6.31).

So $[q - \mathrm d/\mathrm dq]$ is an energy raising operator — one of the pair of raising and lowering operators this section is built on. Starting from the ground state and applying it repeatedly:

\epsilon_1 = \tfrac32\qquad\text{and}\qquad \psi_1(q) = A_1\left[q - \frac{\mathrm d}{\mathrm dq}\right]e^{-q^2/2}\tag{6.35}

\epsilon_2 = \tfrac52\qquad\text{and}\qquad \psi_2(q) = A_2\left[q - \frac{\mathrm d}{\mathrm dq}\right]^2e^{-q^2/2}\tag{6.36}

Equation (6.36) is printed with the wrong subscript

Page 127 gives Eq. (6.36) as $\epsilon_1 = \frac52$ , paired with $\psi_2$ . It should be $\epsilon_2$ .

It contradicts the book on its own page in two directions: Eq. (6.35), three lines above, sets $\epsilon_1 = \frac32$ ; and Eq. (6.37), immediately below, gives $\epsilon_n = n + \frac12$ , so $\frac52$ belongs to $n = 2$ . Verified from page-144.png.

This is the third error found in chapter 6’s printed text, after the “Section 4.8” cross-reference on page 114 and the $\frac12k^2x^2$ of Eq. (6.18). It is the mildest of the three — a subscript, with the correct value beside it — but it is the one most likely to be copied into notes without noticing.

and so on, without end:

\epsilon_n = n + \tfrac12\qquad\text{and}\qquad \psi_n(q) = A_n\left[q - \frac{\mathrm d}{\mathrm dq}\right]^ne^{-q^2/2}\tag{6.37}

Converting back with Eq. (6.28):

E_n = \left(n + \tfrac12\right)\hbar\omega\tag{6.38}

which is Eq. (6.12), now earned. And expressing $q$ in terms of $x$ ,

\psi_n(x) = N_nH_n\!\left(\frac{x}{a}\right)e^{-x^2/2a^2},\qquad a = \sqrt{\frac{\hbar}{m\omega}}\tag{6.39}

where $H_n$ is a polynomial of order $n$ — a Hermite polynomial .

ψ1 — what went in
ψ2 — what it should be
the operator’s output

start at n1

polynomial in	2q	ε = 1.5
polynomial out	4q² − 2	ε = 2.5
\|aₙ\|² = 2(n+1)	4.0000	expected 4

The solid curve is what the operator produced; the thick pale one is ψ₂ drawn independently. They coincide, which is Eq. (6.37): applying [q − d/dq] to a solution of energy ε gives a solution of energy ε + 1, so one state and one operator generate the whole spectrum.

Watch the polynomial row. Starting from 1 and raising repeatedly gives 2q, then 4q² − 2, then 8q³ − 12q — the Hermite polynomials, appearing on their own rather than being looked up. Those are exactly the brackets in Table 6.1. The operators here act on the polynomial alone — raising is p ↦ 2q·p − p′ and lowering is p ↦ p′ — so every coefficient shown is exact integer arithmetic, with no numerical differentiation anywhere.

Where the Hermite polynomials come from

They are not looked up. Applying $[q - \mathrm d/\mathrm dq]$ to $p(q)e^{-q^2/2}$ gives $(2qp - p')e^{-q^2/2}$ — so the operator acts on the polynomial alone, by the rule $p \mapsto 2qp - p'$ .

Start with $p = 1$ and turn the handle: $1 \to 2q \to 4q^2 - 2 \to 8q^3 - 12q$ . Those are $H_0$ through $H_3$ , and they are exactly the brackets printed in Table 6.1. The widget above shows the coefficients appearing as you climb.

The lowering operator is even simpler: $[q + \mathrm d/\mathrm dq]$ acting on $p\,e^{-q^2/2}$ gives $p'\,e^{-q^2/2}$ — it just differentiates the polynomial. Which makes Eq. (6.40) obvious: the ground state has $p = 1$ , and the derivative of a constant is zero.

Why the ladder has a bottom

One item of unfinished business: is $E_0 = \frac12\hbar\omega$ really the lowest energy? Two independent arguments say yes.

From uncertainty. Problem 1 shows the Heisenberg principle alone forbids any state below $\frac12\hbar\omega$ , without solving anything.

From the ladder itself. $[q + \mathrm d/\mathrm dq]$ is an energy lowering operator — problem 9 proves it by the mirror image of the argument above, and it takes $\psi_n$ to $\psi_{n-1}$ . But applied to the ground state it gives

\left[q + \frac{\mathrm d}{\mathrm dq}\right]\psi_0(q) = 0\tag{6.40}

Not a small state — no state at all. The ladder stops because there is nothing left to lower.

Engineer's bridge — these are matrices, and the whole spectrum is one commutator

Give the operators their usual names. Up to a factor of $\sqrt2$ , $[q - \mathrm d/\mathrm dq]$ is the creation operator $a^\dagger$ and $[q + \mathrm d/\mathrm dq]$ is the annihilation operator $a$ . In the basis of the states they generate, they are matrices with a single off-diagonal:

a = \begin{pmatrix}0&\sqrt1&&\\&0&\sqrt2&\\&&0&\sqrt3\\&&&\ddots\end{pmatrix},\qquad a^\dagger = a^{\mathsf T}

and the Hamiltonian is $\hat H = \hbar\omega\left(a^\dagger a + \frac12\right)$ — diagonal, with entries $n + \frac12$ . The code below builds those three matrices and calls eigh; the spectrum falls out with no calculus anywhere.

The commutator is the whole theory. Everything above compresses to $[a, a^\dagger] = 1$ . That single relation forces the spectrum: it makes $a^\dagger$ raise, $a$ lower, and the ladder terminate. A physicist would say the harmonic oscillator is that commutator — and quantum field theory takes the same line, with one oscillator per mode of a field and $a^\dagger$ creating a particle rather than a rung.

You have met this as a shift. On the vector of coefficients, $a^\dagger$ is a weighted shift-down operator and $a$ a shift-up — the same structure as a delay line $z^{-1}$ , or the shift matrix that turns polynomial multiplication into a convolution. The oscillator’s algebra is a shift register with $\sqrt{n}$ weights.

Where it breaks, and it is a theorem rather than an inconvenience. No finite matrices anywhere satisfy $[A, B] = I$ — take the trace of both sides: the left is always 0, the right is $N$ . So the commutation relation this whole section rests on is only satisfiable in infinite dimensions. Truncate at $N$ states and the discrepancy has to go somewhere; the code below prints where. The diagonal of $[a, a^\dagger]$ comes out as $N-1$ ones followed by a single $-(N-1)$ , summing dutifully to zero. Every numerical oscillator calculation lives with that, and the states near the top of the basis are the ones to distrust.

Reproduce it in NumPy

Every operator here acts on the POLYNOMIAL part exactly — raising is p ↦ 2q·p − p′, lowering is p ↦ p′ — so the Hermite polynomials and εₙ = n + ½ come out as integer arithmetic. A first draft differentiated numerically and had drifted to ε = 9.46 instead of 4.5 by n = 4; nested numerical derivatives are not the right tool here. Note the commutator check excludes the truncated last row, where it necessarily fails.

import numpy as np
from numpy.polynomial import Polynomial as P

# Write psi(q) = p(q) exp(-q^2/2).  Then the two operators act on the POLYNOMIAL
# alone, exactly -- no numerical differentiation anywhere:
#   [q - d/dq] (p e) = (2q p - p') e      <- raising
#   [q + d/dq] (p e) = (p') e             <- lowering
UP = lambda p: 2*P([0, 1])*p - p.deriv()
DN = lambda p: p.deriv()
show = lambda p: str(p.convert().coef.round(6).tolist())

print("the operator identity, done in exact polynomial arithmetic")
print("=" * 70)
p = P([1, -2, 0, 0.5])                       # an arbitrary polynomial
lhs, rhs = DN(UP(p)), UP(DN(p))
print(f"  [q+d/dq][q-d/dq] - [q-d/dq][q+d/dq]  applied to p:")
print(f"    difference coefficients = {show(lhs - rhs)}")
print(f"    2p coefficients         = {show(2*p)}")
print(f"    equal? {np.allclose((lhs-rhs).convert().coef, (2*p).convert().coef)}")
print("  The two orderings differ by exactly 2, for EVERY p.  Eqs. (6.30) and")
print("  (6.31) are that one fact, and evenly spaced rungs are its consequence.")

print("\nclimbing from the ground state: Eq. (6.37) generates Hermite polynomials")
print("=" * 70)
H = [P([1])]                                  # psi_0: p = 1, Eq. (6.32)
for n in range(5):
    H.append(UP(H[-1]))
print(f"  {'n':>3}  polynomial p_n(q) coefficients (ascending powers)")
for n, h in enumerate(H[:5]):
    print(f"  {n:3d}  {show(h)}")
print("  These are the Hermite polynomials: H0 = 1, H1 = 2q, H2 = 4q^2 - 2,")
print("  H3 = 8q^3 - 12q -- which is Eq. (6.39), and Table 6.1's brackets.")

print("\n  the lowering operator kills the ground state -- Eq. (6.40)")
print(f"  [q + d/dq] psi_0  ->  polynomial {show(DN(H[0]))}   (identically zero)")
print("  Nothing below E_0 exists, because there is nothing left to lower.")

print("\nthe eigenvalue, checked exactly for each state")
print("=" * 70)
print("  Substituting psi = p e^(-q^2/2) into Eq. (6.29) leaves")
print("      -p'' + 2q p' + p = 2 epsilon p")
hdr = "-p'' + 2q p' + p"
print(f"  {'n':>3} {hdr:>34} {'2n+1':>6} {'epsilon':>9} {'n+1/2':>8}")
for n, h in enumerate(H[:4]):
    res = -h.deriv().deriv() + 2*P([0, 1])*h.deriv() + h
    ratio = res.convert().coef[np.abs(h.convert().coef).argmax()] / \
            h.convert().coef[np.abs(h.convert().coef).argmax()]
    print(f"  {n:3d} {show(res):>34} {2*n+1:6d} {ratio/2:9.4f} {n+0.5:8.4f}")
print("  Every one is exactly (2n+1) times the polynomial it started from, so")
print("  epsilon_n = n + 1/2 -- Eq. (6.37) -- with no differential equation solved.")

print("\nproblem 10's normalization: [q - d/dq] psi_n = a_n psi_(n+1), |a_n|^2 = 2(n+1)")
q = np.linspace(-12, 12, 240001)
g = np.exp(-q*q/2)
nrm = lambda p: np.trapezoid((p(q)*g)**2, q)
print(f"  {'n':>3} {'|a_n|^2':>10} {'2(n+1)':>8}")
for n in range(5):
    print(f"  {n:3d} {nrm(UP(H[n]))/nrm(H[n]):10.4f} {2*(n+1):8d}")

print("\nthe same spectrum from LINEAR ALGEBRA alone -- no calculus at all")
print("=" * 70)
N = 12
a  = np.diag(np.sqrt(np.arange(1, N)), 1)     # lowering matrix
ad = a.T                                       # raising matrix
Hm = ad @ a + 0.5*np.eye(N)                    # H / hbar omega
print(f"  a is {N}x{N} with sqrt(n) on the first superdiagonal; a-dagger its transpose.")
print(f"  [a, a-dagger] = I ?     max deviation {np.abs(a@ad - ad@a - np.eye(N))[:-1,:-1].max():.1e}")
print(f"  H = a-dagger a + 1/2 diagonal ?  max off-diagonal "
      f"{np.abs(Hm - np.diag(np.diag(Hm))).max():.1e}")
print(f"  eigenvalues: {np.round(np.linalg.eigvalsh(Hm)[:6], 6)}")
print("  n + 1/2, exactly, from three lines of matrix algebra.  This is what")
print("  section 6.6 is really doing, and what chapter 8 will do again.")
print("\n  and where a finite matrix must lie about it:")
C = a@ad - ad@a
print(f"    diagonal of [a, a-dagger] = {np.round(np.diag(C), 3).tolist()}")
print(f"    trace = {np.trace(C):.1f}, and it MUST be 0 for any finite matrices,")
print(f"    while the trace of the identity is {N}.  So [a, a-dagger] = I has no")
print(f"    finite-dimensional solution at all -- the truncation dumps the whole")
print(f"    discrepancy into the last entry, {C[N-1, N-1]:.0f}.  Distrust the top of the basis.")

printsthe operator identity, done in exact polynomial arithmetic ====================================================================== [q+d/dq][q-d/dq] - [q-d/dq][q+d/dq] applied to p: difference coefficients = [2.0, -4.0, 0.0, 1.0] 2p coefficients = [2.0, -4.0, 0.0, 1.0] equal? True The two orderings differ by exactly 2, for EVERY p. Eqs. (6.30) and (6.31) are that one fact, and evenly spaced rungs are its consequence. climbing from the ground state: Eq. (6.37) generates Hermite polynomials ====================================================================== n polynomial p_n(q) coefficients (ascending powers) 0 [1.0] 1 [0.0, 2.0] 2 [-2.0, 0.0, 4.0] 3 [0.0, -12.0, 0.0, 8.0] 4 [12.0, 0.0, -48.0, 0.0, 16.0] These are the Hermite polynomials: H0 = 1, H1 = 2q, H2 = 4q^2 - 2, H3 = 8q^3 - 12q -- which is Eq. (6.39), and Table 6.1's brackets. the lowering operator kills the ground state -- Eq. (6.40) [q + d/dq] psi_0 -> polynomial [0.0] (identically zero) Nothing below E_0 exists, because there is nothing left to lower. the eigenvalue, checked exactly for each state ====================================================================== Substituting psi = p e^(-q^2/2) into Eq. (6.29) leaves -p'' + 2q p' + p = 2 epsilon p n -p'' + 2q p' + p 2n+1 epsilon n+1/2 0 [1.0] 1 0.5000 0.5000 1 [0.0, 6.0] 3 1.5000 1.5000 2 [-10.0, 0.0, 20.0] 5 2.5000 2.5000 3 [0.0, -84.0, 0.0, 56.0] 7 3.5000 3.5000 Every one is exactly (2n+1) times the polynomial it started from, so epsilon_n = n + 1/2 -- Eq. (6.37) -- with no differential equation solved. problem 10's normalization: [q - d/dq] psi_n = a_n psi_(n+1), |a_n|^2 = 2(n+1) n |a_n|^2 2(n+1) 0 2.0000 2 1 4.0000 4 2 6.0000 6 3 8.0000 8 4 10.0000 10 the same spectrum from LINEAR ALGEBRA alone -- no calculus at all ====================================================================== a is 12x12 with sqrt(n) on the first superdiagonal; a-dagger its transpose. [a, a-dagger] = I ? max deviation 1.8e-15 H = a-dagger a + 1/2 diagonal ? max off-diagonal 0.0e+00 eigenvalues: [0.5 1.5 2.5 3.5 4.5 5.5] n + 1/2, exactly, from three lines of matrix algebra. This is what section 6.6 is really doing, and what chapter 8 will do again. and where a finite matrix must lie about it: diagonal of [a, a-dagger] = [1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, -11.0] trace = 0.0, and it MUST be 0 for any finite matrices, while the trace of the identity is 12. So [a, a-dagger] = I has no finite-dimensional solution at all -- the truncation dumps the whole discrepancy into the last entry, -11. Distrust the top of the basis.

Put a number on it — CO through the dimensionless variables

Equation (6.28) says every oscillator is the same problem. Here is a real one going through it and coming back out.

Carbon monoxide, from §6.4: $\mu = 6.85$ amu and $k = 1907\ \mathrm{N\,m^{-1}}$ , so $\omega = \sqrt{k/\mu} = 4.09\times10^{14}\ \mathrm{s^{-1}}$ .

The two conversion factors.

\hbar\omega = 0.2695\ \mathrm{eV},\qquad a = \sqrt{\hbar/\mu\omega} = 4.76\ \mathrm{pm}

Now read the dimensionless answers off and convert. $\epsilon_n = n + \frac12$ is the entire solution; multiplying by $\hbar\omega$ gives energies and $\sqrt{n+\frac12}\,a$ gives spreads:

$n$	$\epsilon_n$	$E_n$	$\Delta x$
0	$\frac12$	$0.135\ \mathrm{eV}$	$3.37\ \mathrm{pm}$
1	$\frac32$	$0.404\ \mathrm{eV}$	$5.83\ \mathrm{pm}$
2	$\frac52$	$0.674\ \mathrm{eV}$	$7.52\ \mathrm{pm}$

And a sanity check. CO’s bond is $113\ \mathrm{pm}$ long, so even in $n = 2$ the nuclei wander only about 7% of it. The harmonic approximation is being asked to describe a genuinely small excursion — which is why §6.4 found it good to a few percent for the lowest transitions, and why it fails only far up the well.

That is what non-dimensionalizing bought: one solved problem, and two multiplications per molecule.

The eigenfunctions as a basis

Like the eigenfunctions of any Hamiltonian, the $\psi_n$ form a complete orthonormal basis :

\int_{-\infty}^{+\infty}\psi_m^*(x)\psi_n(x)\,\mathrm dx = \delta_{m,n}\tag{6.41}

so any function can be expanded in them —

f(x) = \sum_{n=0,1,\dots}c_n\psi_n(x),\qquad c_n = \int_{-\infty}^{+\infty}\psi_n^*(x)f(x)\,\mathrm dx\tag{6.42}

— and the general solution of the time-dependent Schrödinger equation for the oscillator is

\Psi(x,t) = \sum_{n=0,1,2\dots}c_n\psi_n(x)e^{-iE_nt/\hbar}\tag{6.43}

which is Eq. (6.15), the starting point of §6.3’s non-stationary states. The chapter closes its own loop.

Where this goes next

Problems 6 has eleven, and the last three — 9, 10 and 11 — are this section’s algebra: the lowering operator, the normalization constants $|a_n|^2 = 2(n+1)$ and $|b_n|^2 = 2n$ , and the selection rule $x_{m,n} = 0$ unless $|m-n| = 1$ that §6.3 and §6.4 both leaned on.

Then chapter 7 makes operators the subject rather than the tool, and chapter 8 runs this exact construction again for angular momentum — a lowering operator that annihilates the bottom state, a raising operator that steps by one, and a ladder whose length is fixed by where it terminates. If §6.6 makes sense, chapter 8 will feel like a rerun.

Check yourself

0 / 6 answered

1.Equations (6.30) and (6.31) are the same two operators in opposite orders, and they differ by 2. What does that constant buy?
2.Solving Eq. (6.30) with the bracket annihilating $ψ$ gives $ϵ = - \frac{1}{2}$ and $ψ = A e^{+ q^{2} /2}$ . Why is that not a state of energy $- \frac{1}{2} ℏ ω$ ?
Set n = 0 in the widget and choose the lowering operator.
3.The output is identically zero. Why does that single fact fix the bottom of the spectrum?
Raise repeatedly from n = 0 and watch the polynomial row.
4.The coefficients go 1, then 2q, then 4q² − 2, then 8q³ − 12q. What are you watching?
5.In matrix language $\hat{H} = ℏ ω (a^{†} a + \frac{1}{2})$ , and the code gets the spectrum from three matrices. Why can no *finite* matrix version be exactly right?
6.The book marks §6.6 optional. What is lost by skipping it?