3.11 Rewriting earlier chapters in Dirac notation

The toolkit is complete. This section settles the accounts: translate every result of chapter 2 into the new language, then tally up what was saved.

This is not revision. Some proofs will shrink from half a page to three lines; others will turn from “solve a partial differential equation” into “do some algebra” — and the second kind is the real prize.

The infinite square well, written as matrices

The picture

Chapter 2’s version

\psi_n(x)=\sqrt{\frac2L}\sin\frac{n\pi x}{L}\tag{3.11.1}

E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}\tag{3.11.2}

\int_0^L\psi_m^*\psi_n\dd x=\delta_{mn}\tag{3.11.3}

\Psi(x,0)=\sum_nc_n\psi_n(x)\tag{3.11.4}

c_n=\int_0^L\psi_n^*\Psi\,\dd x\tag{3.11.5}

The mathematics

The new version

\hat H\ket{n}=E_n\ket{n}\tag{3.11.6}

\braket{m}{n}=\delta_{mn},\qquad \sum_n\ket{n}\bra{n}=\hat 1\tag{3.11.7}

\ket{\Psi}=\sum_nc_n\ket{n},\qquad c_n=\braket{n}{\Psi}\tag{3.11.8}

The wavefunctions are just components: $\psi_n(x)=\braket{x}{n}$ .

In the energy basis, $\hat H$ is a diagonal matrix:

H=\begin{pmatrix}E_1&&&\\&E_2&&\\&&E_3&\\&&&\ddots\end{pmatrix}\tag{3.11.9}

Time evolution: three linesbasic~3 min

\ket{\Psi(t)}=\ee^{-\ii\hat Ht/\hbar}\ket{\Psi(0)} =\sum_nc_n\ee^{-\ii\hat Ht/\hbar}\ket{n} =\sum_nc_n\ee^{-\ii E_nt/\hbar}\ket{n}\tag{3.11.10}

Done. The third equality is section 3.4’s rule " $f(\hat A)\ket{a}=f(a)\ket{a}$ ".

Where does the sloshing of the probability density come from?

|\Psi(x,t)|^2=\Big|\sum_nc_n\ee^{-\ii E_nt/\hbar}\braket{x}{n}\Big|^2\tag{3.11.11}

Expand, and the cross terms carry a factor $\ee^{-\ii(E_n-E_m)t/\hbar}$ . In the $n=m$ terms the phases cancel (nothing changes with time); the $n\ne m$ terms oscillate at the Bohr frequencies $\omega_{nm}=(E_n-E_m)/\hbar$ .

Section 2.5 spent a whole section on “stationary states sit still, superpositions slosh”. Here it is two lines — and you can read off at a glance that the sloshing frequency is set entirely by the level spacing.

The infinite square well

Drag the width L and the quantum number n and watch the energy, the waveform and the probability distribution move together. Units: ħ = m = 1.

Energy Eₙ (n=1): 4.935
relative to E₁(L=1): 1.00 ×
⟨x⟩ / L: 0.500
Δx / L: 0.180

Well width L1.00

Shrink the box and every level is pushed up together: E ∝ 1/L²

Quantum number n1

nodes = n − 1 = 0 (endpoints excluded)

State

Show ψShow |ψ|²

Try this

Drag L from 2.5 down to 0.6 and keep your eye on the dashed line in the level diagram (pinned at the reference energy E₁(L=1)): it sinks all the way to the bottom, meaning every level has risen far above it. The "relative to E₁(L=1)" readout climbs from 0.16 to 2.78 — "the tighter you confine a particle, the more kinetic energy it has", a direct consequence of ΔxΔp ≥ ħ/2.
Change n in the stationary mode and note that |ψ|² never changes with time (the curve stands still); switch to a superposition, press play, and |ψ|² immediately starts sloshing. That is exactly what "stationary" refers to.
Leave only c₁ in the superposition (drag the rest to 0) and press play — the probability density stops moving again. However long a single eigenstate evolves, it only picks up an overall phase e^(−iEₙt/ħ), which no observable can see.
Push n above 10 and look at |ψ|²: the fringes get so fine that the distribution is nearly uniform — the classical picture of a particle equally likely to be anywhere in the box. The correspondence principle, in view.

This is an interactive simulation (infinite-well) and needs JavaScript to run. The derivations and conclusions in the text do not depend on it, so feel free to read on.

The harmonic oscillator: a purely algebraic solution

Section 2.9 already used ladder operators, but Dirac notation was not yet available then. Rewritten in the new language, the structure comes out far more clearly.

The complete algebraic solution, without solving a single differential equationadvanced~10 min

The only starting point: $[\hat x,\hat p]=\ii\hbar$ . Define

\hat a=\sqrt{\frac{m\omega}{2\hbar}}\left(\hat x+\frac{\ii\hat p}{m\omega}\right),\qquad \hat a^\dagger=\sqrt{\frac{m\omega}{2\hbar}}\left(\hat x-\frac{\ii\hat p}{m\omega}\right)\tag{3.11.12}

(Writing $\hat a^\dagger$ rather than $\hat a_\pm$ is the standard convention; note that $\hat a$ is not Hermitian — $\hat a^\dagger$ is its adjoint.)

Step one: the commutator.

[\hat a,\hat a^\dagger]=\frac{m\omega}{2\hbar}\cdot\frac{2\ii}{m\omega}[\hat p,\hat x]\cdot(-1)=1\tag{3.11.13}

Step two: the Hamiltonian. Define the number operator $\hat N=\hat a^\dagger\hat a$ ; then

\hat H=\hbar\omega\left(\hat N+\frac12\right)\tag{3.11.14}

So the problem becomes: find the eigenvalues of $\hat N$ .

Step three: the eigenvalues of $\hat N$ are non-negative. Let $\hat N\ket{\nu}=\nu\ket{\nu}$ with $\braket{\nu}{\nu}=1$ :

\nu=\bra{\nu}\hat a^\dagger\hat a\ket{\nu}=\big\|\hat a\ket{\nu}\big\|^2\ge0\tag{3.11.15}

Step four: the ladder property. Using $[\hat N,\hat a]=-\hat a$ and $[\hat N,\hat a^\dagger]=+\hat a^\dagger$ (both follow directly from $[\hat a,\hat a^\dagger]=1$ ):

\hat N\big(\hat a\ket{\nu}\big)=(\nu-1)\big(\hat a\ket{\nu}\big),\qquad \hat N\big(\hat a^\dagger\ket{\nu}\big)=(\nu+1)\big(\hat a^\dagger\ket{\nu}\big)\tag{3.11.16}

Step five: the ladder must have a bottom rung. Applying $\hat a$ repeatedly produces eigenvalues $\nu-1,\nu-2,\dots$ , but step three says eigenvalues are non-negative. The only way out is for the sequence to terminate somewhere — there must exist a state $\ket{0}$ with

\hat a\ket{0}=0\tag{3.11.17}

From $\hat N\ket{0}=\hat a^\dagger\hat a\ket{0}=0$ , the lowest eigenvalue is $\nu=0$ . The ladder property then gives all the eigenvalues:

\nu=n,\qquad n=0,1,2,\dots\tag{3.11.18}

(If some $\nu$ were not an integer, repeated lowering would step past 0 and produce a negative eigenvalue — a contradiction.)

Step six: the energy levels.

E_n=\hbar\omega\left(n+\frac12\right)\tag{3.11.19}

Step seven: the normalised ladder. From $\|\hat a^\dagger\ket{n}\|^2=\bra{n}\hat a\hat a^\dagger\ket{n}=\bra{n}(\hat N+1)\ket{n}=n+1$ :

\hat a^\dagger\ket{n}=\sqrt{n+1}\,\ket{n+1},\qquad \hat a\ket{n}=\sqrt{n}\,\ket{n-1}\tag{3.11.20}

\ket{n}=\frac{(\hat a^\dagger)^n}{\sqrt{n!}}\ket{0}\tag{3.11.21}

Every result came from a single commutation relation. No Hermite polynomial appeared anywhere, and no differential equation was ever solved.

Only if you want the wavefunctions do you return to the position basis: write $\hat a\ket{0}=0$ in $\braket{x}{\cdot}$ form and you get a first-order ordinary differential equation,

\left(x+\frac{\hbar}{m\omega}\frac{\dd}{\dd x}\right)\psi_0(x)=0 \ \Longrightarrow\ \psi_0\propto\ee^{-m\omega x^2/2\hbar}\tag{3.11.22}

A first-order equation — far easier than the original second-order one.

The one-dimensional harmonic oscillator

Equally spaced levels, zero-point energy, tails reaching into the classically forbidden region, and the one quantum state that really does behave like a classical particle. Units: ħ = m = 1.

Eₙ = (n+½)ħω: 0.500
Level spacing ħω: 1.00
Classical turning point ±A: ±1.00
Zero-point energy E₀: 0.500

Angular frequency ω1.00

A steeper well → larger spacing ħω and a narrower wavefunction

Quantum number n0

nodes = n = 0

View

Show ψ

Try this

In the "level ladder" view, push n up to 8: the levels stay exactly ħω apart. This is unique to the oscillator — look back at the infinite well, where the spacing grows as you climb.
Turn ω down to 0.4 and note that the ground-state energy E₀ = ħω/2 shrinks but never reaches zero. Making E₀ = 0 would require the particle to sit exactly at x = 0 with p = 0, and ΔxΔp ≥ ħ/2 forbids it.
Switch to "single eigenstate" with ψ shown and look at the n = 0 curve beyond the two dashed lines (the classical turning points): it has not gone to zero. The particle has a definite probability of being where classical energy conservation forbids — the very same root as tunnelling.
Show |ψ|², turn on "compare with classical", and take n from 0 to 20: the envelope of the quantum oscillation hugs the classical curve ever more closely (the ends are favoured because a classical particle moves slowest at the turning points).
Switch to "coherent state" and press play: the packet oscillates back and forth as a whole, keeps its shape, and its period is exactly the classical T = 2π/ω. Schrödinger found this most classical of quantum states back in 1926; it is also the theoretical description of laser light.

This is an interactive simulation (harmonic-oscillator) and needs JavaScript to run. The derivations and conclusions in the text do not depend on it, so feel free to read on.

Why this method matters so much

This algebra is not just a trick for the harmonic oscillator. It is a template, reused over and over throughout physics:

Angular momentum (chapter 5): from $[\hat L_x,\hat L_y]=\ii\hbar\hat L_z$ define $\hat L_\pm=\hat L_x\pm\ii\hat L_y$ , and the same ladder argument delivers all the allowed values of $l$ and $m$ — including half-integer spin, which the wavefunction method cannot produce at all.
Quantum field theory (section 8.5): $\hat a^\dagger$ becomes the operator that creates a particle, and $\ket{n}$ becomes “there are $n$ particles”. “The $n$ -th excited state” and ” $n$ particles” become one and the same sentence.
Coherent states: the eigenstates $\hat a\ket{\alpha}=\alpha\ket{\alpha}$ are the quantum description of laser light.

The algebraic method reaches far beyond where the differential-equation method can go. This is the most tangible profit of chapter 3.

The uncertainty principle: from derivation to substitution

The picture

Chapter 2

All it could say was “the rigorous proof needs the generalised uncertainty relation”, plus an intuitive Fourier-analysis origin.

For the infinite-well ground state specifically, it still had to grind through two integrals: $\langle x^2\rangle$ by two rounds of integration by parts, and $\langle p^2\rangle$ smuggled in via $2mE_1$ .

The mathematics

Chapter 3

\sigma_A\sigma_B\ge\frac12\big|\langle[\hat A,\hat B]\rangle\big|\tag{3.11.23}

Substitute $[\hat x,\hat p]=\ii\hbar$ :

\sigma_x\sigma_p\ge\frac{\hbar}{2}\tag{3.11.24}

One step. And the inequality holds for any pair of observables, not just $x$ and $p$ .

The oscillator ground state: σₓ and σ_p without a single integraladvanced~6 min

Invert the ladder-operator definitions:

\hat x=\sqrt{\frac{\hbar}{2m\omega}}(\hat a+\hat a^\dagger),\qquad \hat p=\ii\sqrt{\frac{m\omega\hbar}{2}}(\hat a^\dagger-\hat a)\tag{3.11.25}

First moments. On $\ket{0}$ , $\hat a\ket{0}=0$ and $\braket{0}{1}=0$ , so

\langle x\rangle=\bra{0}\hat x\ket{0}\propto\bra{0}(\hat a+\hat a^\dagger)\ket{0}=0+\braket{0}{1}=0\tag{3.11.26}

and likewise $\langle p\rangle=0$ .

Second moments.

\hat x^2=\frac{\hbar}{2m\omega}\big(\hat a^2+\hat a\hat a^\dagger+\hat a^\dagger\hat a+(\hat a^\dagger)^2\big)\tag{3.11.27}

Take the expectation value on $\ket{0}$ : the $\hat a^2$ and $(\hat a^\dagger)^2$ terms send $\ket{0}$ to $\ket{2}$ or annihilate it — orthogonal to $\bra{0}$ either way, so they contribute nothing; $\hat a^\dagger\hat a\ket{0}=0$ ; all that survives is $\hat a\hat a^\dagger\ket{0}=\ket{0}$ :

\langle x^2\rangle=\frac{\hbar}{2m\omega}\tag{3.11.28}

and in the same way

\langle p^2\rangle=\frac{m\omega\hbar}{2}\tag{3.11.29}

Multiply them together:

\sigma_x\sigma_p=\sqrt{\frac{\hbar}{2m\omega}}\cdot\sqrt{\frac{m\omega\hbar}{2}}=\frac{\hbar}{2}\tag{3.11.30}

The bound is saturated exactly. The oscillator ground state is a minimum-uncertainty state — section 2.10 claimed that Gaussian wavepackets achieve $\hbar/2$ exactly, and here is the proof, without doing a single integral.

The zero-point energy falls out for free:

E_0=\frac{\langle p^2\rangle}{2m}+\frac12m\omega^2\langle x^2\rangle =\frac{\hbar\omega}{4}+\frac{\hbar\omega}{4}=\frac{\hbar\omega}{2}\tag{3.11.31}

And you can see that kinetic and potential energy each contribute exactly half — the virial theorem, as it appears for the oscillator.

The Ehrenfest theorem: one line

The picture

Chapter 2’s approach

Differentiate $\langle x\rangle$ with respect to time, substitute the Schrödinger equation, integrate by parts twice, argue the boundary terms away… most of a page.

Then do it all again for $\langle p\rangle$ .

The mathematics

With the Heisenberg equation

\frac{\dd\langle A\rangle}{\dd t}=\frac{\ii}{\hbar}\big\langle[\hat H,\hat A]\big\rangle\tag{3.11.32}

Take $\hat A=\hat x$ and use $[\hat p^2,\hat x]=-2\ii\hbar\hat p$ :

\frac{\dd\langle x\rangle}{\dd t}=\frac{\langle p\rangle}{m}\tag{3.11.33}

Take $\hat A=\hat p$ and use $[V(\hat x),\hat p]=\ii\hbar V'(\hat x)$ :

\frac{\dd\langle p\rangle}{\dd t}=-\big\langle V'(\hat x)\big\rangle\tag{3.11.34}

Note that the right-hand side is still ⟨V′(x̂)⟩, not V′(⟨x⟩)

The new language shortens the derivation, but it changes none of the physics. The distinction that section 2.6 took pains over is still here:

\big\langle V'(\hat x)\big\rangle\ne V'\big(\langle x\rangle\big)\quad\text{in general}\tag{3.11.35}

Only when $V'$ is approximately linear across the width of the wavepacket are the two equal — and only then does the expectation value truly follow a classical trajectory.

This is a general lesson worth remembering: changing language can save you computation, but it cannot save you physics. If a result seems to “disappear” in the new language, you have most likely dropped it.

Settling the accounts

The picture

Pages saved

Result	Chapter 2	Chapter 3
Eigenvalues are real	Half a page of integrals	3 lines
Eigenstates are orthogonal	Verified system by system	4 lines, holds in general
Completeness of expansions	Taken on faith	Spectral theorem
Uncertainty principle	Intuition only	One substitution
Ehrenfest	Most of a page	2 lines
Harmonic oscillator	Power series + Hermite	Pure algebra
Wavepacket spreading	Gaussian integrals by completing the square	A few lines (section 3.9)

The mathematics

New powers bought

Not just saved pages — things chapter 2 simply could not do at all:

Spin: no wavefunction, only a two-dimensional state vector. All of chapter 5 rests on this.
Half-integer angular momentum: the algebraic method naturally yields $j=1/2,3/2,\dots$ ; the wavefunction method can only give integer $l$ .
Entanglement and mixed states: these need the density operator — the entire foundation of chapters 9 and 10.
Quantum field theory: $\hat a^\dagger$ turns directly into a creation operator.

Questions worth asking yourself before leaving this chapter

Chapter 2 said “the electron’s energy in the box is discrete”; chapter 3 says “the eigenvalues of $\hat H$ are discrete”. Are those the same statement?
Why do we say the wavefunction is not the fundamental object? Would switching to the momentum representation lose any physical content?
The measurement postulate and the evolution postulate have completely opposite characters. Why is this not a problem that finer calculation could make go away?
The oscillator solution used nothing but $[\hat x,\hat p]=\ii\hbar$ . Does that mean every system satisfying this commutation relation has equally spaced energy levels?

(The answer to question 4 is “no” — you also need $\hat H$ to happen to be a linear function of $\hat a^\dagger\hat a$ . The commutation relation fixes the algebraic structure; the Hamiltonian fixes the dynamics. Keeping those two apart is the mark of having finished this chapter.)

Key formulas

Evolution (energy basis)
$\ket{\Psi(t)}=\sum_nc_n\ee^{-\ii E_nt/\hbar}\ket{n}$
A whole section of chapter 2, in three lines
Ladder operators
$[\hat a,\hat a^\dagger]=1,\quad\hat H=\hbar\omega(\hat a^\dagger\hat a+\tfrac12)$
The entire oscillator hangs on this one relation
Normalised ladder
$\hat a^\dagger\ket{n}=\sqrt{n+1}\ket{n+1},\quad\hat a\ket{n}=\sqrt{n}\ket{n-1}$
The creation and annihilation operators of chapter 8 are exactly these
x̂ and p̂ in ladder form
$\hat x=\sqrt{\tfrac{\hbar}{2m\omega}}(\hat a+\hat a^\dagger),\ \hat p=\ii\sqrt{\tfrac{m\omega\hbar}{2}}(\hat a^\dagger-\hat a)$
Matrix elements without touching an integral
Ehrenfest (operator form)
$\frac{\dd\langle A\rangle}{\dd t}=\frac{\ii}{\hbar}\langle[\hat H,\hat A]\rangle$
Take Â = x̂ and p̂ to get one classical equation each

Self-check4 questions

End of chapter

The three questions chapter 2 left open now have answers:

Why do observables correspond to operators? Because demanding “measured values are real, distinct outcomes are distinguishable, every state can be expanded” picks out exactly the Hermitian operators (section 3.5).

By what right is $|c_n|^2$ a probability? That one remains a postulate (the Born rule). But the spectral theorem guarantees $\sum_n|c_n|^2=1$ , so it is at least self-consistent — and Gleason’s theorem goes further: in dimension $\ge3$ there is essentially no other choice.

What actually happens in a measurement? The projection postulate describes its effect but does not explain its mechanism (section 3.8). On this there is still no consensus — and it is not the sort of problem that computing more carefully can resolve.

With this language in hand, the pace can pick up considerably. Chapter 4 returns to one-dimensional problems — but this time with structural methods like symmetry and the transfer matrix, and with a numerical solver that handles arbitrary potentials.

Rewriting earlier chapters in Dirac notation

Recommended first

After this section you should be able to

The infinite square well, written as matrices

The picture

The mathematics

The infinite square well

The harmonic oscillator: a purely algebraic solution

The one-dimensional harmonic oscillator

The uncertainty principle: from derivation to substitution

The picture

The mathematics

The Ehrenfest theorem: one line

The picture

The mathematics

Settling the accounts

The picture

The mathematics

Key formulas

Self-check4 questions

End of chapter

The infinite square well, written as matrices

◑The picture

∑The mathematics

The harmonic oscillator: a purely algebraic solution

The uncertainty principle: from derivation to substitution

◑The picture

∑The mathematics

The Ehrenfest theorem: one line

◑The picture

∑The mathematics

Settling the accounts

◑The picture

∑The mathematics

∎Key formulas

End of chapter

The picture

The mathematics

The picture

The mathematics

The picture

The mathematics

The picture

The mathematics

Key formulas