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3.2

The complete rules of Dirac notation

A notation that compresses "vectors, inner products, projections, changes of basis" into a few angle brackets. Learn it, and the derivations of chapter 2 shrink to a fraction of their length.

Recommended first

After this section you should be able to

  • Read and write kets, bras, inner products, outer products and matrix elements of operators correctly
  • Use the completeness relation Σ|n⟩⟨n| = 1 to convert between any two representations
  • Explain the geometric meaning of the projection operator |ψ⟩⟨ψ|

The previous section established that the state is a vector and the wavefunction its components. This section equips that idea with notation that fits it like a glove.

The ket: the state itself

A quantum state is written

∣ψ⟩(3.2.1)\ket{\psi}\tag{3.2.1}

read “ket psi”. It stands for the state itself, carrying no basis information whatsoever — like the arrow of the previous section, not the (3,4,0)(3,4,0).

The bra: the half you use for inner products

Every ket ∣ψ⟩\ket{\psi} comes with a bra ⟨ψ∣\bra{\psi}. Snap the two together and you get the inner product:

⟨ϕ|ψ⟩=∫ϕ∗(x)ψ(x) dx(3.2.3)\braket{\phi}{\psi}=\int\phi^*(x)\psi(x)\,\dd x\tag{3.2.3}

(The word “bra-ket” comes from “bracket” — Dirac split it down the middle.)

The picture

Three properties of the inner product

⟨ϕ|ψ⟩=⟨ψ|ϕ⟩∗(3.2.4)\braket{\phi}{\psi}=\braket{\psi}{\phi}^*\tag{3.2.4}

Swapping the order takes the complex conjugate — because ∫ϕ∗ψ\int\phi^*\psi and ∫ψ∗ϕ\int\psi^*\phi are conjugates of each other.

⟨ψ|ψ⟩≥0,and =0  ⟺  ∣ψ⟩=0(3.2.5)\braket{\psi}{\psi}\ge0,\quad\text{and}\ =0\iff\ket{\psi}=0\tag{3.2.5}

The squared length is non-negative. A normalised state satisfies ⟨ψ|ψ⟩=1\braket{\psi}{\psi}=1.

⟨ϕ∣(a∣ψ1⟩+b∣ψ2⟩)=a⟨ϕ|ψ1⟩+b⟨ϕ|ψ2⟩(3.2.6)\bra{\phi}\big(a\ket{\psi_1}+b\ket{\psi_2}\big)=a\braket{\phi}{\psi_1}+b\braket{\phi}{\psi_2}\tag{3.2.6}

Linear in the right slot, antilinear in the left (coefficients pulled out of the bra pick up a conjugate).

The mathematics

Chapter 2’s formulas, rewritten

Normalisation:

∫∣Ψ∣2dx=1 ⟶ ⟨Ψ|Ψ⟩=1(3.2.7)\int|\Psi|^2\dd x=1 \ \longrightarrow\ \braket{\Psi}{\Psi}=1\tag{3.2.7}

Expansion coefficients:

cn=∫ψn∗Ψ dx ⟶ cn=⟨n|Ψ⟩(3.2.8)c_n=\int\psi_n^*\Psi\,\dd x \ \longrightarrow\ c_n=\braket{n}{\Psi}\tag{3.2.8}

Orthonormality:

∫ψm∗ψndx=δmn ⟶ ⟨m|n⟩=δmn(3.2.9)\int\psi_m^*\psi_n\dd x=\delta_{mn} \ \longrightarrow\ \braket{m}{n}=\delta_{mn}\tag{3.2.9}

Expectation value:

∫Ψ∗A^Ψ dx ⟶ ⟨Ψ∣A^∣Ψ⟩(3.2.10)\int\Psi^*\hat A\Psi\,\dd x \ \longrightarrow\ \bra{\Psi}\hat A\ket{\Psi}\tag{3.2.10}

An operator sandwiched in the middle

A^∣ψ⟩\hat A\ket{\psi} is another ket. Take its inner product with ⟨ϕ∣\bra{\phi} and you get a matrix element:

⟨ϕ∣A^∣ψ⟩(3.2.11)\bra{\phi}\hat A\ket{\psi}\tag{3.2.11}

The deliberate ambiguity of this notation is a feature: you may read it as “the inner product of ⟨ϕ∣\bra{\phi} with (A^∣ψ⟩)(\hat A\ket{\psi})”, or as “the inner product of (⟨ϕ∣A^)(\bra{\phi}\hat A) with ∣ψ⟩\ket{\psi}”. Both readings give the same number, so no parentheses are needed.

That is where the name matrix element comes from: in a basis {∣n⟩}\{\ket{n}\}, Amn=⟨m∣A^∣n⟩A_{mn}=\bra{m}\hat A\ket{n} is precisely the matrix representation of the operator.

The outer product: snapped together the other way round

Put the ket on the left and the bra on the right:

∣ϕ⟩⟨ψ∣(3.2.12)\ket{\phi}\bra{\psi}\tag{3.2.12}

This is not a number but an operator — acting on any ket, it returns a ket:

(∣ϕ⟩⟨ψ∣)∣χ⟩=∣ϕ⟩⟨ψ|χ⟩⏟a number(3.2.13)\big(\ket{\phi}\bra{\psi}\big)\ket{\chi}=\ket{\phi}\underbrace{\braket{\psi}{\chi}}_{\text{a number}}\tag{3.2.13}

The completeness relation: the most useful identity in the chapter

Let {∣n⟩}\{\ket{n}\} be an orthonormal, complete basis. “Complete” means any state can be expanded:

∣ψ⟩=∑ncn∣n⟩,cn=⟨n|ψ⟩(3.2.18)\ket{\psi}=\sum_n c_n\ket{n},\qquad c_n=\braket{n}{\psi}\tag{3.2.18}

Substitute the cnc_n back in:

∣ψ⟩=∑n∣n⟩⟨n|ψ⟩=(∑n∣n⟩⟨n∣)∣ψ⟩(3.2.19)\ket{\psi}=\sum_n \ket{n}\braket{n}{\psi}=\left(\sum_n\ket{n}\bra{n}\right)\ket{\psi}\tag{3.2.19}

This holds for every ∣ψ⟩\ket{\psi}, so the operator in parentheses is the identity:

 ∑n∣n⟩⟨n∣=1^ (3.2.20)\boxed{\ \sum_n\ket{n}\bra{n}=\hat 1\ }\tag{3.2.20}

The continuous-basis version swaps the sum for an integral:

∫∣x⟩⟨x∣ dx=1^,∫∣p⟩⟨p∣ dp=1^(3.2.21)\int\ket{x}\bra{x}\,\dd x=\hat 1,\qquad \int\ket{p}\bra{p}\,\dd p=\hat 1\tag{3.2.21}

Key formulas

  • State and components

    ψ(x)=⟨x|ψ⟩,cn=⟨n|ψ⟩\psi(x)=\braket{x}{\psi},\quad c_n=\braket{n}{\psi}

    Whichever components you want, dot the state with that basis

  • Inner product

    ⟨ϕ|ψ⟩=⟨ψ|ϕ⟩∗,⟨ψ|ψ⟩=1\braket{\phi}{\psi}=\braket{\psi}{\phi}^*,\quad\braket{\psi}{\psi}=1

    Linear on the right, antilinear on the left

  • Projection operator

    P^ψ=∣ψ⟩⟨ψ∣,P^2=P^=P^†\hat P_\psi=\ket{\psi}\bra{\psi},\qquad \hat P^2=\hat P=\hat P^\dagger

    Idempotent + Hermitian: the algebraic definition of projection

  • Completeness relation

    ∑n∣n⟩⟨n∣=1^,∫∣x⟩⟨x∣dx=1^\sum_n\ket{n}\bra{n}=\hat 1,\qquad \int\ket{x}\bra{x}\dd x=\hat 1

    To change basis, insert a 1

  • Change-of-basis matrix element

    ⟨x|p⟩=12πℏeipx/ℏ\braket{x}{p}=\frac{1}{\sqrt{2\pi\hbar}}\ee^{\ii px/\hbar}

    Fourier transform = change of basis between position and momentum

Self-check4 questions

  1. 1.

    Which of the following equations is correct?

  2. 2.

    What is |φ⟩⟨ψ|?

  3. 3.

    The main use of the completeness relation Σₙ|n⟩⟨n| = 1 is:

  4. 4.

    For the projection operator P̂ = |ψ⟩⟨ψ| (with |ψ⟩ normalised), which statements are correct? (Select all that apply.)

    Select all that apply

What comes next

The notation is in hand. But a few mathematical questions have been quietly stepped around: in what sense does ∑ncn∣n⟩\sum_n c_n\ket{n} converge? ∣x⟩\ket{x} cannot be normalised — what entitles it to serve as a basis at all?

The next section settles these — the only mathematics-first section in the chapter, and it will stick as closely as possible to the parts that actually matter for the physics.

Section 21 of 106 · use ← → to turn the page