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Dirac Notation

Source lecture(s): PHY621 Ch. 1

Intuition

Dirac's bra–ket notation is not a new theory — it is the same inner-product linear algebra written so that the manipulations become almost automatic. Its real virtue is that it makes basis-independence visible: a ket is the vector, and components appear only when you deliberately project.

\[|\psi\rangle \ \text{ket (vector)},\qquad \langle\phi| \ \text{bra (linear functional)},\qquad \langle\phi|\psi\rangle \ \text{inner product (number)}\]

The three objects

Object Type In components
$ \psi\rangle$ vector
$\langle\phi $ dual vector — a machine that eats a ket and returns a number
$\langle\phi \psi\rangle$ scalar
$ \psi\rangle\langle\phi $

That last row is the one that does the work.

Completeness, and why it is the whole trick

For an orthonormal basis \(\{|n\rangle\}\):

\[\sum_n |n\rangle\langle n| = \hat{1}\]

The resolution of the identity. You may insert it anywhere, for free, and every basis expansion in the course is one application of it:

\[|\psi\rangle = \hat{1}|\psi\rangle = \sum_n|n\rangle\underbrace{\langle n|\psi\rangle}_{\text{component }c_n}\]
\[\langle\phi|\hat{A}|\psi\rangle = \sum_{m,n}\langle\phi|m\rangle\langle m|\hat{A}|n\rangle\langle n|\psi\rangle = \sum_{m,n}\phi_m^*A_{mn}\psi_n\]

The second line derives "matrix element" from nothing but inserting the identity twice. Change of basis, matrix representation, Parseval's theorem and the Fourier expansion are all this move.

Continuous bases

The same machinery with sums replaced by integrals:

\[\int dx\,|x\rangle\langle x| = \hat{1},\qquad \langle x|x'\rangle = \delta(x - x'),\qquad \psi(x) = \langle x|\psi\rangle\]

The wavefunction is a component, not the state — the state is \(|\psi\rangle\), and \(\psi(x)\) is its component in the position basis, exactly as \(\tilde\psi(k) = \langle k|\psi\rangle\) is its component in the momentum basis. The Fourier transform then falls out of inserting the identity:

\[\psi(x) = \langle x|\psi\rangle = \int dk\,\langle x|k\rangle\langle k|\psi\rangle = \frac{1}{\sqrt{2\pi}}\int dk\,e^{ikx}\tilde\psi(k)\]

The Fourier transform is a change of basis. Not an analogy — the same operation as rotating coordinates in \(\mathbb{R}^3\), in an infinite-dimensional space with a continuous basis label. Seeing that once makes the rest of the course cohere.

Projection operators

\(\hat{P}_n = |n\rangle\langle n|\) satisfies \(\hat{P}_n^2 = \hat{P}_n\) (projecting twice does nothing new) and picks out a component. In quantum mechanics measurement is projection; in data analysis the same operator is principal-component analysis; in normal modes it extracts a mode amplitude.

Common mistakes

  • Treating a bra as "the vector written sideways". It is a different object — an element of the dual space — and the correspondence involves complex conjugation.
  • Forgetting that \(\delta(x-x')\) is not a function. \(|x\rangle\) is not normalisable and lives in a rigged Hilbert space. Physicists use it anyway, correctly, and mathematicians have supplied the justification.
  • Confusing \(|\psi\rangle\langle\phi|\) with \(\langle\phi|\psi\rangle\). One is an operator, the other a number. The order is the whole difference.

Knowledge graph position

Prerequisites: vector spaces, inner product spaces. Leads to: quantum mechanics, transform methods, Green's functions.

Quiz

Q1 (conceptual). What does inserting \(\sum_n|n\rangle\langle n|\) accomplish?

Answer

It is the identity, so it changes nothing mathematically — but it introduces a basis, turning an abstract statement into components. Every basis expansion, matrix representation and change of basis in the course is one insertion of the identity.

Q2 (conceptual). In what sense is the Fourier transform a change of basis?

Answer

\(\psi(x) = \langle x|\psi\rangle\) and \(\tilde\psi(k) = \langle k|\psi\rangle\) are components of the same state in two different bases. Inserting \(\int dk|k\rangle\langle k|\) and using \(\langle x|k\rangle = e^{ikx}/\sqrt{2\pi}\) produces the transform. It is a rotation in an infinite-dimensional space.

Q3 (MCQ). \(|\psi\rangle\langle\psi|\) (for normalised \(|\psi\rangle\)) is:

  • (a) a scalar equal to 1
  • (b) a projection operator onto the direction of \(|\psi\rangle\)
  • (c) the norm of the state
  • (d) undefined
Answer

(b). It is an outer product, hence an operator, and it satisfies \(P^2 = P\). The scalar \(\langle\psi|\psi\rangle = 1\) is the inner product — reversing the order changes the object entirely.