Part 1 · Mathematical tools

1.4 Hermitian operators and Dirac notation

Source: Centini’s whiteboards of 2 Oct, p. 6–9, and 6 Oct, p. 1–2

Objectives

Understand $\int f^{*} g d^{3} r$ as the dot product of two functions.
Understand why observables are Hermitian: real eigenvalues, orthogonal eigenfunctions, a complete basis.
Recipe: expand a state in a basis, $b_{n} = ⟨ u_{n} ∣ ψ ⟩$ .
Picture: components as projections that change with the basis while the length does not.
Theorem: the three properties of a Hermitian operator; completeness, $\sum_{n} ∣ n ⟩ ⟨ n ∣ = \hat{I}$ ; $\overset{p}{^}_{x}$ is Hermitian.
Vocabulary: orthonormal basis, Kronecker delta, Hilbert space, bra, ket, projector, Hermitian adjoint.
Essential vocabulary: Hermitian operator.

Hermitian operators

Definition 1.4.1Hermitian operatorEssential

$\hat{A}$ is Hermitian if, for every pair of functions $f$ , $g$ , $\int f^{*} (\overset{r}{ˉ}) \hat{A} g (\overset{r}{ˉ}) d^{3} r = \int [\hat{A} f (\overset{r}{ˉ})]^{*} g (\overset{r}{ˉ}) d^{3} r .$

It makes no difference whether $\hat{A}$ acts on the right-hand function or, conjugated, on the left-hand one. Hermitian operators are the ones that represent physical quantities (6 Oct), and this is why:

Theorem 1.4.2Properties of a Hermitian operator

Its eigenvalues are real.
Eigenfunctions with different eigenvalues are orthogonal: $\int u_{n}^{*} u_{m} d^{3} r = 0$ for $a_{n} \neq = a_{m}$ .
Its eigenfunctions are a complete orthogonal basis: any wavefunction is a sum of them.

The board states 1 and 2; the proofs are beyond the boards.

Proof of 1 and 2beyond the boards

Let $\hat{A} u = a u$ and take $f = g = u$ . The left side is $\int u^{*} \hat{A} u = a \int ∣ u ∣^{2}$ ; the right side is $\int (a u)^{*} u = a^{*} \int ∣ u ∣^{2}$ , since conjugating $a u$ conjugates $a$ . So $(a - a^{*}) \int ∣ u ∣^{2} = 0$ , and the integral is positive: $a = a^{*}$ .
Let $\hat{A} u_{n} = a_{n} u_{n}$ , $\hat{A} u_{m} = a_{m} u_{m}$ , and take $f = u_{n}$ , $g = u_{m}$ . The left side is $a_{m} \int u_{n}^{*} u_{m}$ ; the right side is $a_{n}^{*} \int u_{n}^{*} u_{m} = a_{n} \int u_{n}^{*} u_{m}$ , by 1. Subtracting, $(a_{m} - a_{n}) \int u_{n}^{*} u_{m} = 0$ , and $a_{m} \neq = a_{n}$ .

A measurement gives a real number, and that is property 1. Properties 2 and 3 are what make probabilities work (Theorem 2.2.4).

Functions as vectors

Two vectors are perpendicular when their dot product vanishes, $\overset{v}{ˉ} \cdot \overset{w}{ˉ} = ∣ \overset{v}{ˉ} ∣∣ \overset{w}{ˉ} ∣ cos θ = 0$ at $θ = π /2$ .

Key observation 1.4.3The dot product of two functions

$⟨ f ∣ g ⟩ = \int f^{*} (\overset{r}{ˉ}) g (\overset{r}{ˉ}) d^{3} r$ is the dot product of two functions: multiply them point by point, conjugating the first, and add up. Orthogonal functions are those for which it is zero.

The unit vectors $\overset{x}{^}, \overset{y}{^}, \overset{z}{^}$ are a basis: $\overset{ˉ}{V} = V_{x} \overset{x}{^} + V_{y} \overset{y}{^} + V_{z} \overset{z}{^} = \sum_{i} V_{i} \overset{u}{^}_{i}$ . The same for functions:

Definition 1.4.4Orthonormal basis

Functions $u_{n}$ are an orthonormal basis if each is normalised and they are pairwise orthogonal, $⟨ u_{n} ∣ u_{m} ⟩ = δ_{nm} = {10 n = m n \neq = m$ ( $δ_{nm}$ is the Kronecker delta), and complete: every allowed $ψ$ is $ψ = \sum_{n} b_{n} u_{n}$ .

The difference from 3D is that there are infinitely many $u_{n}$ . ==The eigenfunctions of a Hermitian operator are a basis of an infinite-dimensional space of functions, the Hilbert space==, whose members are the functions of finite length, $\int f^{*} f d^{3} r < \infty$ (square-integrable). A physical wavefunction must be one, since $∣ ψ ∣^{2}$ is a probability density that adds up to 1.

Expanding a function in the basis

Recipe 1.4.5Components in a basis

In an orthonormal basis ${u_{n}}$ , $ψ = n \sum b_{n} u_{n}, b_{n} = ⟨ u_{n} ∣ ψ ⟩ = \int u_{n}^{*} ψ d^{3} r .$

The board’s 3D analogy: $\overset{ˉ}{V} = 3 \overset{u}{^}_{x} + 4 \overset{u}{^}_{y} + 6 \overset{u}{^}_{z}$ has components $\overset{u}{^}_{x} \cdot \overset{ˉ}{V} = 3$ , $\overset{u}{^}_{y} \cdot \overset{ˉ}{V} = 4$ , $\overset{u}{^}_{z} \cdot \overset{ˉ}{V} = 6$ . $b_{n}$ is the same thing: how much of $u_{n}$ there is in $ψ$ .

Whybeyond the boards

Multiply $ψ = \sum_{n} b_{n} u_{n}$ by $u_{m}^{*}$ and integrate. Orthonormality kills every term but one: $\int u_{m}^{*} ψ = \sum_{n} b_{n} \int u_{m}^{*} u_{n} = \sum_{n} b_{n} δ_{mn} = b_{m}$ .

Loading the demo…

Figure 1.4.1 A state

ψ

with real components, in two dimensions, and an orthonormal basis

{u_{1}, u_{2}}

. Drag

ψ

, and drag

u_{1}

to turn the basis. The components

b_{n}

are the projections; they change as the basis turns, but

b_{1}^{2} + b_{2}^{2} = ⟨ ψ ∣ ψ ⟩

does not. Put

ψ

on the unit circle: the squares become the probabilities of the next chapter.

Dirac notation

Definition 1.4.6Bra, ket, bracket

$∣ g ⟩$ , a ket, is a function; $⟨ f ∣$ , a bra, is a function that will be conjugated.
$⟨ f ∣ g ⟩ = \int f^{*} g d^{3} r$ , the bracket, is a number; so is $⟨ ψ ∣ ψ ⟩$ .
$\hat{A} ∣ g ⟩$ is another ket, and $⟨ g ∣ \hat{A} ∣ f ⟩ = \int g^{*} \hat{A} f d^{3} r$ is a number: first $\hat{A}$ acts on $f$ , then the dot product with $g$ .

Two rules follow from the integral (beyond the boards): swapping conjugates, $⟨ f ∣ g ⟩^{*} = ⟨ g ∣ f ⟩$ ; and a constant comes out of a bra conjugated, $⟨ c f ∣ g ⟩ = c^{*} ⟨ f ∣ g ⟩$ , but out of a ket as it is, $⟨ f ∣ c g ⟩ = c ⟨ f ∣ g ⟩$ .

A ket before a bra is an operator. $∣ f ⟩ ⟨ g ∣$ , fed $∣ ψ ⟩$ , returns $∣ f ⟩$ times the number $⟨ g ∣ ψ ⟩$ . With $∣ f ⟩$ normalised, $∣ f ⟩ ⟨ f ∣$ is a projector: it keeps the component of $∣ ψ ⟩$ along $∣ f ⟩$ and throws the rest away, like a shadow on an axis (beyond the boards). In the figure, $b_{1} u_{1} = ∣ u_{1} ⟩ ⟨ u_{1} ∣ ψ ⟩$ .

Theorem 1.4.7Completeness

For an orthonormal basis $∣ n ⟩ = u_{n}$ , $n \sum ∣ n ⟩ ⟨ n ∣ = \hat{I} .$

Proof

$∣ ψ ⟩ = \sum_{n} b_{n} ∣ n ⟩ = \sum_{n} ⟨ n ∣ ψ ⟩ ∣ n ⟩ = \sum_{n} ∣ n ⟩ ⟨ n ∣ ψ ⟩ = (\sum_{n} ∣ n ⟩ ⟨ n ∣) ∣ ψ ⟩$ : the number $⟨ n ∣ ψ ⟩$ can be moved behind $∣ n ⟩$ . The bracket returns every $∣ ψ ⟩$ unchanged, so it is the identity.

The projections onto all the basis states add up to everything. In 3D, $\overset{x}{^} \overset{x}{^}^{T} + \overset{y}{^} \overset{y}{^}^{T} + \overset{z}{^} \overset{z}{^}^{T}$ is the $3 \times 3$ identity (beyond the boards).

Warning

Order matters. In general $\hat{A} \hat{B} ∣ f ⟩ \neq = \hat{B} \hat{A} ∣ f ⟩$ . With $\hat{A} = x$ and $\hat{B} = d / d x$ on $f = x$ : $\hat{A} \hat{B} f = x$ , but $\hat{B} \hat{A} f = \frac{d}{d x} x^{2} = 2 x$ (beyond the boards). Whether two operators commute decides whether two quantities can be known together (Definition 2.3.1).

The Hermitian adjoint

In Dirac notation the definition of Hermitian reads $⟨ f ∣ \hat{A} ∣ g ⟩ = ⟨ \hat{A} f ∣ g ⟩$ : $\hat{A}$ can act to the right, on $g$ , or to the left, on $f$ , with the same result.

Definition 1.4.8Hermitian adjoint

For any $\hat{A}$ , the Hermitian adjoint $\hat{A}^{+}$ is the operator that does $\hat{A}$ 's job from the other side of the bracket: $\int f^{*} \hat{A} g d^{3} r = \int [\hat{A}^{+} f]^{*} g d^{3} r .$ $\hat{A}$ is Hermitian exactly when $\hat{A}^{+} = \hat{A}$ .

Example 1.4.9

\hat{G} = \hat{A} + i \hat{B}

On the board: for Hermitian $\hat{A}$ and $\hat{B}$ , $\hat{G} = \hat{A} + i \hat{B}$ has $\hat{G}^{+} = \hat{A} - i \hat{B}$ . Why (beyond the boards): a constant moved out of the bra is conjugated, $(c \hat{B})^{+} = c^{*} \hat{B}^{+}$ , so $(\hat{A} + i \hat{B})^{+} = \hat{A} - i \hat{B}$ . Multiplying a Hermitian operator by $i$ breaks Hermiticity; multiplying by a real number does not.

For matrices (beyond the boards), $\hat{A}^{+}$ is the conjugate transpose, $(\hat{A}^{+})_{ij} = A_{ji}^{*}$ ; a real symmetric matrix, like $(2112)$ , is Hermitian.

Theorem 1.4.10Momentum is Hermitian

$\overset{p}{^}_{x} = - i ℏ \partial_{x}$ is Hermitian on functions that vanish at $\pm \infty$ .

Proof

In 1D, with $\frac{\partial g}{\partial x} d x = d g$ and the product rule $f^{*} d g = d (f^{*} g) - g d f^{*}$ : $\int f^{*} (- i ℏ \frac{\partial g}{\partial x}) d x = - i ℏ [f^{*} g]_{- \infty}^{+ \infty} + i ℏ \int \frac{\partial f ^{*}}{\partial x} g d x .$ ==The boundary term vanishes==, because a physical wavefunction goes to zero at $\pm \infty$ . What remains is $\int (- i ℏ \partial_{x} f)^{*} g d x$ , since $(- i)^{*} = i$ .

Remark

Why the $- i$ . The same integration by parts without $- i ℏ$ gives $\int f^{*} g^{'} = - \int (f^{'})^{*} g$ : a bare derivative is anti-Hermitian, $(d / d x)^{+} = - d / d x$ . The factor $- i$ , conjugated to $+ i$ , fixes the sign. A real momentum needs it. (Beyond the boards.)

Exercises

Exercise 1.4.1Sample questions, Q2 (ii)

Discuss the main properties of a Hermitian operator.

Solution

Definition: $\int f^{*} \hat{A} g = \int (\hat{A} f)^{*} g$ , i.e. $\hat{A}^{+} = \hat{A}$ . Real eigenvalues; orthogonal eigenfunctions for different eigenvalues; the eigenfunctions are a complete orthonormal basis of Hilbert space (Theorem 1.4.2, with the proofs of the first two). Physical meaning: Hermitian operators represent observables, whose measured values are real (Theorem 2.2.4).

Exercise 1.4.2beyond the boards

Is $\overset{x}{^}$ , multiplication by $x$ , Hermitian?

Solution

$\int f^{*} (xg) d x = \int (x f)^{*} g d x$ , because $x$ is real. Yes.

Exercise 1.4.3beyond the boards

Show that $d^{2} / d x^{2}$ is Hermitian on functions vanishing at $\pm \infty$ .

Solution

Integrate by parts twice: $\int f^{*} g^{''} = [f^{*} g^{'}] - \int f^{*'} g^{'} = - [f^{*'} g] + \int f^{*''} g = \int (f^{''})^{*} g$ . The sign flip of $d / d x$ happens twice and cancels. So the kinetic energy $- \frac{ℏ ^{2}}{2 m} \frac{d ^{2}}{d x ^{2}}$ is Hermitian.

Exercise 1.4.4beyond the boards

Show that $⟨ f ∣ \hat{A}^{+} \hat{A} ∣ f ⟩ \geq 0$ for any $\hat{A}$ , and that $\hat{A}^{+} \hat{A}$ is Hermitian.

Solution

$⟨ f ∣ \hat{A}^{+} \hat{A} ∣ f ⟩ = ⟨ \hat{A} f ∣ \hat{A} f ⟩ = \int ∣ \hat{A} f ∣^{2} \geq 0$ , moving $\hat{A}^{+}$ to the bra as $\hat{A}$ . And $(\hat{A}^{+} \hat{A})^{+} = \hat{A}^{+} (\hat{A}^{+})^{+} = \hat{A}^{+} \hat{A}$ . The oscillator’s $\hat{N} = \overset{a}{^}^{+} \overset{a}{^}$ is of this kind (Definition 3.5.5).