Part 1 · Mathematical tools

1.3 Operators and eigenvalues

Source: Centini’s whiteboard of 2 Oct, p. 5–6; the 3×3 eigenvalue problem and diagonalisation from the 2024/25 board #1

Objectives

Understand an operator as a machine that turns a vector, or a function, into another.
Understand an eigenvector as one the operator only rescales.
Recipe: eigenvalues from $det (\hat{A} - λ \hat{I}) = 0$ , then eigenvectors, then the diagonal form.
Picture: hunting for the directions an operator does not turn.
Theorem: the eigenfunctions of $\overset{p}{^}_{x} = - i ℏ \partial_{x}$ are the plane waves $e^{i p x /ℏ}$ .
Vocabulary: linear operator, characteristic equation, diagonalisation, Planck’s constant.
Essential vocabulary: eigenvalue, eigenfunction.

In quantum mechanics every measurable quantity is an operator, and the values a measurement can give are its eigenvalues (Theorem 2.2.4). This section is the mathematics of that sentence.

What an operator is

Definition 1.3.1Operator

An operator $\hat{O}$ takes an object, a vector or a function, and returns another of the same kind: $\hat{O} f = g$ .

On vectors of the plane, $\overset{c}{ˉ} = c_{x} \overset{u}{^}_{x} + c_{y} \overset{u}{^}_{y}$ , an operator is a matrix, applied row by column: $\hat{A} \overset{c}{ˉ} = (a_{11} a_{21} a_{12} a_{22}) (c_{x} c_{y}) = (a_{11} c_{x} + a_{12} c_{y} a_{21} c_{x} + a_{22} c_{y}) .$ With numbers (2024/25 board #1): $(2435) (13) = (1119)$ . A matrix moves every arrow of the plane: $(01 - 1 0)$ turns each by $9 0^{\circ}$ , $(2001)$ stretches along $x$ only (beyond the boards).

On functions, the board’s example is $\hat{A} = 1 + \frac{\partial}{\partial x}, \hat{A} f (x) = f (x) + \frac{\partial f}{\partial x} .$ It sends $x^{2}$ to $x^{2} + 2 x$ , $sin x$ to $sin x + cos x$ , and $e^{- x}$ to $0$ (beyond the boards).

Definition 1.3.2Linear operator

$\hat{A}$ is linear if constants come out and sums split: $\hat{A} [c f] = c \hat{A} f, \hat{A} [f + g] = \hat{A} f + \hat{A} g .$

Matrices and derivatives are linear, and so is every operator of quantum mechanics. $f \mapsto f^{2}$ is not: $(2 f)^{2} = 4 f^{2} \neq = 2 f^{2}$ . Linearity is what lets an operator act on each piece of a superposition separately (Expanding a function in the basis).

Eigenvalues and eigenfunctions

Definition 1.3.3Eigenvalue and eigenfunctionEssential

If $\hat{A} u_{n} = a_{n} u_{n}$ with $u_{n} \neq = 0$ and $a_{n}$ a number, $u_{n}$ is an eigenvector of $\hat{A}$ (an eigenfunction, in a space of functions) and $a_{n}$ is its eigenvalue.

==The operator does not change $u_{n}$ 's direction or shape; it only scales it by $a_{n}$ .== Most vectors are turned into something else; eigenvectors come back on their own line.

Loading the demo…

Figure 1.3.1 Hunt for the eigenvectors. Turn

x

and watch

\hat{A} x

; when the two lie on one line,

x

snaps and the

orange

line stays. For the Hermitian matrices you will find two lines at right angles. The non-Hermitian ones give lines at an angle, or only one; the rotation gives none.

Example 1.3.4A symmetric matrixbeyond the boards

$\hat{A} = (2112)$ (the first matrix of the figure).

$\hat{A} (1, 1) = (3, 3) = 3 (1, 1)$ : eigenvalue $3$ .
$\hat{A} (1, - 1) = (1, - 1)$ : eigenvalue $1$ .
$\hat{A} (1, 0) = (2, 1)$ is not a multiple of $(1, 0)$ : not an eigenvector.

The two eigenvectors are perpendicular and the eigenvalues real. That is no accident: the matrix is Hermitian (Theorem 1.4.2).

Example 1.3.5An eigenfunctionbeyond the boards

$\hat{A} = 1 + \partial_{x}$ on $e^{αx}$ gives $(1 + α) e^{αx}$ : every exponential is an eigenfunction, with eigenvalue $1 + α$ . The $e^{- x}$ above has eigenvalue $0$ .

Finding them: the characteristic equation

Write $λ \overset{a}{ˉ}$ as $λ \hat{I} \overset{a}{ˉ}$ with the identity matrix and move it across: $\hat{A} \overset{a}{ˉ} = λ \overset{a}{ˉ}$ becomes $(\hat{A} - λ \hat{I}) \overset{a}{ˉ} = 0$ . A non-zero $\overset{a}{ˉ}$ is sent to zero, so $\hat{A} - λ \hat{I}$ squashes a direction flat, and ==a matrix that flattens a direction has determinant zero== (the determinant is the factor by which it scales volumes) (beyond the boards).

Recipe 1.3.6Eigenvalues, eigenvectors, diagonal form2024/25 board #1

Solve the characteristic equation $det (\hat{A} - λ \hat{I}) = 0$ : subtract $λ$ from the diagonal and expand. An $n \times n$ matrix gives a polynomial of degree $n$ , whose roots are the eigenvalues.
For each root, solve $(\hat{A} - λ \hat{I}) \overset{a}{ˉ} = 0$ . One equation is always redundant (that is what $det = 0$ says), so one number stays free: an eigenvalue fixes a direction, not a length.
Normalise: $\overset{a}{^} = \overset{a}{ˉ} /∣ \overset{a}{ˉ} ∣$ .
With the eigenvectors as columns of $\hat{P}$ , $\hat{D} = \hat{P}^{- 1} \hat{A} \hat{P}$ is diagonal, with the eigenvalues on the diagonal.

Example 1.3.7The board’s 3×3 matrix2024/25 board #1

$\hat{A} = 112 - 1 21 013 .$ Picture first. $\hat{A} (1, 1, 0) = (0, 3, 3)$ points somewhere else: $(1, 1, 0)$ is not an eigenvector.

Eigenvalues. Expanding along the first row, $(1 - λ) [(2 - λ) (3 - λ) - 1] + [(3 - λ) - 2] = (1 - λ) [λ^{2} - 5 λ + 6] = (1 - λ) (λ - 2) (λ - 3),$ since the second bracket is $1 - λ$ and factors out. So $λ_{1} = 1$ , $λ_{2} = 2$ , $λ_{3} = 3$ .

Eigenvector of $λ_{1} = 1$ . $\hat{A} (x, y, z) = (x, y, z)$ gives $y = 0$ and $x + z = 0$ twice over. Choose $x = - 1$ : $\overset{a}{ˉ}_{1} = (- 1, 0, 1)$ , normalised $\overset{a}{^}_{1} = \frac{1}{2} (- 1, 0, 1)$ . Any multiple, like $(- 2, 0, 2)$ , would do as well. (The board ends its check with $λ_{1} \overset{a}{ˉ}_{1}$ where it means $λ_{1} \overset{a}{^}_{1}$ .)

The other two (beyond the boards): $\overset{a}{ˉ}_{2} = (1, - 1, - 1)$ , with $\hat{A} \overset{a}{ˉ}_{2} = (2, - 2, - 2)$ ; and $\overset{a}{ˉ}_{3} = (1, - 2, - 3)$ , with $\hat{A} \overset{a}{ˉ}_{3} = (3, - 6, - 9)$ .

Diagonal form. In the frame of $\overset{a}{ˉ}_{1}, \overset{a}{ˉ}_{2}, \overset{a}{ˉ}_{3}$ the operator only stretches: $\hat{D} = 100020003 = \hat{P}^{- 1} \hat{A} \hat{P}, \hat{P} = - 1 01 1 - 1 - 1 1 - 2 - 3 .$

Warning

That frame is skewed: $\overset{a}{ˉ}_{1} \cdot \overset{a}{ˉ}_{2} = - 2 \neq = 0$ . This $\hat{A}$ is not Hermitian, not even symmetric ( $A_{12} = - 1$ , $A_{21} = 1$ ), so nothing promised perpendicular eigenvectors, or even real eigenvalues. The matrices of quantum mechanics are Hermitian, and for them the frame is always orthonormal (Theorem 1.4.2).

Diagonalising means finding the frame in which an operator only stretches. Column $i$ of $\hat{A} \hat{P}$ is $\hat{A} \overset{a}{ˉ}_{i} = λ_{i} \overset{a}{ˉ}_{i}$ , which is column $i$ of $\hat{P} \hat{D}$ ; so $\hat{A} \hat{P} = \hat{P} \hat{D}$ (beyond the boards).

The momentum operator

The operator that will stand for momentum along $x$ (Definition 2.4.1): $\overset{p}{^}_{x} = - i ℏ \frac{\partial}{\partial x}, ℏ = \frac{h}{2 π} = 1.054 \times 1 0^{- 34} J s,$ with $h = 6.62 \times 1 0^{- 34}$ J s Planck’s constant and $ℏ$ the reduced Planck constant.

Theorem 1.3.8Eigenfunctions of momentum

The eigenfunctions of $\overset{p}{^}_{x}$ are the plane waves $u (x) = C e^{i p x /ℏ},$ one for every real eigenvalue $p$ .

Proof

Solve $- i ℏ u^{'} = p u$ . Dividing by $- i ℏ$ , and since $1/ (- i) = i$ , $\frac{d u}{d x} = \frac{i p}{ℏ} u$ . Separate and integrate: $ln u = \frac{i p}{ℏ} x + const$ , so $u = C e^{i p x /ℏ}$ . Check: $- i ℏ \frac{d}{d x} C e^{i p x /ℏ} = - i ℏ \cdot \frac{i p}{ℏ} C e^{i p x /ℏ} = p u$ , as $- i \cdot i = 1$ .

Read backwards: ==a plane wave $e^{ik x}$ is a state of definite momentum $p = ℏ k$ ==, de Broglie’s relation. Nothing yet restricts $p$ ; a box will (§1.5 Momentum in a box, Fourier and the Dirac delta).

Exercises

Exercise 1.3.1Sample questions, Q2 (i)

What is an eigenvalue? What is an eigenvector?

Solution

If $\hat{A} u = a u$ with $a$ a number and $u \neq = 0$ , $u$ is an eigenvector (eigenfunction) of $\hat{A}$ and $a$ its eigenvalue: $\hat{A}$ only rescales $u$ . Example: $- i ℏ \frac{d}{d x} e^{ik x} = ℏ k e^{ik x}$ , so a plane wave is an eigenfunction of momentum with eigenvalue $ℏ k$ .

Exercise 1.3.2beyond the boards

Find the eigenvalues and eigenvectors of $(0110)$ . Are the eigenvectors orthogonal?

Solution

$det = λ^{2} - 1$ : $λ = \pm 1$ , with $(1, 1)$ for $+ 1$ and $(1, - 1)$ for $- 1$ , normalised $\frac{1}{2} (1, \pm 1)$ . Orthogonal, as for any real symmetric matrix. (This is $\frac{2}{ℏ} \hat{S}_{x}$ of the spin chapter.)

Exercise 1.3.3beyond the boards

Eigenvalues and eigenvectors of $(20 5 - 1)$ . Are they orthogonal?

Solution

$(2 - λ) (- 1 - λ) = 0$ : $λ = 2, - 1$ (a triangular matrix has its eigenvalues on the diagonal). $λ = 2$ : $5 y = 0$ , eigenvector $(1, 0)$ . $λ = - 1$ : $3 x + 5 y = 0$ , eigenvector $(5, - 3)$ ; check, $(10 - 15, 3) = - (5, - 3)$ . Not orthogonal, $(1, 0) \cdot (5, - 3) = 5$ : the matrix is not symmetric.

Exercise 1.3.4beyond the boards

Diagonalise $(2112)$ .

Solution

With the normalised eigenvectors as columns, $\hat{P} = \frac{1}{2} (11 1 - 1)$ and $\hat{D} = (3001)$ . The columns are orthonormal, so $\hat{P}^{- 1} = \hat{P}^{T} = \hat{P}$ and $\hat{P} \hat{A} \hat{P} = \hat{D}$ .

Exercise 1.3.5beyond the boards

Is $sin k x$ an eigenfunction of $\overset{p}{^}_{x}$ ? Of $\overset{p}{^}_{x}^{2} = - ℏ^{2} \frac{d ^{2}}{d x ^{2}}$ ?

Solution

$\overset{p}{^}_{x} sin k x = - i ℏ k cos k x$ , not a multiple of $sin k x$ : no. $\overset{p}{^}_{x}^{2} sin k x = ℏ^{2} k^{2} sin k x$ : yes, eigenvalue $ℏ^{2} k^{2}$ . $sin k x = \frac{1}{2 i} (e^{ik x} - e^{- ik x})$ mixes $+ ℏ k$ and $- ℏ k$ ; squaring makes the sign irrelevant.