3.4 Linear operators and the eigenvalue problem

The previous section settled where the states live. This section asks: what can be done on that space?

An operator = a map from states to states

A linear operator $\hat A$ sends each ket to another ket:

\hat A\ket{\psi}=\ket{\phi}\tag{3.4.1}

Linearity means

\hat A\big(a\ket{\psi_1}+b\ket{\psi_2}\big)=a\hat A\ket{\psi_1}+b\hat A\ket{\psi_2}\tag{3.4.2}

Why only linear operators

Because the Schrödinger equation is linear.

If $\ket{\psi_1}$ and $\ket{\psi_2}$ evolve into $\ket{\psi_1(t)}$ and $\ket{\psi_2(t)}$ respectively, the superposition $a\ket{\psi_1}+b\ket{\psi_2}$ must evolve into $a\ket{\psi_1(t)}+b\ket{\psi_2(t)}$ . Only a linear operator can do that.

Introduce a nonlinear term and the superposition principle collapses — and the superposition principle is the entire source of double-slit interference and of quantum computing’s parallelism. Every experiment to date is consistent with strict linearity — dedicated high-precision tests have been run (searching, for instance, for a logarithmic nonlinearity in the Schrödinger equation), and the bounds are extremely tight.

In a basis, an operator is a matrix

Take an orthonormal complete basis $\{\ket{n}\}$ and define

A_{mn}=\bra{m}\hat A\ket{n}\tag{3.4.3}

Arrange these numbers into a matrix. It determines everything about $\hat A$ :

Derivation: how the matrix elements fix the operator's actionbasic~4 min

Let $\ket{\psi}=\sum_n c_n\ket{n}$ and $\ket{\phi}=\hat A\ket{\psi}=\sum_m d_m\ket{m}$ . To find $d_m$ :

d_m=\braket{m}{\phi}=\bra{m}\hat A\ket{\psi}\tag{3.4.4}

Insert a completeness relation $\sum_n\ket{n}\bra{n}=\hat 1$ to the right of $\hat A$ :

d_m=\sum_n\bra{m}\hat A\ket{n}\braket{n}{\psi}=\sum_n A_{mn}c_n\tag{3.4.5}

\boxed{\ d_m=\sum_n A_{mn}c_n\ }\tag{3.4.6}

This is exactly a matrix multiplying a column vector. An operator acting on a state is, in component language, matrix multiplication.

Likewise, a product of two operators corresponds to a product of matrices:

(AB)_{mn}=\bra{m}\hat A\hat B\ket{n}=\sum_k A_{mk}B_{kn}\tag{3.4.7}

(again by inserting a $\hat 1$ ).

The eigenvalue problem

\hat A\ket{a}=a\ket{a}\tag{3.4.8}

$\ket{a}$ is called an eigenvector, and $a$ an eigenvalue.

The geometric meaning is intuitive: most vectors, when hit by $\hat A$ , change both length and direction; the eigenvectors are the special directions that only get rescaled, never turned.

The picture

How to solve it (finite dimensions)

Write the equation as $(\hat A-a\hat 1)\ket{a}=0$ . For a non-zero solution the matrix must be singular:

\det(A-a I)=0\tag{3.4.9}

This is a degree- $N$ polynomial in $a$ (the characteristic equation) with $N$ roots (counted with multiplicity). Substitute each root back and solve the homogeneous system for the eigenvectors.

The mathematics

Example: the x component of spin

\hat S_x=\frac{\hbar}{2}\begin{pmatrix}0&1\\1&0\end{pmatrix}\tag{3.4.10}

Characteristic equation $a^2-(\hbar/2)^2=0$ , eigenvalues $a=\pm\hbar/2$ .

Substituting back gives the eigenvectors

\ket{\pm_x}=\frac{1}{\sqrt2}\begin{pmatrix}1\\ \pm1\end{pmatrix}\tag{3.4.11}

In the $\ket{\pm_z}$ basis, these are precisely the states “spin up/down along x”. Chapter 5 will use them, and these are the very directions the Bloch-sphere scene draws.

Chapter 2's equations were eigenvalue problems all along

The time-independent Schrödinger equation

-\frac{\hbar^2}{2m}\frac{\dd^2\psi}{\dd x^2}+V\psi=E\psi\tag{3.4.12}

reads, in the new language,

\hat H\ket{\psi}=E\ket{\psi}\tag{3.4.13}

Energy levels = eigenvalues of the Hamiltonian; stationary states = its eigenvectors.

What section 2.3 phrased as “only special values of $E$ admit normalisable solutions” here reads “the eigenvalues are discrete”. And the discreteness comes from restricting the operator to $L^2$ (the boundary conditions) — automatic for a finite matrix (an $N\times N$ matrix has at most $N$ eigenvalues), but something that must be proved in infinite dimensions.

Degeneracy

If a single eigenvalue $a$ corresponds to several linearly independent eigenvectors, the eigenvalue is called degenerate, and the degeneracy is the dimension of the space those eigenvectors span.

The picture

Eigenspaces

Suppose $\ket{a_1},\dots,\ket{a_g}$ all belong to the eigenvalue $a$ . Any linear combination of them is again an eigenvector:

\hat A\big(\textstyle\sum_i c_i\ket{a_i}\big)=a\big(\textstyle\sum_i c_i\ket{a_i}\big)\tag{3.4.14}

So they span a $g$ -dimensional subspace $\mathcal H_a$ , called the eigenspace.

Note: with degeneracy, “the eigenvector belonging to $a$ ” is not unique — there is not even a natural way to choose one. You are free to pick any basis inside $\mathcal H_a$ . That freedom becomes crucial in perturbation theory (section 7.2).

The mathematics

Degeneracy always travels with symmetry

The $n=2$ level of hydrogen holds 4 states ( $2s$ and three $2p$ ), all with the same energy.

Why? Because the Hamiltonian has symmetries:

spherical symmetry → degeneracy in $m$ ( $2p_x,2p_y,2p_z$ share one energy)
an extra symmetry peculiar to the Coulomb potential → degeneracy in $l$ ( $2s$ and $2p$ share one energy)

The second is unique to the Coulomb potential; swap in any other central potential (say the screened potential in a many-electron atom) and the $l$ degeneracy breaks — which is precisely where the structure of the periodic table comes from. See section 6.4.

Break the symmetry and the degeneracy lifts: a magnetic field breaks spherical symmetry, the $m$ degeneracy lifts, spectral lines split — the Zeeman effect.

Key formulas

Matrix elements
$A_{mn}=\bra{m}\hat A\ket{n}$
The operator's bookkeeping in one basis; changes when the basis does
Operator action = matrix multiplication
$d_m=\sum_n A_{mn}c_n$
Derived by inserting a single completeness relation
Eigenvalue equation
$\hat A\ket{a}=a\ket{a}$
Eigenvectors = directions that get rescaled, never turned
Characteristic equation
$\det(A-aI)=0$
The standard route to eigenvalues in finite dimensions
Eigenvalues of a function
$f(\hat A)\ket{a}=f(a)\ket{a}$
Used directly to build e^{−iHt/ħ}

Self-check4 questions

What comes next

There are infinitely many kinds of linear operator, but an observable can only belong to a very small class — it must guarantee real measurement results, and distinct results must be distinguishable.

The next section pins down exactly which class that is — and, along the way, redoes in three lines the proof that took section 2.3 the better part of a page.

Linear operators and the eigenvalue problem

Recommended first

After this section you should be able to

An operator = a map from states to states

In a basis, an operator is a matrix

The eigenvalue problem

The picture

The mathematics

Degeneracy

The picture

The mathematics

Key formulas

Self-check4 questions

What comes next

An operator = a map from states to states

In a basis, an operator is a matrix

The eigenvalue problem

◑The picture

∑The mathematics

Degeneracy

◑The picture

∑The mathematics

∎Key formulas

What comes next

The picture

The mathematics

The picture

The mathematics

Key formulas