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3.5

Hermitian operators and observables

In chapter 2, "observables correspond to operators" was a bare decree. This section turns it into a conclusion: only a Hermitian operator can be an observable.

Recommended first

After this section you should be able to

  • Write down the definition of the Hermitian conjugate, and decide whether a given operator is Hermitian
  • Prove that Hermitian operators have real eigenvalues and that eigenvectors of distinct eigenvalues are orthogonal
  • Use the spectral theorem to explain why a Hermitian operator's eigenvectors form a complete basis
  • Articulate the physical reasons why an observable must be Hermitian

The previous section noted that linear operators come in infinitely many kinds. This section asks: which kind earns the title of “observable”?

Working backwards from the physics

Set the mathematics aside for a moment and list the three things physics cannot do without:

The picture

Requirement one: measurement results are real numbers.

Measure a length, an energy, a spin — the readout is always real. So the eigenvalues of A^\hat A must be real.

Requirement two: distinct results can be told apart.

If the result is a1a_1, the state should land among “the states belonging to a1a_1”, with no overlap with “the states belonging to a2a_2”. Mathematically: eigenvectors of distinct eigenvalues are orthogonal.

Requirement three: every state can be resolved by this measurement.

For ∑n∣cn∣2=1\sum_n|c_n|^2=1 to hold, the eigenvectors must form a complete basis, so that any state can be expanded.

The mathematics

The answer: Hermitian operators

Taken together, the three requirements single out exactly one class of operator. Define the Hermitian conjugate A^†\hat A^\dagger of A^\hat A as the operator satisfying

⟨ϕ∣A^†∣ψ⟩=⟨ψ∣A^∣ϕ⟩∗(3.5.1)\bra{\phi}\hat A^\dagger\ket{\psi}=\bra{\psi}\hat A\ket{\phi}^*\tag{3.5.1}

If

A^†=A^(3.5.2)\hat A^\dagger=\hat A\tag{3.5.2}

then A^\hat A is Hermitian (self-adjoint).

In matrix language: A†=(AT)∗A^\dagger=(A^{\mathsf T})^*, and Hermitian means Amn=Anm∗A_{mn}=A_{nm}^* — transpose-and-conjugate equals itself. The diagonal elements must therefore be real.

Cashing in the three requirements, one by one

Proof one: Hermitian operators have real eigenvaluesbasic~3 min

Let A^∣a⟩=a∣a⟩\hat A\ket{a}=a\ket{a} with ⟨a|a⟩=1\braket{a}{a}=1. Multiply on the left by ⟨a∣\bra{a}:

⟨a∣A^∣a⟩=a(3.5.3)\bra{a}\hat A\ket{a}=a\tag{3.5.3}

Take the complex conjugate and use Hermiticity, A^†=A^\hat A^\dagger=\hat A:

a∗=⟨a∣A^∣a⟩∗=⟨a∣A^†∣a⟩=⟨a∣A^∣a⟩=a(3.5.4)a^*=\bra{a}\hat A\ket{a}^*=\bra{a}\hat A^\dagger\ket{a}=\bra{a}\hat A\ket{a}=a\tag{3.5.4}

So a=a∗a=a^*: the eigenvalue is real. ■\blacksquare

Compare: proving the same thing took section 2.3 half a page — conjugate, subtract, massage into a total derivative, integrate by parts, kill the boundary term via normalisation.

That half page was not wasted: it was verifying, in the concrete, the Hermiticity of H^\hat H (integration by parts shifts the derivative from one side to the other with vanishing boundary term — exactly the content of ⟨ϕ∣H^∣ψ⟩=⟨H^ϕ∣∣ψ⟩\bra{\phi}\hat H\ket{\psi}=\bra{\hat H\phi}\ket{\psi}).

Switching languages is not a magic trick: it abstracts that half page once and for all, so it never has to be repeated.

Proof two: eigenvectors of distinct eigenvalues are orthogonalbasic~4 min

Let A^∣a1⟩=a1∣a1⟩\hat A\ket{a_1}=a_1\ket{a_1} and A^∣a2⟩=a2∣a2⟩\hat A\ket{a_2}=a_2\ket{a_2} with a1≠a2a_1\ne a_2.

Evaluate ⟨a1∣A^∣a2⟩\bra{a_1}\hat A\ket{a_2} two ways:

Acting to the right: ⟨a1∣A^∣a2⟩=a2⟨a1|a2⟩\bra{a_1}\hat A\ket{a_2}=a_2\braket{a_1}{a_2}

Acting to the left: by Hermiticity, A^\hat A acting on the bra gives a1∗=a1a_1^*=a_1 (already proved real), so =a1⟨a1|a2⟩=a_1\braket{a_1}{a_2}

Subtract:

(a1−a2)⟨a1|a2⟩=0(3.5.5)(a_1-a_2)\braket{a_1}{a_2}=0\tag{3.5.5}

Since a1≠a2a_1\ne a_2, we get ⟨a1|a2⟩=0\braket{a_1}{a_2}=0. ■\blacksquare

The degenerate case: if a1=a2a_1=a_2 the equation holds trivially and orthogonality does not follow. But inside the eigenspace you can manufacture an orthogonal basis by Gram-Schmidt. So an orthonormal set of eigenvectors can always be chosen.

Some examples

The picture

Hermitian

  • x^\hat x (position): in the position basis, multiplication by the real number xx
  • p^=−iℏ∂x\hat p=-\ii\hbar\partial_x: the i\ii and the derivative each contribute one minus sign, which cancel
  • H^=p^2/2m+V(x^)\hat H=\hat p^2/2m+V(\hat x) (with VV real)
  • the projection operator ∣ψ⟩⟨ψ∣\ket{\psi}\bra{\psi}
  • the Pauli matrices σx,σy,σz\sigma_x,\sigma_y,\sigma_z

Not Hermitian

  • a^±\hat a_\pm (ladder operators): a^+†=a^−\hat a_+^\dagger=\hat a_- — conjugates of each other, not of themselves
  • x^p^\hat x\hat p: see the right-hand column
  • ∂x\partial_x by itself (missing that i\ii)

The mathematics

Why p^\hat p must carry the i\ii

⟨ϕ∣∂x∣ψ⟩=∫ϕ∗ψ′ dx=−∫ϕ′∗ψ dx=−⟨∂xϕ∣∣ψ⟩(3.5.8)\bra{\phi}\partial_x\ket{\psi}=\int\phi^*\psi'\,\dd x =-\int\phi'^*\psi\,\dd x=-\bra{\partial_x\phi}\ket{\psi}\tag{3.5.8}

(integration by parts; the boundary term vanishes by normalisation). So ∂x†=−∂x\partial_x^\dagger=-\partial_x — anti-Hermitian.

Multiply by −iℏ-\ii\hbar and conjugation sends i→−i\ii\to-\ii, contributing a second minus sign; the two cancel:

p^†=(−iℏ∂x)†=(+iℏ)(−∂x)=−iℏ∂x=p^(3.5.9)\hat p^\dagger=(-\ii\hbar\partial_x)^\dagger=(+\ii\hbar)(-\partial_x)=-\ii\hbar\partial_x=\hat p\tag{3.5.9}

Section 2.6 said “that i\ii makes p^\hat p Hermitian” — this is the full verification.

Key formulas

  • Hermitian conjugate

    ⟨ϕ∣A^†∣ψ⟩=⟨ψ∣A^∣ϕ⟩∗\bra{\phi}\hat A^\dagger\ket{\psi}=\bra{\psi}\hat A\ket{\phi}^*

    Matrix language: transpose, then complex-conjugate

  • Hermitian operator

    A^†=A^ ⟺ Amn=Anm∗\hat A^\dagger=\hat A\ \Longleftrightarrow\ A_{mn}=A_{nm}^*

    Diagonal elements must be real

  • Real eigenvalues

    a=⟨a∣A^∣a⟩=a∗a=\bra{a}\hat A\ket{a}=a^*

    Three lines; chapter 2 needed half a page of integrals

  • Orthogonal eigenvectors

    (a1−a2)⟨a1|a2⟩=0(a_1-a_2)\braket{a_1}{a_2}=0

    Distinct eigenvalues ⟹ orthogonal; degenerate ones can be orthogonalised

  • Spectral decomposition

    A^=∑nan∣an⟩⟨an∣,∑n∣an⟩⟨an∣=1^\hat A=\sum_n a_n\ket{a_n}\bra{a_n},\quad\sum_n\ket{a_n}\bra{a_n}=\hat 1

    The origin of the completeness relation

  • Order reversal under conjugation

    (A^B^)†=B^†A^†(\hat A\hat B)^\dagger=\hat B^\dagger\hat A^\dagger

    A product of two Hermitian operators is generally not Hermitian

Self-check4 questions

  1. 1.

    Why must observables be represented by Hermitian operators? (Select all that apply.)

    Select all that apply

  2. 2.

    Why is the momentum operator −iħ∂/∂x rather than ħ∂/∂x?

  3. 3.

    If  and B̂ are both Hermitian, then ÂB̂ is:

  4. 4.

    The direct physical consequence of the spectral theorem is:

What comes next

There can be many Hermitian operators: x^\hat x, p^\hat p, H^\hat H, L^z\hat L_z…

A natural question: can two observables be measured sharply at the same time? The answer hinges on whether their product cares about the order. The next section upgrades the uncertainty principle from “a corollary of Fourier analysis” to a theorem that holds for any pair of observables.

Section 24 of 106 · use ← → to turn the page