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Theme 2 — Hilbert Spaces and Dirac Notation

Self-adjoint operators

Keywords: self-adjoint operator · Hermitian matrix · real eigenvalues · orthogonal eigenvectors · spectral decomposition

Exercise 1 : Self-adjoint operators

In a finite-dimensional Hilbert space H\H, an operator A^\hat A is called self-adjoint, or Hermitian, if A^†=A^\hat A^\dagger=\hat A, that is, if ⟨u|A^v⟩=⟨A^u|v⟩\braket{u}{\hat Av}=\braket{\hat Au}{v} for all ∣u⟩,∣v⟩\ket u,\ket v.

Question 1
Show that, for every vector ∣v⟩\ket v, the quantity ⟨v∣A^∣v⟩\bra v\hat A\ket v is real. Deduce that the eigenvalues of A^\hat A are real.

Solution
By Hermitian symmetry of the inner product, followed by self-adjointness,

⟨v∣A^∣v⟩∗=⟨v|A^v⟩∗=⟨A^v|v⟩=⟨v|A^v⟩=⟨v∣A^∣v⟩.\bra v\hat A\ket v^*=\braket{v}{\hat Av}^*=\braket{\hat Av}{v}=\braket{v}{\hat Av}=\bra v\hat A\ket v .

This number is equal to its complex conjugate, so it is real. If A^∣v⟩=λ∣v⟩\hat A\ket v=\lambda\ket v with ∣v⟩≠0\ket v\neq0, then ⟨v∣A^∣v⟩=λ∥v∥2\bra v\hat A\ket v=\lambda\norm v^2, and λ=⟨v∣A^∣v⟩/∥v∥2\lambda=\bra v\hat A\ket v/\norm v^2 is real. This is essential if the eigenvalues of an observable are to represent measurement outcomes.

Question 2
Show that two eigenvectors associated with distinct eigenvalues are orthogonal.

Solution
Let A^∣u⟩=λ∣u⟩\hat A\ket u=\lambda\ket u and A^∣v⟩=μ∣v⟩\hat A\ket v=\mu\ket v, with λ≠μ\lambda\neq\mu, both real by the preceding question. We calculate ⟨u|A^v⟩\braket{u}{\hat Av} in two ways:

⟨u|A^v⟩=μ⟨u|v⟩,⟨u|A^v⟩=⟨A^u|v⟩=λ∗⟨u|v⟩=λ⟨u|v⟩.\braket{u}{\hat Av}=\mu\braket uv, \qquad \braket{u}{\hat Av}=\braket{\hat Au}{v}=\lambda^*\braket uv=\lambda\braket uv .

Thus (μ−λ)⟨u|v⟩=0(\mu-\lambda)\braket uv=0, and since λ≠μ\lambda\neq\mu, we have ⟨u|v⟩=0\braket uv=0.

Question 3
Consider the matrix

A=(11−i1+i2).\begin{aligned} A=\begin{pmatrix}1&1-i\\1+i&2\end{pmatrix}. \end{aligned}

Verify that it is Hermitian, then determine its eigenvalues and normalised eigenvectors. Verify their orthogonality.

Solution
The conjugate transpose of AA is equal to AA: the diagonal entries are real, and the off-diagonal entries are complex conjugates of each other. The characteristic polynomial is

det⁡(A−λ1)=(1−λ)(2−λ)−(1−i)(1+i)=λ2−3λ+2−2=λ(λ−3).\det(A-\lambda\mathbf{1})=(1-\lambda)(2-\lambda)-(1-i)(1+i)=\lambda^2-3\lambda+2-2=\lambda(\lambda-3).

The eigenvalues are 00 and 33, real as expected. For λ=0\lambda=0, the equation x+(1−i)y=0x+(1-i)y=0 gives the vector (−1+i, 1)(-1+i,\,1), whose norm is 3\sqrt3. For λ=3\lambda=3, the equation −2x+(1−i)y=0-2x+(1-i)y=0 gives the vector (1−i, 2)(1-i,\,2), whose norm is 6\sqrt6. The normalised eigenvectors are therefore

∣a0⟩=13(−1+i1),∣a3⟩=16(1−i2).\ket{a_0}=\frac{1}{\sqrt3}\begin{pmatrix}-1+i\\1\end{pmatrix}, \qquad \ket{a_3}=\frac{1}{\sqrt6}\begin{pmatrix}1-i\\2\end{pmatrix}.

Their inner product is 118((−1−i)(1−i)+1⋅2)=118(−2+2)=0\frac{1}{\sqrt{18}}\bigl((-1-i)(1-i)+1\cdot2\bigr)=\frac{1}{\sqrt{18}}(-2+2)=0, using (−1−i)(1−i)=−1+i−i+i2=−2(-1-i)(1-i)=-1+i-i+i^2=-2.

Question 4
Write down the spectral decomposition of AA and verify it.

Solution
The spectral decomposition is A=0⋅∣a0⟩⟨a0∣+3∣a3⟩⟨a3∣=3∣a3⟩⟨a3∣A=0\cdot\ket{a_0}\bra{a_0}+3\ket{a_3}\bra{a_3}=3\ket{a_3}\bra{a_3}. Let us verify it:

3∣a3⟩⟨a3∣=36(1−i2)(1+i2)=12(∣1−i∣22(1−i)2(1+i)4)=(11−i1+i2)=A.\begin{aligned} 3\ket{a_3}\bra{a_3}=\frac36\begin{pmatrix}1-i\\2\end{pmatrix}\begin{pmatrix}1+i&2\end{pmatrix}=\frac12\begin{pmatrix}|1-i|^2&2(1-i)\\2(1+i)&4\end{pmatrix}=\begin{pmatrix}1&1-i\\1+i&2\end{pmatrix}=A . \end{aligned}

The matrix AA is therefore equal to 33 times a rank-one projector. This is consistent with the trace, tr⁡A=3=0+3\tr A=3=0+3, and the determinant, det⁡A=0=0×3\det A=0=0\times3.