1.1.3 Hermitian Operators#

Worked solutions for the homework problems in the 1.1.3 Hermitian Operators lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

1. Spectral decomposition of a two-level Hamiltonian. Consider the Hermitian operator \(\hat{H} = \omega\,\hat{X} + \Delta\,\hat{Z}\), with \(\omega,\Delta\in\mathbb{R}\). This represents a generic two-level system in which \(\hat Z\) encodes an energy splitting \(2\Delta\) between \(\vert 0\rangle\) and \(\vert 1\rangle\) and \(\hat X\) provides a coupling of strength \(\omega\) between them.

(a) Verify that \(\hat H\) is Hermitian, and write its matrix in the \(\{\vert 0\rangle,\vert 1\rangle\}\) basis.

(b) Find the eigenvalues \(E_\pm\) and the corresponding eigenstates \(\vert E_\pm\rangle\). Express the result using the mixing angle \(\theta_0\) defined by \(\tan\theta_0 = \omega/\Delta\) (with \(\theta_0\in[0,\pi]\)).

(c) Write the spectral decomposition \(\hat H = E_+\vert E_+\rangle\langle E_+\vert + E_-\vert E_-\rangle\langle E_-\vert\) and use it to compute \(\hat H^2\) without doing any matrix multiplication. Show that \(\hat H^2 = (\omega^2+\Delta^2)\,\hat I\).

(d) Use the spectral form to write an integer power \(\hat H^n\) in closed form for any \(n\ge 1\).

Solution.

(a) The matrix is

\[\begin{split} \hat H = \omega\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix} + \Delta\begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix} = \begin{pmatrix} \Delta & \omega \\ \omega & -\Delta\end{pmatrix}. \end{split}\]

The diagonal entries \(\pm\Delta\) are real, and the off-diagonal entries are an equal pair \(\omega = \omega^*\) (since \(\omega\) is real). So \(\hat H = \hat H^\dagger\).

(b) Eigenvalues come from \(\det(\hat H - E\hat I) = 0\):

\[ (\Delta - E)(-\Delta - E) - \omega^2 = E^2 - \Delta^2 - \omega^2 = 0 \quad\Longrightarrow\quad E_\pm = \pm\sqrt{\omega^2 + \Delta^2}. \]

For the eigenstates, set \(\sqrt{\omega^2+\Delta^2} = \Omega\) and \(\tan\theta_0 = \omega/\Delta\) so that \(\cos\theta_0 = \Delta/\Omega\) and \(\sin\theta_0 = \omega/\Omega\). Solving \((\hat H - E_+\hat I)v = 0\),

\[\begin{split} \begin{pmatrix} \Delta - \Omega & \omega \\ \omega & -\Delta - \Omega\end{pmatrix}\begin{pmatrix} a\\ b\end{pmatrix} = 0 \;\Longrightarrow\; \frac{b}{a} = \frac{\Omega - \Delta}{\omega}. \end{split}\]

Apply half-angle identities: \(\Omega - \Delta = \Omega(1 - \cos\theta_0) = 2\Omega\sin^2(\theta_0/2)\) and \(\omega = \Omega\sin\theta_0 = 2\Omega\sin(\theta_0/2)\cos(\theta_0/2)\), giving \(b/a = \tan(\theta_0/2)\). Choosing \(a = \cos(\theta_0/2)\),

\[ \vert E_+\rangle = \cos(\theta_0/2)\,\vert 0\rangle + \sin(\theta_0/2)\,\vert 1\rangle. \]

The orthogonal \(\vert E_-\rangle\) follows from the antipodal-on-Bloch result of 1.1.2 Problem 5:

\[ \vert E_-\rangle = \sin(\theta_0/2)\,\vert 0\rangle - \cos(\theta_0/2)\,\vert 1\rangle. \]

Both are real-amplitude states (no relative phase), as expected since \(\hat H\) has no \(\hat Y\) component.

\[ \hat H = \Omega\,\vert E_+\rangle\langle E_+\vert + (-\Omega)\,\vert E_-\rangle\langle E_-\vert. \]

Squaring uses the orthogonality \(\langle E_+\vert E_-\rangle = 0\) and idempotence of each projector:

\[ \hat H^2 = \Omega^2\,\vert E_+\rangle\langle E_+\vert + \Omega^2\,\vert E_-\rangle\langle E_-\vert = \Omega^2\bigl(\vert E_+\rangle\langle E_+\vert + \vert E_-\rangle\langle E_-\vert\bigr) = \Omega^2\,\hat I, \]

using completeness \(\vert E_+\rangle\langle E_+\vert + \vert E_-\rangle\langle E_-\vert = \hat I\). So \(\hat H^2 = (\omega^2 + \Delta^2)\hat I\). Crucially, no \(2\times 2\) matrix multiplication was needed — the spectral form turns operator powers into scalar powers of eigenvalues.

(d) For any integer \(n\ge 1\),

\[ \hat H^n = (+\Omega)^n\,\vert E_+\rangle\langle E_+\vert + (-\Omega)^n\,\vert E_-\rangle\langle E_-\vert. \]

The same spectral structure extends to any function of \(\hat H\): \(f(\hat H) = f(\Omega)\vert E_+\rangle\langle E_+\vert + f(-\Omega)\vert E_-\rangle\langle E_-\vert\). This is the operator-function machinery that powers unitary time evolution \(\mathrm{e}^{-\mathrm{i}\hat H t/\hbar}\) — developed in Section 1.3.

2. Hermitian and anti-Hermitian decomposition. Any operator \(\hat A\) on a Hilbert space admits a unique decomposition \(\hat A = \hat H + \mathrm{i}\hat K\) with both \(\hat H\) and \(\hat K\) Hermitian.

(a) Show that \(\hat H = \tfrac{1}{2}(\hat A + \hat A^\dagger)\) and \(\hat K = \tfrac{1}{2\mathrm{i}}(\hat A - \hat A^\dagger)\) are both Hermitian, and that \(\hat A = \hat H + \mathrm{i}\hat K\).

(b) Apply to \(\hat A = \begin{pmatrix} 1 & \mathrm{i} \\ \mathrm{i} & 1\end{pmatrix}\). Compute \(\hat H\) and \(\hat K\) explicitly, and express each in terms of the Pauli operators \(\hat I, \hat X, \hat Y, \hat Z\).

(c) Draw the analogy with the complex-number decomposition \(z = x + \mathrm{i}y\): which operator plays the role of the real part of \(z\), and which the imaginary part? Explain in one sentence why this analogy makes the Hermitian operators the natural “real” subspace of the operator algebra.

Solution.

(a) Hermiticity of \(\hat H\). Using \((\hat A^\dagger)^\dagger = \hat A\):

\[ \hat H^\dagger = \left[\tfrac{1}{2}(\hat A + \hat A^\dagger)\right]^\dagger = \tfrac{1}{2}(\hat A^\dagger + \hat A) = \hat H. \quad\checkmark \]

Hermiticity of \(\hat K\). Recall that \((\mathrm{i})^\dagger = -\mathrm{i}\) when distributed through a dagger (the dagger conjugates scalars). So

\[ \hat K^\dagger = \left[\tfrac{1}{2\mathrm{i}}(\hat A - \hat A^\dagger)\right]^\dagger = \tfrac{1}{-2\mathrm{i}}(\hat A^\dagger - \hat A) = \tfrac{1}{2\mathrm{i}}(\hat A - \hat A^\dagger) = \hat K. \quad\checkmark \]

Decomposition holds. Plug back:

\[ \hat H + \mathrm{i}\hat K = \tfrac{1}{2}(\hat A + \hat A^\dagger) + \mathrm{i}\cdot\tfrac{1}{2\mathrm{i}}(\hat A - \hat A^\dagger) = \tfrac{1}{2}\hat A + \tfrac{1}{2}\hat A^\dagger + \tfrac{1}{2}\hat A - \tfrac{1}{2}\hat A^\dagger = \hat A. \quad\checkmark \]

(b) Compute \(\hat A^\dagger\) by transposing and conjugating: \(\hat A^\dagger = \begin{pmatrix} 1 & -\mathrm{i} \\ -\mathrm{i} & 1\end{pmatrix}\). Then

\[\begin{split} \hat H = \tfrac{1}{2}(\hat A + \hat A^\dagger) = \tfrac{1}{2}\begin{pmatrix} 2 & 0 \\ 0 & 2\end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} = \hat I, \end{split}\]

\[\begin{split} \hat K = \tfrac{1}{2\mathrm{i}}(\hat A - \hat A^\dagger) = \tfrac{1}{2\mathrm{i}}\begin{pmatrix} 0 & 2\mathrm{i} \\ 2\mathrm{i} & 0\end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix} = \hat X. \end{split}\]

So \(\hat A = \hat I + \mathrm{i}\hat X\) — a clean Pauli decomposition. Verification: \(\hat I + \mathrm{i}\hat X = \begin{pmatrix}1 & 0\\0 & 1\end{pmatrix} + \begin{pmatrix}0 & \mathrm{i}\\ \mathrm{i} & 0\end{pmatrix} = \begin{pmatrix}1 & \mathrm{i}\\ \mathrm{i} & 1\end{pmatrix} = \hat A\).

(c) The Hermitian part \(\hat H\) corresponds to the real part of \(z = x + \mathrm{i}y\); the anti-Hermitian part \(\mathrm{i}\hat K\) (with \(\hat K\) Hermitian) corresponds to the imaginary part. The analogy is exact: just as every complex number splits uniquely into real + imaginary parts via \(x = \tfrac{1}{2}(z+z^*)\) and \(y = \tfrac{1}{2\mathrm{i}}(z-z^*)\), every operator splits uniquely into Hermitian + (i times) Hermitian parts via the formulas above — with \(\dagger\) playing the role of complex conjugation. Hermitian operators are therefore the natural “real” subspace of the algebra, and observables — being measurable real-valued — are restricted to it. (Forward pointer: the unitary operators \(\hat U = \mathrm{e}^{\mathrm{i}\hat O}\) — to be introduced in Section 1.3 — play the role of phase factors \(\mathrm{e}^{\mathrm{i}\theta}\) for real \(\theta\), completing the analogy: Hermitian ↔ real, unitary ↔ unit-modulus.)

3. Bloch vector from amplitudes. A qubit is in the state \(\vert\psi\rangle = \tfrac{1}{\sqrt 3}\vert 0\rangle + \sqrt{\tfrac{2}{3}}\,\vert 1\rangle\). Using the lecture’s amplitude formulas

\[ \langle\hat X\rangle = 2\operatorname{Re}(\alpha^*\beta),\quad \langle\hat Y\rangle = 2\operatorname{Im}(\alpha^*\beta),\quad \langle\hat Z\rangle = \vert\alpha\vert^2 - \vert\beta\vert^2, \]

compute the Bloch vector \(\boldsymbol n = (\langle\hat X\rangle, \langle\hat Y\rangle, \langle\hat Z\rangle)\). Verify \(\vert\boldsymbol n\vert = 1\), and explain in one sentence why this had to come out unit-length.

Solution.

With \(\alpha = 1/\sqrt 3\) and \(\beta = \sqrt{2/3}\) (both real and positive), the product \(\alpha^*\beta = (1/\sqrt 3)\cdot\sqrt{2/3} = \sqrt{2}/3\) is real, so \(\operatorname{Im}(\alpha^*\beta) = 0\). Hence

\[ \langle\hat X\rangle = 2\cdot\frac{\sqrt 2}{3} = \frac{2\sqrt 2}{3}, \qquad \langle\hat Y\rangle = 0, \qquad \langle\hat Z\rangle = \tfrac{1}{3} - \tfrac{2}{3} = -\tfrac{1}{3}. \]

The Bloch vector is \(\boldsymbol n = \bigl(\tfrac{2\sqrt 2}{3},\, 0,\, -\tfrac{1}{3}\bigr)\). Sum of squares:

\[ \vert\boldsymbol n\vert^2 = \left(\tfrac{2\sqrt 2}{3}\right)^{\!2} + 0 + \left(-\tfrac{1}{3}\right)^{\!2} = \tfrac{8}{9} + \tfrac{1}{9} = 1. \]

So \(\vert\boldsymbol n\vert = 1\), exactly as required for any pure qubit state — pure states sit on the surface of the Bloch sphere (1.1.2 Problem 4). The vanishing \(\langle\hat Y\rangle\) reflects the absence of a relative phase between the two real amplitudes: the state lies in the \(xz\)-plane of the Bloch sphere.

4. Eigenstate of spin along an arbitrary axis. Continuing from 1.1.2 Problem 7, the spin observable along a unit axis \(\boldsymbol{m}\) is \(\boldsymbol{m}\cdot\hat{\boldsymbol\sigma} = m_x\hat X + m_y\hat Y + m_z\hat Z\). Find the \(+1\) eigenstate explicitly for the axis \(\boldsymbol{m} = (\sin\theta_0, 0, \cos\theta_0)\) in the \(xz\)-plane.

(a) Write \(\boldsymbol{m}\cdot\hat{\boldsymbol\sigma}\) as an explicit \(2\times 2\) matrix.

(b) Solve the eigenvalue equation \((\boldsymbol{m}\cdot\hat{\boldsymbol\sigma})\vert\psi\rangle = +\vert\psi\rangle\). Show that the normalized \(+1\) eigenstate is \(\vert\psi\rangle = \cos(\theta_0/2)\vert 0\rangle + \sin(\theta_0/2)\vert 1\rangle\).

(c) Compute the Bloch vector of this \(\vert\psi\rangle\) from the formulas in Problem 3 and verify it equals \(\boldsymbol{m}\) itself — consistent with the result of 1.1.2 Problem 7 that the Bloch axis is the unique direction of maximal spin alignment.

Solution.

(a) With \(n_y = 0\),

\[\begin{split} \boldsymbol{m}\cdot\hat{\boldsymbol\sigma} = \sin\theta_0\,\hat X + \cos\theta_0\,\hat Z = \begin{pmatrix} \cos\theta_0 & \sin\theta_0 \\ \sin\theta_0 & -\cos\theta_0\end{pmatrix}. \end{split}\]

(b) The \(+1\) eigenvalue equation reads \((\boldsymbol{m}\cdot\hat{\boldsymbol\sigma} - \hat I)v = 0\), i.e.

\[\begin{split} \begin{pmatrix} \cos\theta_0 - 1 & \sin\theta_0 \\ \sin\theta_0 & -\cos\theta_0 - 1\end{pmatrix}\begin{pmatrix} a\\b\end{pmatrix} = 0. \end{split}\]

The first row gives \((\cos\theta_0 - 1)a + \sin\theta_0\,b = 0\), so \(b/a = (1-\cos\theta_0)/\sin\theta_0\). Applying the half-angle identities \(1-\cos\theta_0 = 2\sin^2(\theta_0/2)\) and \(\sin\theta_0 = 2\sin(\theta_0/2)\cos(\theta_0/2)\):

\[ \frac{b}{a} = \frac{2\sin^2(\theta_0/2)}{2\sin(\theta_0/2)\cos(\theta_0/2)} = \tan(\theta_0/2). \]

The natural normalised choice is \(a = \cos(\theta_0/2)\), \(b = \sin(\theta_0/2)\), satisfying \(a^2 + b^2 = 1\). Hence

\[ \vert\psi\rangle = \cos(\theta_0/2)\,\vert 0\rangle + \sin(\theta_0/2)\,\vert 1\rangle. \]

This is exactly the Bloch parametrization \(\vert\psi\rangle = \cos(\theta/2)\vert 0\rangle + \mathrm{e}^{\mathrm{i}\varphi}\sin(\theta/2)\vert 1\rangle\) with \(\theta = \theta_0\) and \(\varphi = 0\) — consistent with \(\boldsymbol{m}\) lying in the \(xz\)-plane (no azimuth).

\[ \langle\hat X\rangle = 2\cdot\tfrac12\sin\theta_0 = \sin\theta_0, \quad \langle\hat Y\rangle = 0, \quad \langle\hat Z\rangle = \cos^2(\theta_0/2) - \sin^2(\theta_0/2) = \cos\theta_0. \]

So \(\boldsymbol n = (\sin\theta_0, 0, \cos\theta_0) = \boldsymbol{m}\). The Bloch vector of the \(+1\) eigenstate of \(\boldsymbol{m}\cdot\hat{\boldsymbol\sigma}\) points along \(\boldsymbol{m}\) itself — saturating the alignment bound of 1.1.2 Problem 7. Geometrically, the \(+1\) eigenstate of spin along \(\boldsymbol{m}\) is the state whose Bloch arrow is \(\boldsymbol{m}\).

5. Pauli decomposition of a 2x2 Hermitian operator. Any Hermitian operator on the qubit Hilbert space can be written uniquely as

\[ \hat O = a_0\,\hat I + a_x\,\hat X + a_y\,\hat Y + a_z\,\hat Z, \]

with \(a_0, a_x, a_y, a_z\in\mathbb{R}\) (lecture’s “\(4\) real degrees of freedom” statement).

(a) Using the lecture’s trace identities \(\operatorname{Tr}(\hat\sigma^i) = 0\), \(\operatorname{Tr}(\hat I) = 2\), and \(\operatorname{Tr}(\hat\sigma^i\hat\sigma^j) = 2\delta_{ij}\), show that the coefficients are recovered by

\[ a_0 = \tfrac{1}{2}\operatorname{Tr}(\hat O), \qquad a_i = \tfrac{1}{2}\operatorname{Tr}(\hat O\,\hat\sigma^i)\quad (i\in\{x,y,z\}). \]

(b) Apply to \(\hat O = \begin{pmatrix} 3 & 2-\mathrm{i} \\ 2+\mathrm{i} & 1\end{pmatrix}\). Compute \(a_0, a_x, a_y, a_z\) and write the result.

(c) For an observable of the form \(\hat O = a_0\,\hat I + \boldsymbol a\cdot\hat{\boldsymbol\sigma}\) with \(\boldsymbol a = (a_x, a_y, a_z)\), identify (i) what \(a_0\) represents, (ii) what \(\vert\boldsymbol a\vert\) controls, and (iii) what the direction \(\boldsymbol{e}_a = \boldsymbol a/\vert\boldsymbol a\vert\) encodes. (Hint: use the result of Problem 4 and the spectral structure of \(\boldsymbol{n}\cdot\hat{\boldsymbol\sigma}\) from Problem 7.)

Solution.

(a) Apply \(\operatorname{Tr}\) to the decomposition. The Pauli operators are traceless and the identity has trace \(2\), so

\[ \operatorname{Tr}(\hat O) = a_0\,\operatorname{Tr}(\hat I) + \sum_i a_i\,\operatorname{Tr}(\hat\sigma^i) = 2a_0 + 0 \;\Longrightarrow\; a_0 = \tfrac{1}{2}\operatorname{Tr}(\hat O). \]

For each Pauli component, multiply by \(\hat\sigma^j\) on the right and take the trace. Using \(\hat\sigma^i\hat\sigma^j = \delta_{ij}\hat I + \mathrm{i}\epsilon_{ijk}\hat\sigma^k\) (lecture eq-pauli-multiplication), \(\operatorname{Tr}(\hat I) = 2\), and \(\operatorname{Tr}(\hat\sigma^k) = 0\):

\[ \operatorname{Tr}(\hat O\,\hat\sigma^j) = a_0\operatorname{Tr}(\hat\sigma^j) + \sum_i a_i\operatorname{Tr}(\hat\sigma^i\hat\sigma^j) = 0 + \sum_i a_i\cdot 2\delta_{ij} = 2 a_j \;\Longrightarrow\; a_j = \tfrac{1}{2}\operatorname{Tr}(\hat O\,\hat\sigma^j). \]

(b) Compute each trace.

\(\operatorname{Tr}(\hat O) = 3 + 1 = 4\), so \(a_0 = 2\).

\(\operatorname{Tr}(\hat O\,\hat X)\): multiply \(\hat O\) by \(\hat X = \begin{pmatrix}0&1\\1&0\end{pmatrix}\), which swaps columns:

\[\begin{split} \hat O\,\hat X = \begin{pmatrix} 3 & 2-\mathrm{i} \\ 2+\mathrm{i} & 1\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix} = \begin{pmatrix} 2-\mathrm{i} & 3 \\ 1 & 2+\mathrm{i}\end{pmatrix}, \qquad \operatorname{Tr}(\hat O\hat X) = (2-\mathrm{i}) + (2+\mathrm{i}) = 4. \end{split}\]

So \(a_x = 2\).

\(\operatorname{Tr}(\hat O\,\hat Y)\): with \(\hat Y = \begin{pmatrix}0 & -\mathrm{i}\\ \mathrm{i} & 0\end{pmatrix}\),

\[\begin{split} \hat O\,\hat Y = \begin{pmatrix} 3 & 2-\mathrm{i} \\ 2+\mathrm{i} & 1\end{pmatrix}\begin{pmatrix}0 & -\mathrm{i}\\ \mathrm{i} & 0\end{pmatrix} = \begin{pmatrix} \mathrm{i}(2-\mathrm{i}) & -3\mathrm{i} \\ \mathrm{i} & -\mathrm{i}(2+\mathrm{i})\end{pmatrix} = \begin{pmatrix} 1 + 2\mathrm{i} & -3\mathrm{i} \\ \mathrm{i} & 1 - 2\mathrm{i}\end{pmatrix}, \end{split}\]

\[ \operatorname{Tr}(\hat O\hat Y) = (1 + 2\mathrm{i}) + (1 - 2\mathrm{i}) = 2 \;\Longrightarrow\; a_y = 1. \]

\(\operatorname{Tr}(\hat O\,\hat Z)\): with \(\hat Z = \begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}\), \(\hat O\hat Z\) flips the sign of the second column, leaving the diagonal as \(3\) and \(-1\):

\[ \operatorname{Tr}(\hat O\,\hat Z) = 3 + (-1) = 2 \;\Longrightarrow\; a_z = 1. \]

So \(\hat O = 2\hat I + 2\hat X + \hat Y + \hat Z\). (Sanity: rebuild the matrix — \(2\hat X = \begin{pmatrix}0&2\\2&0\end{pmatrix}\), \(\hat Y = \begin{pmatrix}0&-\mathrm{i}\\ \mathrm{i}&0\end{pmatrix}\), \(\hat Z = \begin{pmatrix}1&0\\0&-1\end{pmatrix}\). Summing with \(2\hat I\) gives \(\begin{pmatrix}3 & 2-\mathrm{i}\\ 2+\mathrm{i} & 1\end{pmatrix}\). ✓)

\(a_0\) is the trace-average of \(\hat O\). It is an overall shift of every measured eigenvalue (an additive constant in the spectrum): replacing \(\hat O\) by \(\hat O + c\hat I\) adds \(c\) to each outcome without changing eigenstates or any differences between measurement results (splittings, commutators, etc.). Only when \(\hat O\) is the Hamiltonian does this constant read as an energy offset; for a general observable it is simply an unobservable overall shift of reported values.
\(\vert\boldsymbol a\vert\) sets the eigenvalue spread. By Problem 7, \((\boldsymbol{e}_a\cdot\hat{\boldsymbol\sigma})^2 = \hat I\), so \(\boldsymbol{e}_a\cdot\hat{\boldsymbol\sigma}\) has eigenvalues \(\pm 1\). Therefore \(\hat O = a_0\hat I + \vert\boldsymbol a\vert(\boldsymbol{e}_a\cdot\hat{\boldsymbol\sigma})\) has eigenvalues \(a_0 \pm \vert\boldsymbol a\vert\), separated by \(2\vert\boldsymbol a\vert\).
\(\boldsymbol{e}_a = \boldsymbol a/\vert\boldsymbol a\vert\) encodes the Bloch axis of the eigenstates: by Problem 4, the \(+1\) eigenstate of \(\boldsymbol{e}_a\cdot\hat{\boldsymbol\sigma}\) has Bloch vector \(\boldsymbol{e}_a\), and the \(-1\) eigenstate has Bloch vector \(-\boldsymbol{e}_a\) (antipodal, 1.1.2 Problem 5). The direction of \(\boldsymbol a\) thus determines the spatial axis of the observable; its magnitude determines the eigenvalue spread; the additive \(a_0\) is an unobservable overall shift of measured eigenvalues.

For the example above, \(\boldsymbol a = (2, 1, 1)\), \(\vert\boldsymbol a\vert = \sqrt{6}\), \(\boldsymbol{e}_a = (2,1,1)/\sqrt 6\), and eigenvalues \(2 \pm \sqrt 6\).

6. Non-Hermitian eigenstates. The lecture proves that distinct eigenvalues of a Hermitian operator yield orthogonal eigenvectors. Investigate what happens when the operator is not Hermitian.

Consider the (non-Hermitian) operator

\[\begin{split} \hat A = \begin{pmatrix} 1 & 1 \\ 0 & 2\end{pmatrix}. \end{split}\]

(a) Verify that \(\hat A\) is not Hermitian.

(b) Find the two eigenvalues by solving the characteristic equation.

(c) Find the corresponding eigenvectors \(\vert v_1\rangle, \vert v_2\rangle\) (normalised). Compute the inner product \(\langle v_1\vert v_2\rangle\) and check whether it vanishes.

(d) Identify the step in the lecture’s proof of “orthogonal eigenvectors for distinct eigenvalues” that required Hermiticity. Why does the argument fail for \(\hat A\)?

Solution.

(a) The transpose-conjugate is \(\hat A^\dagger = \begin{pmatrix} 1 & 0 \\ 1 & 2\end{pmatrix} \neq \hat A\) (the upper-right \(1\) versus the lower-left \(0\) already differ). Not Hermitian.

(b) Eigenvalues from \(\det(\hat A - \lambda\hat I) = 0\):

\[ (1 - \lambda)(2 - \lambda) - (1)(0) = 0 \;\Longrightarrow\; \lambda_1 = 1,\ \lambda_2 = 2. \]

(For an upper-triangular matrix, the eigenvalues are simply the diagonal entries.)

(c) Eigenvector for \(\lambda_1 = 1\): \((\hat A - \hat I)v = 0\) gives \(\begin{pmatrix} 0 & 1 \\ 0 & 1\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix} = 0\), so \(b = 0\) and \(a\) is free. Normalised:

\[\begin{split} \vert v_1\rangle = \begin{pmatrix} 1\\ 0\end{pmatrix} = \vert 0\rangle. \end{split}\]

Eigenvector for \(\lambda_2 = 2\): \((\hat A - 2\hat I)v = 0\) gives \(\begin{pmatrix} -1 & 1 \\ 0 & 0\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix} = 0\), so \(a = b\). Normalised:

\[\begin{split} \vert v_2\rangle = \frac{1}{\sqrt 2}\begin{pmatrix} 1\\ 1\end{pmatrix} = \vert+\rangle. \end{split}\]

Inner product:

\[ \langle v_1\vert v_2\rangle = \langle 0\vert+\rangle = \frac{1}{\sqrt 2} \neq 0. \]

The eigenvectors are not orthogonal, despite belonging to distinct eigenvalues.

(d) The lecture’s proof writes

\[ \langle v_1\vert\hat O\vert v_2\rangle = \lambda_2\langle v_1\vert v_2\rangle \quad\text{(from }\hat O\vert v_2\rangle = \lambda_2\vert v_2\rangle\text{)}. \]

To get a second expression with \(\lambda_1\), one uses Hermiticity to act \(\hat O\) to the left:

\[ \langle v_1\vert\hat O\vert v_2\rangle = \langle\hat O^\dagger v_1\vert v_2\rangle = \langle\hat O v_1\vert v_2\rangle = \lambda_1^*\langle v_1\vert v_2\rangle. \]

The last step uses \(\hat O^\dagger = \hat O\) (Hermiticity) and \(\lambda_1^* = \lambda_1\) (which itself requires Hermiticity). Equating the two expressions gives \((\lambda_2 - \lambda_1)\langle v_1\vert v_2\rangle = 0\), hence orthogonality for distinct eigenvalues.

For our \(\hat A\), \(\hat A^\dagger \neq \hat A\), so the “act to the left” step does not turn \(\hat A^\dagger\) into \(\hat A\). The chain of equalities breaks, and the conclusion fails. Hermiticity is the load-bearing assumption. This is the operational reason quantum mechanics promotes observables to Hermitian operators: not just to ensure real eigenvalues, but to guarantee an orthonormal eigenbasis — without which the spectral decomposition and the Born rule would have no clean home.

7. Pauli square along an arbitrary axis. Use the Pauli multiplication law \(\hat\sigma^i\hat\sigma^j = \delta_{ij}\hat I + \mathrm{i}\epsilon_{ijk}\hat\sigma^k\) to prove that, for any unit vector \(\boldsymbol{n} = (n_x, n_y, n_z)\),

\[ (\boldsymbol{n}\cdot\hat{\boldsymbol\sigma})^2 = \hat I. \]

(a) Expand \((\boldsymbol{n}\cdot\hat{\boldsymbol\sigma})^2 = \sum_{i,j} n_i n_j\,\hat\sigma^i\hat\sigma^j\) using the Pauli multiplication law. Identify the two contributions (symmetric and antisymmetric in \(i,j\)).

(b) Use the symmetry of \(n_i n_j\) and the antisymmetry of \(\epsilon_{ijk}\) to argue that the \(\hat\sigma^k\) term vanishes after summation.

(c) Use the result to deduce that every eigenvalue of \(\boldsymbol{n}\cdot\hat{\boldsymbol\sigma}\) satisfies \(\lambda^2 = 1\), so \(\lambda = \pm 1\). Why does this confirm that “every spin observable along any axis has the same two-point spectrum \(\{+1, -1\}\)” — and not, say, more outcomes when \(\boldsymbol{n}\) is far from one of the coordinate axes?

Solution.

(a) Expanding the product and applying the multiplication law:

\[ (\boldsymbol{n}\cdot\hat{\boldsymbol\sigma})^2 = \sum_{i,j} n_i n_j\,\hat\sigma^i\hat\sigma^j = \sum_{i,j} n_i n_j\bigl(\delta_{ij}\hat I + \mathrm{i}\epsilon_{ijk}\hat\sigma^k\bigr) = \underbrace{\sum_i n_i^2\,\hat I}_{\text{diagonal part}} + \mathrm{i}\underbrace{\sum_{i,j,k} n_i n_j\,\epsilon_{ijk}\,\hat\sigma^k}_{\text{off-diagonal part}}. \]

The diagonal part collapses via \(\delta_{ij}\) to \(\sum_i n_i^2\), which equals \(\vert\boldsymbol{n}\vert^2 = 1\). So this contribution is \(\hat I\).

(b) The off-diagonal part contracts the symmetric tensor \(n_i n_j\) (since \(n_i n_j = n_j n_i\)) with the antisymmetric tensor \(\epsilon_{ijk}\). Any such contraction vanishes:

\[ \sum_{i,j} n_i n_j\,\epsilon_{ijk} = \tfrac{1}{2}\sum_{i,j}(n_i n_j - n_j n_i)\,\epsilon_{ijk} = 0, \]

(making the contracted pair manifestly antisymmetric/symmetric). Hence the off-diagonal term is zero, and

\[ (\boldsymbol{n}\cdot\hat{\boldsymbol\sigma})^2 = \hat I. \]

(c) If \(\lambda\) is an eigenvalue with eigenvector \(\vert v\rangle\), applying the operator twice gives \((\boldsymbol{n}\cdot\hat{\boldsymbol\sigma})^2\vert v\rangle = \lambda^2\vert v\rangle\). But the square equals \(\hat I\), so \(\lambda^2\vert v\rangle = \vert v\rangle\), forcing

\[ \lambda^2 = 1 \quad\Longrightarrow\quad \lambda = \pm 1. \]

There are no additional eigenvalues: the spin observable along any axis returns exactly two possible outcomes, \(+1\) and \(-1\). This is a special, geometry-independent statement — the \(z\)-axis carries no privileged spectrum, and a “tilted” spin observable along an off-axis \(\boldsymbol{n}\) still gives the same dichotomous outcome set. (Misconception check: one might guess that tilting \(\boldsymbol{n}\) produces intermediate eigenvalues like \(\pm n_z\). Problem 7 rules this out: the spectrum is invariant under rotation of the measurement axis; only the eigenstates rotate.)

8. Pauli multiplication from commutator and anti-commutator. The lecture gives two relations: the commutator \([\hat\sigma^i, \hat\sigma^j] = 2\mathrm{i}\epsilon_{ijk}\hat\sigma^k\) and the anti-commutator \(\{\hat\sigma^i, \hat\sigma^j\} = 2\delta_{ij}\hat I\). Use them to derive the product \(\hat\sigma^i\hat\sigma^j\) without computing any matrices.

(a) Write the defining identities for the operator product:

\[ \hat\sigma^i\hat\sigma^j = \tfrac{1}{2}\bigl(\{\hat\sigma^i,\hat\sigma^j\} + [\hat\sigma^i,\hat\sigma^j]\bigr). \]

Substitute the lecture’s commutator and anti-commutator to recover the Pauli multiplication law \(\hat\sigma^i\hat\sigma^j = \delta_{ij}\hat I + \mathrm{i}\epsilon_{ijk}\hat\sigma^k\).

(b) Specialise to \((i,j) = (1,2)\) to compute \(\hat X\hat Y\) and \(\hat Y\hat X\) directly. Verify \(\hat X\hat Y = \mathrm{i}\hat Z\) and \(\hat Y\hat X = -\mathrm{i}\hat Z\).

(c) The decomposition “product = (anti-commutator + commutator)/2” works for any pair of operators. Explain in one sentence why this identity is structural: every operator product splits into a symmetric (anti-commutator) and an antisymmetric (commutator) part, independent of whether the operators are Hermitian, unitary, or arbitrary.

Solution.

(a) Add and subtract the two identities. For any operators \(\hat A, \hat B\),

\[ \{\hat A,\hat B\} + [\hat A,\hat B] = (\hat A\hat B + \hat B\hat A) + (\hat A\hat B - \hat B\hat A) = 2\hat A\hat B, \]

so \(\hat A\hat B = \tfrac{1}{2}(\{\hat A,\hat B\} + [\hat A,\hat B])\). Applied to Pauli operators and substituting the lecture’s data,

\[ \hat\sigma^i\hat\sigma^j = \tfrac{1}{2}\bigl(2\delta_{ij}\hat I + 2\mathrm{i}\epsilon_{ijk}\hat\sigma^k\bigr) = \delta_{ij}\hat I + \mathrm{i}\epsilon_{ijk}\hat\sigma^k. \]

This is the lecture’s multiplication law — recovered without computing any \(2\times 2\) matrices.

(b) For \((i,j) = (1,2)\) (i.e. \((\hat X,\hat Y)\)), \(\delta_{ij} = 0\) and \(\epsilon_{ijk} = \epsilon_{12k}\) is nonzero only at \(k = 3\), with \(\epsilon_{123} = +1\). So

\[ \hat X\hat Y = 0\cdot\hat I + \mathrm{i}\cdot(+1)\cdot\hat Z = \mathrm{i}\hat Z. \]

Swapping order, \(\epsilon_{21k}\) has \(\epsilon_{213} = -1\), so

\[ \hat Y\hat X = \mathrm{i}\cdot(-1)\cdot\hat Z = -\mathrm{i}\hat Z. \]

(Cross-check via the anti-commutator: \(\hat X\hat Y + \hat Y\hat X = \mathrm{i}\hat Z - \mathrm{i}\hat Z = 0\), agreeing with \(\{\hat X,\hat Y\} = 0\) from anti-commutation.)

\[ \hat A\hat B = \underbrace{\tfrac12(\hat A\hat B + \hat B\hat A)}_{\text{symmetric, anti-commutator}} + \underbrace{\tfrac12(\hat A\hat B - \hat B\hat A)}_{\text{antisymmetric, commutator}}. \]

The split into symmetric + antisymmetric parts under exchange \(\hat A \leftrightarrow \hat B\) is purely algebraic — it makes no reference to Hermiticity, unitarity, or the dimension of the Hilbert space. It is the operator-theoretic analogue of writing a matrix as the sum of its symmetric and antisymmetric parts under transposition. The Pauli multiplication law is the special case where both the commutator and anti-commutator are particularly simple (proportional to Pauli operators and to the identity, respectively), but the structure of the derivation is universal.