Mathematical Foundations

Solution:

(a) By Euler: $e^{i\theta} = \cos\theta + i\sin\theta$, so $|e^{i\theta}| = \sqrt{\cos^2\theta + \sin^2\theta} = 1$. Every point on the unit circle has unit modulus. Thus $|e^{i\pi/3}| = 1$.

(b) $\text{Re}(e^{i\pi/3}) = \cos(\pi/3) = \tfrac{1}{2}$.

(c) $e^{i\pi/3}\cdot e^{-i\pi/3} = e^{i(\pi/3 - \pi/3)} = e^0 = 1$. In general $z \cdot z^* = |z|^2$; since $|e^{i\theta}| = 1$ we get $e^{i\theta}(e^{i\theta})^* = e^{i\theta}e^{-i\theta} = 1$.

QM connection: Time-evolution operators $e^{-i\hat{H}t/\hbar}$ are unitary precisely because $|e^{-iEt/\hbar}| = 1$ — evolution preserves the norm (total probability).

Solution:

(a) $z_1^* = 3 - 4i$.

(b) $|z_1| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$.

(c) $z_1 z_2 = (3+4i)(1-2i) = 3 - 6i + 4i - 8i^2 = 3 - 2i + 8 = 11 - 2i$.

(d) $z_1^* z_2 = (3-4i)(1-2i) = 3 - 6i - 4i + 8i^2 = 3 - 10i - 8 = -5 - 10i$. And $z_1 z_2^* = (3+4i)(1+2i) = 3+6i+4i+8i^2 = 3+10i-8 = -5+10i$. Sum: $(-5-10i) + (-5+10i) = -10 \in \mathbb{R}$. The imaginary parts cancel because $z_1^*z_2 + z_1 z_2^* = 2\,\text{Re}(z_1^* z_2)$ always.

Solution:

First, $\langle\psi| = \tfrac{1}{\sqrt{2}}(1^*, (i)^*) = \tfrac{1}{\sqrt{2}}(1, -i)$.

$\langle\psi|\psi\rangle = \tfrac{1}{2}(1\cdot 1 + (-i)(i)) = \tfrac{1}{2}(1 + (-i^2)) = \tfrac{1}{2}(1+1) = 1$. Normalised ✓

$\langle\phi|\phi\rangle$: $\langle\phi| = \tfrac{1}{\sqrt{2}}(1, i)$. So $\langle\phi|\phi\rangle = \tfrac{1}{2}(1 + i(-i)) = \tfrac{1}{2}(1+1) = 1$. Normalised ✓

$\langle\psi|\phi\rangle = \tfrac{1}{2}(1\cdot 1 + (-i)(-i)) = \tfrac{1}{2}(1 + i^2) = \tfrac{1}{2}(1-1) = 0$. These are orthogonal.

QM connection: $|\psi\rangle$ and $|\phi\rangle$ form an orthonormal basis for $\mathbb{C}^2$ — they could represent the two eigenstates of a spin operator (e.g., eigenstates of $\hat{S}_y$).

Hermitian check: $\hat{\sigma}_z^\dagger = (\hat{\sigma}_z^*)^T = \hat{\sigma}_z$ since all entries are real. Hermitian ✓

Eigenvalues: $\det(\hat{\sigma}_z - \lambda\mathbf{1}) = (1-\lambda)(-1-\lambda) = -(1-\lambda^2) = \lambda^2 - 1 = 0$, so $\lambda = \pm 1$.

Eigenvectors:

$\lambda = +1$: $(\hat{\sigma}_z - \mathbf{1})\mathbf{v} = 0 \Rightarrow \begin{pmatrix}0&0\\0&-2\end{pmatrix}\mathbf{v} = 0 \Rightarrow v_2 = 0$. Eigenstate: $|{+}\rangle = \begin{pmatrix}1\\0\end{pmatrix}$ (spin up).

$\lambda = -1$: $(\hat{\sigma}_z + \mathbf{1})\mathbf{v} = 0 \Rightarrow \begin{pmatrix}2&0\\0&0\end{pmatrix}\mathbf{v} = 0 \Rightarrow v_1 = 0$. Eigenstate: $|{-}\rangle = \begin{pmatrix}0\\1\end{pmatrix}$ (spin down).

Measurement outcomes: The Spectral Theorem guarantees the outcomes are the eigenvalues: $\pm 1$ (in units of $\hbar/2$ for actual spin). This is the spin-$\tfrac{1}{2}$ measurement in the $z$-direction.

(a) Normalization:

$\int_{-\infty}^{\infty}|A|^2 e^{-x^2}dx = |A|^2\sqrt{\pi} = 1 \Rightarrow A = \pi^{-1/4}$.

(b) Expectation value of position:

$\langle x\rangle = \int_{-\infty}^{\infty}\psi^*(x)\cdot x\cdot\psi(x)\,dx = |A|^2\int_{-\infty}^{\infty}x\,e^{-x^2}dx$.

The integrand $x\,e^{-x^2}$ is an odd function (product of odd $x$ and even $e^{-x^2}$), integrated over a symmetric interval. Therefore $\langle x\rangle = 0$.

QM connection: This is the ground state of the quantum harmonic oscillator (with $\hbar = m = \omega = 1$). The particle is most likely found at the origin, and on average is found there: $\langle x\rangle = 0$.

(a) Normalization:

$|c_+|^2 + |c_-|^2 = \left(\tfrac{\sqrt{3}}{2}\right)^2 + \left(\tfrac{1}{2}\right)^2 = \tfrac{3}{4} + \tfrac{1}{4} = 1$ ✓

(b) Probabilities (Born Rule):

$P(+1) = |\langle{+}|\psi\rangle|^2 = |c_+|^2 = \tfrac{3}{4}$

$P(-1) = |\langle{-}|\psi\rangle|^2 = |c_-|^2 = \tfrac{1}{4}$

(c) Expectation value:

$\langle\hat{\sigma}_z\rangle = (+1)\cdot P(+1) + (-1)\cdot P(-1) = \tfrac{3}{4} - \tfrac{1}{4} = \tfrac{1}{2}$

Alternatively: $\langle\hat{\sigma}_z\rangle = \langle\psi|\hat{\sigma}_z|\psi\rangle = c_+^* c_+ (+1) + c_-^* c_- (-1) = \tfrac{3}{4} - \tfrac{1}{4} = \tfrac{1}{2}$.

Compute $\langle\hat{\sigma}_z^2\rangle$:

Since eigenvalues are $\pm 1$, we have $\hat{\sigma}_z^2|{\pm}\rangle = (\pm 1)^2|{\pm}\rangle = |{\pm}\rangle$, so $\hat{\sigma}_z^2 = \mathbf{1}$.

$\langle\hat{\sigma}_z^2\rangle = \langle\psi|\mathbf{1}|\psi\rangle = \langle\psi|\psi\rangle = 1$.

Uncertainty:

$\Delta\sigma_z = \sqrt{1 - \left(\tfrac{1}{2}\right)^2} = \sqrt{1 - \tfrac{1}{4}} = \sqrt{\tfrac{3}{4}} = \tfrac{\sqrt{3}}{2} \approx 0.866$.

Interpretation: $\Delta\Omega = 0$ if and only if $|\psi\rangle$ is an eigenstate of $\hat{\Omega}$. In an eigenstate, measurement always returns the same eigenvalue with certainty — no spread.

Here $|\psi\rangle$ is a superposition, so $\Delta\sigma_z > 0$: there is genuine spread in the measurement outcomes (+1 with probability $\tfrac{3}{4}$, $-1$ with probability $\tfrac{1}{4}$).

Solution:

$|{+}\rangle\langle{+}| = \begin{pmatrix}1\\0\end{pmatrix}(1\;0) = \begin{pmatrix}1&0\\0&0\end{pmatrix}$

$|{-}\rangle\langle{-}| = \begin{pmatrix}0\\1\end{pmatrix}(0\;1) = \begin{pmatrix}0&0\\0&1\end{pmatrix}$

Sum: $\begin{pmatrix}1&0\\0&0\end{pmatrix} + \begin{pmatrix}0&0\\0&1\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix} = \mathbf{1}$ ✓

Application: Insert this identity into any expression to expand in the $\hat{\sigma}_z$ basis: $|\psi\rangle = \mathbf{1}|\psi\rangle = (|{+}\rangle\langle{+}| + |{-}\rangle\langle{-}|)|\psi\rangle = \langle{+}|\psi\rangle|{+}\rangle + \langle{-}|\psi\rangle|{-}\rangle$.

(a) Hermitian conjugate:

For an outer product $(|a\rangle\langle b|)^\dagger = |b\rangle\langle a|$. So $\hat{\sigma}_+^\dagger = (|{+}\rangle\langle{-}|)^\dagger = |{-}\rangle\langle{+}| = \hat{\sigma}_-$ (lowering operator).

(b) Matrix representation:

$\hat{\sigma}_+ = |{+}\rangle\langle{-}| = \begin{pmatrix}1\\0\end{pmatrix}(0\;1) = \begin{pmatrix}0&1\\0&0\end{pmatrix}$

$\hat{\sigma}_- = |{-}\rangle\langle{+}| = \begin{pmatrix}0\\1\end{pmatrix}(1\;0) = \begin{pmatrix}0&0\\1&0\end{pmatrix}$

(c) Product:

$\hat{\sigma}_+^\dagger\hat{\sigma}_+ = \hat{\sigma}_-\hat{\sigma}_+ = |{-}\rangle\langle{+}|\,|{+}\rangle\langle{-}| = |{-}\rangle(\langle{+}|{+}\rangle)\langle{-}| = |{-}\rangle\cdot 1\cdot\langle{-}| = |{-}\rangle\langle{-}|$ ✓

This is the projection operator onto spin-down — as expected, $\hat{\sigma}_+^\dagger\hat{\sigma}_+$ counts the probability of the state being in $|{-}\rangle$.

Solution:

$\hat{\sigma}_+\hat{\sigma}_- = \begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}0&0\\1&0\end{pmatrix} = \begin{pmatrix}1&0\\0&0\end{pmatrix} = |{+}\rangle\langle{+}|$

$\hat{\sigma}_-\hat{\sigma}_+ = \begin{pmatrix}0&0\\1&0\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix} = \begin{pmatrix}0&0\\0&1\end{pmatrix} = |{-}\rangle\langle{-}|$

$[\hat{\sigma}_+, \hat{\sigma}_-] = \hat{\sigma}_+\hat{\sigma}_- - \hat{\sigma}_-\hat{\sigma}_+ = \begin{pmatrix}1&0\\0&0\end{pmatrix} - \begin{pmatrix}0&0\\0&1\end{pmatrix} = \begin{pmatrix}1&0\\0&-1\end{pmatrix} = \hat{\sigma}_z$ ✓

QM connection: This is one of the $\mathfrak{su}(2)$ Lie algebra relations. The three Pauli matrices $\hat{\sigma}_{x,y,z}$ satisfy $[\hat{\sigma}_i, \hat{\sigma}_j] = 2i\varepsilon_{ijk}\hat{\sigma}_k$, which is the fundamental commutation relation of angular momentum.

Step 1 — Normalize $|v_1\rangle$:

$\|v_1\| = \sqrt{1^2 + 1^2} = \sqrt{2}$. So $|e_1\rangle = \tfrac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}$.

Step 2 — Remove component of $|v_2\rangle$ along $|e_1\rangle$:

$\langle e_1|v_2\rangle = \tfrac{1}{\sqrt{2}}(1\cdot 1 + 1\cdot 0) = \tfrac{1}{\sqrt{2}}$

$|v_2'\rangle = |v_2\rangle - \langle e_1|v_2\rangle|e_1\rangle = \begin{pmatrix}1\\0\end{pmatrix} - \tfrac{1}{\sqrt{2}}\cdot\tfrac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix} - \begin{pmatrix}1/2\\1/2\end{pmatrix} = \begin{pmatrix}1/2\\-1/2\end{pmatrix}$

Step 3 — Normalize:

$\|v_2'\| = \sqrt{(1/2)^2 + (-1/2)^2} = \tfrac{1}{\sqrt{2}}$. So $|e_2\rangle = \tfrac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}$.

Verification: $\langle e_1|e_2\rangle = \tfrac{1}{\sqrt{2}}\cdot\tfrac{1}{\sqrt{2}}(1\cdot 1 + 1\cdot(-1)) = \tfrac{1}{2}(0) = 0$ ✓

(a) State after measuring $+1$:

By the Projection Postulate, the state collapses to the eigenstate corresponding to the measured eigenvalue: $|\psi'\rangle = |{+}\rangle$.

(b) Immediately measure again:

$P(+1\,|\,\psi') = |\langle{+}|{+}\rangle|^2 = 1$ and $P(-1\,|\,\psi') = |\langle{-}|{+}\rangle|^2 = 0$.

Conclusion: Measuring twice in immediate succession always yields the same result. This is a fundamental feature of quantum measurement.

(c) If we wait:

$|\psi'(t)\rangle = e^{-i\hat{H}t/\hbar}|{+}\rangle$. If the Hamiltonian doesn't commute with $\hat{\sigma}_z$ (e.g., $\hat{H} \propto \hat{\sigma}_x$), the state will evolve away from $|{+}\rangle$, and a subsequent $\hat{\sigma}_z$ measurement will no longer be certain to yield $+1$.

$\hat{A}$ is Hermitian: $\hat{A}^* = \begin{pmatrix}0 & -i \\ i & 0\end{pmatrix}$, then $(\hat{A}^*)^T = \begin{pmatrix}0 & i \\ -i & 0\end{pmatrix} = \hat{A}$. So $\hat{A}^\dagger = \hat{A}$ ✓. This is $\hat{\sigma}_y$ (the $y$-Pauli matrix).

$\hat{B}$ is anti-Hermitian: $\hat{B}^* = \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$, so $(\hat{B}^*)^T = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix} = -\hat{B}$ ✓.

Eigenvalues of $\hat{B}$: $\det(\hat{B} - \lambda\mathbf{1}) = \lambda^2 + 1 = 0 \Rightarrow \lambda = \pm i$. Purely imaginary — consistent with anti-Hermitian operators having purely imaginary (or zero) eigenvalues.

QM implication: Only Hermitian operators can be observables (real eigenvalues). Anti-Hermitian operators do not correspond to measurable quantities; they generate unitary transformations via $e^{\hat{B}}$.

$\langle\psi| = \tfrac{1}{\sqrt{2}}(1\;1)$

$\langle\hat{\sigma}_x\rangle$: $\hat{\sigma}_x|\psi\rangle = \tfrac{1}{\sqrt{2}}\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix} = \tfrac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix} = |\psi\rangle$. So $\langle\hat{\sigma}_x\rangle = \langle\psi|\psi\rangle = 1$.

Physical meaning: $|\psi\rangle$ is the eigenstate of $\hat{\sigma}_x$ with eigenvalue $+1$! So measurement of $\sigma_x$ always gives $+1$ with certainty.

$\langle\hat{\sigma}_y\rangle$: $\hat{\sigma}_y|\psi\rangle = \tfrac{1}{\sqrt{2}}\begin{pmatrix}-i\\i\end{pmatrix}$. Then $\langle\hat{\sigma}_y\rangle = \tfrac{1}{\sqrt{2}}\cdot\tfrac{1}{\sqrt{2}}(1\cdot(-i) + 1\cdot i) = \tfrac{1}{2}(0) = 0$.

$\langle\hat{\sigma}_z\rangle$: $\hat{\sigma}_z|\psi\rangle = \tfrac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}$. Then $\langle\hat{\sigma}_z\rangle = \tfrac{1}{2}(1\cdot 1 + 1\cdot(-1)) = 0$.

Summary: The spin vector points in the $+x$ direction: $(\langle\sigma_x\rangle, \langle\sigma_y\rangle, \langle\sigma_z\rangle) = (1, 0, 0)$.

Time evolution operator:

$\hat{U}(t) = e^{-i\hat{H}t/\hbar} = e^{-i\omega t\hat{\sigma}_z/2}$

Since $|{+}\rangle$ is an eigenstate of $\hat{\sigma}_z$ with eigenvalue $+1$:

$e^{-i\omega t\hat{\sigma}_z/2}|{+}\rangle = e^{-i\omega t/2}|{+}\rangle$

Time-evolved state: $|\psi(t)\rangle = e^{-i\omega t/2}|{+}\rangle$

Probabilities:

$P(+1, t) = |\langle{+}|\psi(t)\rangle|^2 = |e^{-i\omega t/2}|^2 = 1$

$P(-1, t) = |\langle{-}|\psi(t)\rangle|^2 = 0$

Interpretation: An energy eigenstate acquires only a global phase $e^{-iEt/\hbar}$ under time evolution. The probabilities of any measurement outcome are time-independent — this is a stationary state. The phase is physical only when the state is a superposition of different energy eigenstates.