Lecture 04 · Dynamics

Time Evolution & the Schrödinger Equation

The time-evolution operator, unitarity, the Hamiltonian, the Schrödinger equation, expectation values, commutators, and the bridge to classical mechanics.

Theory Exercises

Part I — Recap of the Four Postulates

Lecture 3 established the formal structure of quantum mechanics through four postulates. Here is the complete list:

State Space. The state of a quantum system is represented by a normalized vector $|\psi\rangle$ in a complex Hilbert space.
Observables. Every measurable physical quantity corresponds to a Hermitian (self-adjoint) operator $L = L^\dagger$ on the state space.
Measurement Outcomes. The only possible results of a measurement of $L$ are its eigenvalues $\lambda_i$, where $L|\lambda_i\rangle = \lambda_i|\lambda_i\rangle$.
Born Rule. If the system is in state $|\psi\rangle$, the probability of measuring eigenvalue $\lambda_i$ is $P(\lambda_i) = |\langle\lambda_i|\psi\rangle|^2$. After measurement, the state collapses to $|\lambda_i\rangle$.

These four postulates describe the structure of quantum mechanics — what states are, what observables are, and what measurement does. But they say nothing about how a state changes with time when no measurement is made. That requires a fifth principle.

Part II — The Time-Evolution Operator

We want to describe how the state $|s\rangle$ changes between measurements. The fundamental assumption is that time evolution is linear — the same operator $U(t)$ acts on every initial state:

Time-Evolved State

$$|s(t)\rangle = U(t)|s(0)\rangle$$

Linearity is not optional: it follows from the superposition principle. If each basis state evolves by $U(t)$, any superposition must evolve by the same $U(t)$.

Conservation of Information — the Minus-First Law

Susskind calls it the minus-first law of physics: information is never lost. More precisely, distinct quantum states must remain distinct at all times. If $|\psi(0)\rangle \neq |\phi(0)\rangle$, then $|\psi(t)\rangle \neq |\phi(t)\rangle$ for all $t$. Equivalently:

\langle\psi(t)|\phi(t)\rangle = \langle\psi(0)|\phi(0)\rangle \quad \text{for all states and all } t

In particular, normalization is preserved: a normalized state stays normalized. This is a conservation law deeper than energy conservation.

The Unitary Condition

Substituting $|\psi(t)\rangle = U|\psi(0)\rangle$ and $\langle\psi(t)| = \langle\psi(0)|U^\dagger$ into the inner product conservation:

\langle\psi(0)|U^\dagger U|\phi(0)\rangle = \langle\psi(0)|\phi(0)\rangle \quad \text{for all } |\psi(0)\rangle, |\phi(0)\rangle

This must hold for all pairs of states, which forces:

Unitary Condition

$$U^\dagger(t)\,U(t) = \mathbf{1}$$

An operator satisfying $U^\dagger U = \mathbf{1}$ is called unitary. Unitary operators are the quantum mechanical counterparts of classical symmetry transformations that preserve phase space volume.

Part III — From Unitarity to the Hamiltonian

For an infinitesimally small time step $\varepsilon$, $U(\varepsilon)$ must be close to the identity. Write it as a Taylor expansion:

U(\varepsilon) = \mathbf{1} - i\varepsilon H + O(\varepsilon^2)

where $H$ is some operator to be determined. Now apply the unitary condition $U^\dagger U = \mathbf{1}$:

U^\dagger(\varepsilon)\,U(\varepsilon) = (\mathbf{1} + i\varepsilon H^\dagger)(\mathbf{1} - i\varepsilon H) = \mathbf{1} + i\varepsilon(H^\dagger - H) + O(\varepsilon^2) = \mathbf{1}

Matching coefficients of $\varepsilon$ gives $H^\dagger - H = 0$, i.e.:

The Hamiltonian is Hermitian

$$H^\dagger = H$$

The generator of time evolution must be a Hermitian operator — which by Postulate 2 means it is an observable. This operator is called the Hamiltonian, $H$. Its specific form depends on the physical system. For a particle in a potential, $H = p^2/2m + V(x)$. For a spin in a magnetic field, $H = \omega\sigma_z$. The abstract structure is universal.

Part IV — The Schrödinger Equation

Apply the infinitesimal evolution to the state $|s(t)\rangle$:

|s(t+\varepsilon)\rangle = U(\varepsilon)|s(t)\rangle = (\mathbf{1} - i\varepsilon H)|s(t)\rangle

Rearranging:

\frac{|s(t+\varepsilon)\rangle - |s(t)\rangle}{\varepsilon} = -iH|s(t)\rangle

Taking the limit $\varepsilon \to 0$:

Schrödinger Equation (natural units $\hbar=1$)

$$\frac{d}{dt}|s(t)\rangle = -iH|s(t)\rangle$$

Schrödinger Equation (SI units)

$$i\hbar\frac{d}{dt}|s(t)\rangle = H|s(t)\rangle$$

Key properties:

First-order in time — the current state fully determines all future states.
Linear — superpositions of solutions are solutions.
Deterministic — no randomness between measurements.
Reversible — since $U$ is unitary, it is invertible; the past is recoverable from the present.

General Solution

For a time-independent Hamiltonian, the formal solution is the operator exponential:

|s(t)\rangle = e^{-iHt/\hbar}|s(0)\rangle

The energy eigenstates satisfy $H|E_n\rangle = E_n|E_n\rangle$. Because the eigenstates form a complete basis, any state can be expanded:

|s(t)\rangle = \sum_n c_n\, e^{-iE_n t/\hbar}|E_n\rangle, \qquad c_n = \langle E_n|s(0)\rangle

Stationary states: If the system is in a single energy eigenstate $|E_n\rangle$, then $|s(t)\rangle = e^{-iE_n t/\hbar}|E_n\rangle$. The phase $e^{-iE_n t/\hbar}$ is a global phase and cancels in all probability calculations $|\langle\phi|s(t)\rangle|^2$. Thus probability distributions are completely static — hence "stationary." Interference effects and observable dynamics arise only from superpositions of different energy eigenstates.

Part V — Expectation Values

Since quantum measurements are probabilistic, we often care about the average value of an observable $L$ over many measurements. If the system is in state $|a\rangle$ and $L$ has eigenstates $|\ell_i\rangle$ with eigenvalues $\lambda_i$:

\langle L\rangle = \sum_i \lambda_i P(\lambda_i) = \sum_i \lambda_i |\langle\ell_i|a\rangle|^2

Inserting the resolution of the identity $\sum_i |\ell_i\rangle\langle\ell_i| = \mathbf{1}$ and using $L|\ell_i\rangle = \lambda_i|\ell_i\rangle$:

Expectation Value (Sandwich Formula)

$$\langle L\rangle = \langle a|L|a\rangle$$

Important caveat: $\langle L\rangle$ is the mathematical average over many independent measurements. It is not the most probable single result. For a spin in state $|r\rangle = \frac{1}{\sqrt{2}}(|u\rangle+|d\rangle)$, we get $\langle\sigma_z\rangle = 0$ — but you will never actually measure $0$. Every measurement yields either $+1$ or $-1$. The average zero emerges only statistically.

Part VI — Equations of Motion for Expectation Values

How does $\langle L\rangle$ evolve in time? Differentiate the sandwich formula:

\frac{d}{dt}\langle L\rangle = \frac{d}{dt}\langle s(t)|L|s(t)\rangle = \left(\frac{d\langle s|}{dt}\right)L|s\rangle + \langle s|L\left(\frac{d|s\rangle}{dt}\right)

Using the Schrödinger equation $\frac{d|s\rangle}{dt} = -iH|s\rangle$ and its adjoint $\frac{d\langle s|}{dt} = +i\langle s|H$:

\frac{d}{dt}\langle L\rangle = i\langle s|HL|s\rangle - i\langle s|LH|s\rangle = i\langle s|(HL - LH)|s\rangle

Defining the commutator $[H,L] \equiv HL - LH$:

Equation of Motion for Expectation Values ($\hbar=1$)

$$\frac{d}{dt}\langle L\rangle = i\langle[H,L]\rangle$$

With $\hbar$ restored: $\dfrac{d}{dt}\langle L\rangle = \dfrac{i}{\hbar}\langle[H,L]\rangle$. This single equation governs the time evolution of every observable's expectation value. If $[H,L]=0$, then $\langle L\rangle$ is constant — $L$ is a conserved quantity.

Part VII — Commutators & the Bridge to Classical Mechanics

In classical mechanics, the time evolution of any observable $f(q,p)$ is governed by the Poisson bracket with the Hamiltonian:

\frac{df}{dt} = \{f, H\}_\text{PB}

The Poisson bracket is antisymmetric $\{A,B\}=-\{B,A\}$ and satisfies the product (Leibniz) rule $\{AB,C\}=A\{B,C\}+\{A,C\}B$.

Quantum commutators share exactly these algebraic properties:

Antisymmetry: $[A,B] = -[B,A]$
Product rule: $[AB,C] = A[B,C] + [A,C]B$
Jacobi identity: $[A,[B,C]]+[B,[C,A]]+[C,[A,B]]=0$

Dirac recognized this deep structural parallel and identified the two structures:

Dirac's Quantization Rule

$$\{A,B\}_\text{PB} \;\longleftrightarrow\; \frac{1}{i\hbar}[A,B]$$

The most important application: classically, position and momentum satisfy $\{q,p\}_\text{PB} = 1$. Applying Dirac's rule:

Canonical Commutation Relation

$$[\hat{q},\hat{p}] = i\hbar$$

This single relation is the origin of the Heisenberg uncertainty principle $\Delta x\,\Delta p \geq \hbar/2$. It encodes the impossibility of simultaneously knowing position and momentum with perfect precision.

Part VIII — Energy Conservation

Set $L = H$ in the equation of motion:

\frac{d}{dt}\langle H\rangle = i\langle[H,H]\rangle

Any operator commutes with itself: $[H,H] = HH - HH = 0$. Therefore:

Energy Conservation

$$\frac{d}{dt}\langle H\rangle = 0 \quad \text{(for time-independent } H\text{)}$$

Energy is conserved as a direct consequence of time-translation invariance (time-independent $H$). If an external field switches on or off — making $H$ explicitly time-dependent — energy is not conserved. Work is done on or by the system.

Part IX — Solving the Schrödinger Equation

The systematic strategy: expand the initial state in the energy eigenbasis.

Step 1. Find all eigenvalues and eigenstates: $H|E_n\rangle = E_n|E_n\rangle$.

Step 2. Expand $|s(0)\rangle = \sum_n c_n|E_n\rangle$, where $c_n = \langle E_n|s(0)\rangle$.

Step 3. The time-evolved state is:

|s(t)\rangle = \sum_n c_n\,e^{-iE_n t/\hbar}|E_n\rangle

Each energy eigenstate oscillates with its own characteristic frequency $\omega_n = E_n/\hbar$. The coefficients $c_n$ are fixed by initial conditions and never change in magnitude.

Key insight: A single eigenstate has constant probabilities (stationary state). Observable time dependence — spin precession, oscillating dipole moments, quantum beats — arises from interference between terms oscillating at different frequencies. The cross-terms in $|\langle\phi|s(t)\rangle|^2$ contain factors like $e^{i(E_m-E_n)t/\hbar}$, which oscillate at the Bohr frequencies $\omega_{mn} = (E_m-E_n)/\hbar$.

Unitarity $U^\dagger U=1$ — information conservation

Hamiltonian $H=H^\dagger$ — generator of time evolution

Schrödinger equation — first-order, linear, reversible

Expectation value $\langle L\rangle = \langle a|L|a\rangle$

Commutator $[H,L]$ — drives time evolution of $\langle L\rangle$

Canonical commutation $[\hat{q},\hat{p}]=i\hbar$

Energy eigenstates — stationary states

Schrödinger equation

$i\hbar\dfrac{d}{dt}|s\rangle = H|s\rangle$

General solution

$|s(t)\rangle = \sum_n c_n e^{-iE_n t/\hbar}|E_n\rangle$

Expectation value

$\langle L\rangle = \langle a|L|a\rangle$

Commutator EOM

$\dfrac{d\langle L\rangle}{dt} = \dfrac{i}{\hbar}\langle[H,L]\rangle$

Canonical commutation

$[\hat{q},\hat{p}] = i\hbar$

Energy conservation

$\dfrac{d\langle H\rangle}{dt} = 0$

Exercises — Lecture 4

Nine exercises covering unitarity, stationary states, spin precession, commutator algebra, and the uncertainty principle. Click any exercise to reveal the full worked solution.

Unitary Check

EasyUnitarity

A $2\times2$ matrix is given by:

$$U = \frac{1}{\sqrt{2}}\begin{pmatrix}1 & -i \\ -i & 1\end{pmatrix}$$

Verify that $U$ is unitary by computing $UU^\dagger$ and showing it equals the identity $\mathbf{1}$.

Complete Solution

First, find the conjugate transpose $U^\dagger$. Taking complex conjugates of each entry and transposing:

$$U^\dagger = \frac{1}{\sqrt{2}}\begin{pmatrix}1 & i \\ i & 1\end{pmatrix}$$

Compute $UU^\dagger = \dfrac{1}{2}\begin{pmatrix}1 & -i \\ -i & 1\end{pmatrix}\begin{pmatrix}1 & i \\ i & 1\end{pmatrix}$. Work out each entry:

$$(UU^\dagger)_{11} = \frac{1}{2}(1\cdot1 + (-i)\cdot i) = \frac{1}{2}(1 + 1) = 1$$

$$(UU^\dagger)_{12} = \frac{1}{2}(1\cdot i + (-i)\cdot 1) = \frac{1}{2}(i - i) = 0$$

$$(UU^\dagger)_{21} = \frac{1}{2}((-i)\cdot 1 + 1\cdot i) = \frac{1}{2}(-i + i) = 0$$

$$(UU^\dagger)_{22} = \frac{1}{2}((-i)\cdot i + 1\cdot 1) = \frac{1}{2}(1 + 1) = 1$$

Result

$$UU^\dagger = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} = \mathbf{1} \checkmark$$ $U$ is unitary — it preserves all inner products and represents a valid quantum time-evolution step.

Stationary State

EasyEnergy Eigenstates

A spin is prepared in an energy eigenstate $|E_1\rangle$ of some Hamiltonian $H$, with $H|E_1\rangle = E_1|E_1\rangle$.

Show that the probability of measuring any observable outcome $\lambda_i$ at time $t$ is the same as at $t=0$. Why is this called a "stationary state"?

Complete Solution

The time-evolved state for an energy eigenstate is:

$$|s(t)\rangle = e^{-iE_1 t/\hbar}|E_1\rangle$$

The probability of measuring eigenvalue $\lambda_i$ at time $t$ is:

$$P(\lambda_i, t) = |\langle\lambda_i|s(t)\rangle|^2 = \left|e^{-iE_1 t/\hbar}\right|^2|\langle\lambda_i|E_1\rangle|^2$$

Since $\left|e^{-iE_1 t/\hbar}\right|^2 = 1$ (the exponential is a pure phase with unit modulus):

Result

$$P(\lambda_i, t) = |\langle\lambda_i|E_1\rangle|^2 = P(\lambda_i, 0)$$ The global phase $e^{-iEt/\hbar}$ cancels in every squared amplitude. All probability distributions are completely frozen in time — hence the name stationary state.

Expectation Value Time Evolution

MediumSpin Precession

A spin-½ particle has Hamiltonian $H = \omega\sigma_z$. The initial state is $|{+x}\rangle = |r\rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}$.

(a) Identify the energy eigenstates and eigenvalues of $H$.

(b) Expand $|r\rangle$ in the energy eigenbasis and write $|s(t)\rangle$.

(c) Compute $\langle\sigma_x\rangle(t)$ and describe the physical motion.

Complete Solution

Since $H = \omega\sigma_z = \omega\begin{pmatrix}1&0\\0&-1\end{pmatrix}$, the eigenstates are the $z$-spin eigenstates:

$$H|u\rangle = +\omega|u\rangle, \qquad H|d\rangle = -\omega|d\rangle$$

Energy eigenvalues: $E_u = +\omega$ and $E_d = -\omega$.

Expanding: $|r\rangle = \dfrac{1}{\sqrt{2}}|u\rangle + \dfrac{1}{\sqrt{2}}|d\rangle$, so $c_u = c_d = \dfrac{1}{\sqrt{2}}$. The time-evolved state is:

$$|s(t)\rangle = \frac{1}{\sqrt{2}}\left(e^{-i\omega t}|u\rangle + e^{+i\omega t}|d\rangle\right)$$

Use $\sigma_x|u\rangle = |d\rangle$ and $\sigma_x|d\rangle = |u\rangle$:

$$\sigma_x|s(t)\rangle = \frac{1}{\sqrt{2}}\left(e^{-i\omega t}|d\rangle + e^{+i\omega t}|u\rangle\right)$$

$$\langle\sigma_x\rangle = \langle s(t)|\sigma_x|s(t)\rangle = \frac{1}{2}\left(e^{+i\omega t}\langle u| + e^{-i\omega t}\langle d|\right)\left(e^{-i\omega t}|d\rangle + e^{+i\omega t}|u\rangle\right)$$

$$= \frac{1}{2}\left(e^{2i\omega t} + e^{-2i\omega t}\right) = \cos(2\omega t)$$

Result

$\langle\sigma_x\rangle(t) = \cos(2\omega t)$. The spin expectation value oscillates — this is quantum precession around the $z$-axis at angular frequency $2\omega$.

Commutator Algebra

MediumCommutators

(a) Compute $[\sigma_z, \sigma_x]$ directly from the Pauli matrix definitions.

(b) Verify the result equals $2i\sigma_y$.

(c) Using $H = \omega\sigma_z$ and the equation of motion $d\langle\sigma_x\rangle/dt = i\langle[H,\sigma_x]\rangle$, find $d\langle\sigma_x\rangle/dt$.

Complete Solution

With $\sigma_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and $\sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix}$:

$$\sigma_z\sigma_x = \begin{pmatrix}0&1\\-1&0\end{pmatrix}, \qquad \sigma_x\sigma_z = \begin{pmatrix}0&-1\\1&0\end{pmatrix}$$

$$[\sigma_z,\sigma_x] = \sigma_z\sigma_x - \sigma_x\sigma_z = \begin{pmatrix}0&2\\-2&0\end{pmatrix}$$

Recall $\sigma_y = \begin{pmatrix}0&-i\\i&0\end{pmatrix}$, so $2i\sigma_y = 2i\begin{pmatrix}0&-i\\i&0\end{pmatrix} = \begin{pmatrix}0&2\\-2&0\end{pmatrix}$. ✓

$$[\sigma_z,\sigma_x] = 2i\sigma_y \checkmark$$

$$\frac{d\langle\sigma_x\rangle}{dt} = i\langle[H,\sigma_x]\rangle = i\langle\omega[\sigma_z,\sigma_x]\rangle = i\cdot\omega\cdot 2i\langle\sigma_y\rangle = -2\omega\langle\sigma_y\rangle$$

Result

$\dfrac{d\langle\sigma_x\rangle}{dt} = -2\omega\langle\sigma_y\rangle$. This is the quantum precession equation — $\langle\boldsymbol{\sigma}\rangle$ rotates around the $z$-axis like a classical angular momentum vector in a magnetic field.

Canonical Commutation Relation

MediumWave Functions

Let $\hat{q}\psi(x) = x\psi(x)$ (multiplication by $x$) and $\hat{p} = -i\hbar\,d/dx$.

Prove that $[\hat{q},\hat{p}]\psi(x) = i\hbar\psi(x)$ for an arbitrary differentiable function $\psi$. Conclude that $[\hat{q},\hat{p}] = i\hbar$.

Complete Solution

Expand the commutator acting on $\psi$:

$$[\hat{q},\hat{p}]\psi = \hat{q}(\hat{p}\psi) - \hat{p}(\hat{q}\psi) = x\!\left(-i\hbar\frac{d\psi}{dx}\right) - \left(-i\hbar\frac{d}{dx}\right)(x\psi)$$

Apply the product rule to the second term: $\dfrac{d}{dx}(x\psi) = \psi + x\psi'$.

$$[\hat{q},\hat{p}]\psi = -i\hbar x\psi'(x) - (-i\hbar)(\psi(x) + x\psi'(x))$$

$$= -i\hbar x\psi' + i\hbar\psi + i\hbar x\psi' = i\hbar\psi(x)$$

Result

Since $[\hat{q},\hat{p}]\psi = i\hbar\psi$ for every test function $\psi$, we conclude $[\hat{q},\hat{p}] = i\hbar$. This is the canonical commutation relation — the quantum version of Dirac's quantization rule applied to the classical Poisson bracket $\{q,p\}=1$.

Energy Conservation: Direct Check

MediumEnergy Conservation

For $H=\omega\sigma_z$ and the state $|s(t)\rangle = \dfrac{1}{\sqrt{2}}\!\left(e^{-i\omega t}|u\rangle + e^{i\omega t}|d\rangle\right)$ (from Exercise 3), directly compute $\langle H\rangle(t)$ and verify it is constant.

Interpret the result physically.

Complete Solution

Apply $H$ to the state using $H|u\rangle=\omega|u\rangle$, $H|d\rangle=-\omega|d\rangle$:

$$H|s(t)\rangle = \frac{1}{\sqrt{2}}\left(\omega e^{-i\omega t}|u\rangle - \omega e^{i\omega t}|d\rangle\right)$$

Compute the sandwich:

$$\langle H\rangle = \langle s(t)|H|s(t)\rangle = \frac{1}{2}\left(e^{i\omega t}\langle u| + e^{-i\omega t}\langle d|\right)\left(\omega e^{-i\omega t}|u\rangle - \omega e^{i\omega t}|d\rangle\right)$$

$$= \frac{1}{2}\left(\omega\cdot1 + (-\omega)\cdot1\right) = 0$$

Result

$\langle H\rangle = 0$ for all $t$. The state $|r\rangle$ is an equal superposition of energy $+\omega$ and energy $-\omega$, so the average energy is exactly zero — constant, as required by energy conservation. ✓

Product Rule for Commutators

MediumAlgebra

Prove the product rule (Leibniz rule) for commutators:

$$[AB, C] = A[B,C] + [A,C]B$$

Then explain why this is significant for the Dirac quantization program connecting Poisson brackets to commutators.

Complete Solution

Expand directly from the definition $[AB,C] = ABC - CAB$. Add and subtract the term $ACB$:

$$[AB,C] = ABC - CAB = (ABC - ACB) + (ACB - CAB)$$

Factor each group:

$$= A(BC - CB) + (AC - CA)B = A[B,C] + [A,C]B \checkmark$$

Significance

Classical Poisson brackets also satisfy $\{AB,C\} = A\{B,C\} + \{A,C\}B$. The fact that commutators satisfy the same Leibniz rule confirms the structural parallel that motivated Dirac's identification $\{A,B\}_\text{PB} \leftrightarrow \frac{1}{i\hbar}[A,B]$. Both algebras are Lie algebras with a derivation property — quantization preserves algebraic structure.

Conserved Observables and Stationary States

HardEhrenfest

Use the equation of motion $d\langle L\rangle/dt = (i/\hbar)\langle[H,L]\rangle$ to address three related questions:

(a) If $[H,L]=0$, show that $\langle L\rangle$ is constant in time for any state.

(b) Show directly that $[H,H]=0$, and hence energy is always conserved (for time-independent $H$).

(c) If $|\psi\rangle$ is an eigenstate of $H$ with eigenvalue $E$, show that $e^{-iEt/\hbar}|\psi\rangle$ is a stationary state — all probabilities are time-independent.

Complete Solution

If $[H,L]=0$, then $\langle[H,L]\rangle = \langle 0\rangle = 0$, so:

$$\frac{d\langle L\rangle}{dt} = \frac{i}{\hbar}\langle[H,L]\rangle = \frac{i}{\hbar}\cdot 0 = 0$$

Therefore $\langle L\rangle = \text{const}$ for any state. Observables that commute with $H$ are conserved quantities — the quantum analogue of Noether's theorem. ✓

Any operator commutes with itself:

$$[H,H] = HH - HH = 0$$

By part (a) with $L=H$: $d\langle H\rangle/dt = 0$, so $\langle H\rangle$ is constant. Energy is conserved. ✓

Let $|s(t)\rangle = e^{-iEt/\hbar}|\psi\rangle$. For any projector onto a measurement outcome $|\phi\rangle$:

$$P(\phi, t) = |\langle\phi|s(t)\rangle|^2 = \left|e^{-iEt/\hbar}\right|^2|\langle\phi|\psi\rangle|^2 = |\langle\phi|\psi\rangle|^2 = P(\phi, 0)$$

Conclusion

All probabilities are independent of $t$ — the global phase $e^{-iEt/\hbar}$ has unit modulus and cancels in every probability amplitude squared. Energy eigenstates are stationary states. ✓

Spin Precession: Full General Solution

HardSpin Dynamics

A spin-½ particle with $H=\omega\sigma_z$ is prepared in the general state:

$$|\psi(0)\rangle = \cos\!\tfrac{\theta}{2}\,|u\rangle + e^{i\varphi}\sin\!\tfrac{\theta}{2}\,|d\rangle$$

(a) Write $|\psi(t)\rangle$.

(b) Compute $\langle\sigma_x\rangle(t)$, $\langle\sigma_y\rangle(t)$, $\langle\sigma_z\rangle(t)$.

(c) Show that the vector $(\langle\sigma_x\rangle,\langle\sigma_y\rangle,\langle\sigma_z\rangle)$ precesses around the $z$-axis with angular frequency $2\omega$. What is the quantum analogue of this classical phenomenon?

Complete Solution

Each component picks up its energy phase:

$$|\psi(t)\rangle = \cos\!\tfrac{\theta}{2}\,e^{-i\omega t}|u\rangle + e^{i\varphi}\sin\!\tfrac{\theta}{2}\,e^{+i\omega t}|d\rangle$$

Define amplitudes $\alpha_1 = \cos\!\tfrac{\theta}{2}\,e^{-i\omega t}$ and $\alpha_2 = e^{i\varphi}\sin\!\tfrac{\theta}{2}\,e^{i\omega t}$. Using the Pauli matrix sandwich formulas:

$$\langle\sigma_z\rangle = |\alpha_1|^2 - |\alpha_2|^2 = \cos^2\!\tfrac{\theta}{2} - \sin^2\!\tfrac{\theta}{2} = \cos\theta$$

$$\langle\sigma_x\rangle = 2\,\mathrm{Re}(\alpha_1^*\alpha_2) = 2\,\mathrm{Re}\!\left(\cos\!\tfrac{\theta}{2}\,e^{i\omega t}\cdot e^{i\varphi}\sin\!\tfrac{\theta}{2}\,e^{i\omega t}\right)$$

$$= 2\cos\!\tfrac{\theta}{2}\sin\!\tfrac{\theta}{2}\cos(\varphi+2\omega t) = \sin\theta\cos(\varphi+2\omega t)$$

$$\langle\sigma_y\rangle = 2\,\mathrm{Im}(\alpha_1^*\alpha_2) = \sin\theta\sin(\varphi+2\omega t)$$

The $z$-component $\langle\sigma_z\rangle = \cos\theta$ is constant. The $x$ and $y$ components trace a circle of radius $\sin\theta$ in the $xy$-plane with phase angle $\varphi + 2\omega t$:

$$\langle\boldsymbol{\sigma}\rangle(t) = \sin\theta\cos(\varphi+2\omega t)\,\hat{x} + \sin\theta\sin(\varphi+2\omega t)\,\hat{y} + \cos\theta\,\hat{z}$$

Conclusion

The expectation value vector precesses around the $\hat{z}$-axis at angular frequency $2\omega$, tracing a cone of half-angle $\theta$. This is the quantum analogue of Larmor precession — a magnetic moment precessing in a uniform magnetic field. Here the "field" is encoded in $H=\omega\sigma_z$.

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