Lecture 05 · Dynamics

Uncertainty, Energy Eigenstates & Spin Precession

The uncertainty principle for incompatible observables, stationary states and time evolution, the spin Hamiltonian in a magnetic field, Larmor precession of the Bloch vector, Rabi oscillations, the full Pauli algebra, and the Robertson uncertainty relation.

Part I — The Uncertainty Principle (Conceptual)

Two observables $A$ and $B$ are called compatible if their commutator vanishes, $[A,B]=0$, and incompatible if $[A,B]\neq 0$. This algebraic distinction carries deep physical consequences.

Compatible observables can be measured simultaneously without disturbance: knowing the result of $A$ does not affect the statistics of $B$. They share a common set of eigenstates, so a state can be simultaneously an eigenstate of both.

Incompatible observables are fundamentally different: measuring one necessarily disturbs the other. This is not a limitation of the apparatus — it is a theorem of the mathematics.

The key example: $\sigma_z$ and $\sigma_x$ do not commute. Measuring $\sigma_x$ on a state that had a definite value of $\sigma_z$ destroys that definite value. After the $\sigma_x$ measurement, $\sigma_z$ is random — even though no physical disturbance was applied between measurements.

The formal quantitative statement is:

Robertson Uncertainty Relation
$$\Delta A \cdot \Delta B \;\geq\; \tfrac{1}{2}\bigl|\langle[A,B]\rangle\bigr|$$

For the spin operators $\sigma_z$ and $\sigma_x$: since $[\sigma_z,\sigma_x] = 2i\sigma_y$, the relation gives $\Delta\sigma_z\,\Delta\sigma_x \geq |\langle\sigma_y\rangle|$. In an eigenstate of $\sigma_z$ one has $\langle\sigma_y\rangle = 0$, so the bound is trivially zero — but $\sigma_x$ is still completely random with a 50/50 outcome distribution. The inequality is satisfied as $1 \geq 0$; the uncertainty is real even when the bound is not tight.

Part II — Energy Eigenstates and Stationary States

The eigenvalue equation for the Hamiltonian defines the energy eigenstates:

$$H|E_n\rangle = E_n|E_n\rangle$$

Because $H$ is Hermitian, the eigenvalues $E_n$ are real and the eigenstates form an orthonormal basis. The time evolution of an energy eigenstate is particularly simple:

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

The state acquires a time-dependent global phase — but global phases are unobservable. All probabilities are therefore time-independent:

$$|\langle\phi|E_n(t)\rangle|^2 = |e^{-iE_n t/\hbar}|^2|\langle\phi|E_n\rangle|^2 = |\langle\phi|E_n\rangle|^2$$

Similarly, all expectation values $\langle A\rangle$ are constant. This is why energy eigenstates are called stationary states: no measurement statistics change with time.

For a general state expanded in the energy basis:

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

the different energy components accumulate different phases. The amplitude $\langle\phi|\psi(t)\rangle$ is a sum of rotating complex numbers — and their interference makes probabilities oscillate at the beat frequencies $(E_n - E_m)/\hbar$ between levels.

Part III — The Spin Hamiltonian in a Magnetic Field

The simplest dynamical quantum system: a spin-$\tfrac{1}{2}$ particle in a uniform magnetic field $\mathbf{B}$ pointing along the $z$-axis. The Hamiltonian is:

Spin Hamiltonian
$$H = \frac{\omega}{2}\,\sigma_z = \frac{\omega}{2}\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$

where $\omega = eB/mc$ is the Larmor frequency, and we work in natural units with $\hbar = 1$ throughout. The energy eigenvalues are:

$$E_+ = +\tfrac{\omega}{2} \quad (\text{spin-up}), \qquad E_- = -\tfrac{\omega}{2} \quad (\text{spin-down})$$

The energy eigenstates are $|u\rangle = (1,0)^T$ and $|d\rangle = (0,1)^T$ — identical to the $\sigma_z$ eigenstates. This is to be expected: $H$ is proportional to $\sigma_z$, so they share eigenstates.

Part IV — Time Evolution of an Arbitrary Spin State

Starting from a general initial state $|\psi(0)\rangle = \alpha|u\rangle + \beta|d\rangle$ with $|\alpha|^2+|\beta|^2=1$, the Schrödinger equation gives:

$$|\psi(t)\rangle = \alpha\,e^{-i\omega t/2}|u\rangle + \beta\,e^{+i\omega t/2}|d\rangle$$

To see the geometric meaning, write $\alpha = \cos(\theta/2)$ and $\beta = e^{i\varphi}\sin(\theta/2)$ (the Bloch sphere parametrization). After time $t$:

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

Factoring out a global phase $e^{-i\omega t/2}$ (unobservable), the azimuthal angle evolves as $\varphi \to \varphi + \omega t$. This is precisely precession: the Bloch vector rotates around the $z$-axis.

Computing expectation values explicitly confirms this picture:

$$\langle\sigma_z\rangle = \cos\theta = \text{const}, \qquad \langle\sigma_x\rangle(t) = \sin\theta\cos(\varphi+\omega t), \qquad \langle\sigma_y\rangle(t) = \sin\theta\sin(\varphi+\omega t)$$

The $z$-component of the Bloch vector is unchanged; the transverse component rotates at angular frequency $\omega$. The magnitude of the Bloch vector is preserved.

The Bloch vector precesses around the magnetic field axis at the Larmor frequency $\omega$ — identical to classical Larmor precession. Quantum mechanics gives the same average behavior as classical mechanics.

Part V — Transition Probabilities

Consider starting in $|u\rangle$ and measuring $\sigma_x$ at time $t$. The evolved state is $|\psi(t)\rangle = e^{-i\omega t/2}|u\rangle$. The probability of finding $\sigma_x = +1$ is:

$$P(\sigma_x=+1,\,t) = |\langle r|\psi(t)\rangle|^2 = \left|\frac{e^{-i\omega t/2}}{\sqrt{2}}\right|^2 = \frac{1}{2} \quad \text{(constant — no oscillation)}$$

An energy eigenstate shows no oscillating probabilities. Now start instead in the superposition $|\psi(0)\rangle = \tfrac{1}{\sqrt{2}}(|u\rangle+|d\rangle) = |r\rangle$:

$$|\psi(t)\rangle = \frac{1}{\sqrt{2}}\!\left(e^{-i\omega t/2}|u\rangle + e^{+i\omega t/2}|d\rangle\right)$$
$$P(\sigma_x=+1,\,t) = |\langle r|\psi(t)\rangle|^2 = \left|\frac{e^{-i\omega t/2}+e^{i\omega t/2}}{2}\right|^2 = \cos^2\!\!\left(\frac{\omega t}{2}\right)$$

This oscillates between $0$ and $1$ with period $2\pi/\omega$ — these are Rabi oscillations, the quantum beating between two energy levels.

Part VI — The Pauli Matrices: Full Algebra

The three Pauli matrices:

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

satisfy a rich set of algebraic identities:

IdentityFormula
Each squares to identity$\sigma_i^2 = I$
Product of distinct Paulis$\sigma_i\sigma_j = i\varepsilon_{ijk}\sigma_k \quad (i\neq j)$
Anticommutator$\{\sigma_i,\sigma_j\} = 2\delta_{ij}I$
Commutator$[\sigma_i,\sigma_j] = 2i\varepsilon_{ijk}\sigma_k$

All four identities are unified by the single master formula:

Pauli Product Formula
$$\sigma_i\sigma_j = \delta_{ij}\mathbf{1} + i\varepsilon_{ijk}\sigma_k$$

From the Pauli algebra one can derive the rotation formula. The exponential of a Pauli matrix is:

$$e^{i\theta\sigma_z/2} = \cos\!\tfrac{\theta}{2}\,I + i\sin\!\tfrac{\theta}{2}\,\sigma_z$$

This is the operator that rotates the spin state around the $z$-axis by angle $\theta$. The time evolution operator $e^{-iHt} = e^{-i\omega t\sigma_z/2}$ is precisely a rotation by angle $\omega t$ — confirming that time evolution in the magnetic field is nothing but rotation.

Part VII — Commutators and the Uncertainty Principle: Formal Statement

The Robertson uncertainty relation gives a precise lower bound on the product of uncertainties:

Robertson Relation
$$\Delta A \cdot \Delta B \;\geq\; \tfrac{1}{2}\bigl|\langle[A,B]\rangle\bigr|$$

where $\Delta A^2 = \langle A^2\rangle - \langle A\rangle^2$ is the variance. The derivation applies the Cauchy-Schwarz inequality to the vectors $(A - \langle A\rangle)|\psi\rangle$ and $(B - \langle B\rangle)|\psi\rangle$. Since $A$ and $B$ are Hermitian, their commutator $[A,B]$ is anti-Hermitian, so $\langle[A,B]\rangle$ is purely imaginary — and $|\langle[A,B]\rangle|$ is real and well-defined.

For position and momentum: $[x,p] = i\hbar$, so the bound gives the celebrated Heisenberg relation:

$$\Delta x \cdot \Delta p \;\geq\; \frac{\hbar}{2} \checkmark$$

For spin: $[\sigma_x,\sigma_y] = 2i\sigma_z$, so $\Delta\sigma_x\,\Delta\sigma_y \geq |\langle\sigma_z\rangle|$. The uncertainty is state-dependent — it vanishes when $\langle\sigma_z\rangle = 0$ and saturates when the state is an eigenstate of $\sigma_z$.

Exercises — Lecture 5

Eight exercises covering uncertainty, stationary states, precession, Rabi oscillations, Pauli algebra, and the Robertson relation. Click any exercise to reveal the full worked solution.

1
Energy Eigenvalue Equation
EasyHamiltonian

For the spin Hamiltonian $H = \dfrac{\omega}{2}\sigma_z$, verify that $|u\rangle = (1,0)^T$ is an eigenstate with eigenvalue $+\omega/2$ and $|d\rangle = (0,1)^T$ is an eigenstate with eigenvalue $-\omega/2$.

Complete Solution
spin-up

Apply $H$ to $|u\rangle$:

$$H|u\rangle = \frac{\omega}{2}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \frac{\omega}{2}\begin{pmatrix}1\\0\end{pmatrix} = \frac{\omega}{2}|u\rangle \checkmark$$

So $|u\rangle$ is an eigenstate with eigenvalue $E_+ = +\omega/2$.

spin-down

Apply $H$ to $|d\rangle$:

$$H|d\rangle = \frac{\omega}{2}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix} = \frac{\omega}{2}\begin{pmatrix}0\\-1\end{pmatrix} = -\frac{\omega}{2}|d\rangle \checkmark$$
Summary
$E_+ = +\omega/2$ for $|u\rangle$; $E_- = -\omega/2$ for $|d\rangle$. The energy gap is $\Delta E = \omega$ (in natural units $\hbar=1$).
2
Stationary State Probability
EasyStationary States

A spin is in the energy eigenstate $|u\rangle$. Compute $P(\sigma_z=+1)$ and $P(\sigma_x=+1)$ at time $t$, showing both are time-independent even though the state is evolving.

Complete Solution
evolve

The evolved state is $|\psi(t)\rangle = e^{-i\omega t/2}|u\rangle$ — a global phase times the same vector.

$\sigma_z$
$$P(\sigma_z=+1) = |\langle u|\psi(t)\rangle|^2 = |e^{-i\omega t/2}|^2|\langle u|u\rangle|^2 = 1 \quad \text{(constant)}$$
$\sigma_x$

The spin-right state is $|r\rangle = \tfrac{1}{\sqrt{2}}(1,1)^T$, so $\langle r|u\rangle = \tfrac{1}{\sqrt{2}}$:

$$P(\sigma_x=+1) = |\langle r|\psi(t)\rangle|^2 = |e^{-i\omega t/2}|^2\,\tfrac{1}{2} = \frac{1}{2} \quad \text{(constant)}$$
Key Insight
The global phase $e^{-i\omega t/2}$ has modulus 1 and cancels in every probability. For an energy eigenstate, all measurement statistics are time-independent — hence the name "stationary state".
3
Rabi Oscillation
MediumDynamics

Starting from $|\psi(0)\rangle = |r\rangle = \tfrac{1}{\sqrt{2}}(|u\rangle+|d\rangle)$ with $H=\tfrac{\omega}{2}\sigma_z$:

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

(b) Compute $P(\sigma_x=+1,\,t) = |\langle r|\psi(t)\rangle|^2$.

(c) At what time does the spin first reach $|l\rangle$ with certainty? (i.e., $P(\sigma_x=-1)=1$)

Complete Solution
a

Each component picks up its energy phase factor:

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

Using $\langle r| = \tfrac{1}{\sqrt{2}}(1,1)$:

$$\langle r|\psi(t)\rangle = \frac{1}{2}\!\left(e^{-i\omega t/2} + e^{+i\omega t/2}\right) = \cos\!\left(\frac{\omega t}{2}\right)$$
$$P(\sigma_x=+1,\,t) = \cos^2\!\!\left(\frac{\omega t}{2}\right)$$
c

$P(\sigma_x=-1)=1$ requires $P(\sigma_x=+1)=0$, i.e., $\cos^2(\omega t/2)=0$:

$$\frac{\omega t}{2} = \frac{\pi}{2} \implies t = \frac{\pi}{\omega}$$
Answer
The spin first reaches $|l\rangle$ at $t = \pi/\omega$ — half the Larmor period. At this time the probability oscillates to a complete flip.
4
Expectation Values Under Precession
MediumBloch Sphere

With $H=\tfrac{\omega}{2}\sigma_z$ and $|\psi(0)\rangle = \tfrac{1}{\sqrt{2}}(|u\rangle+|d\rangle)$, compute $\langle\sigma_x\rangle(t)$, $\langle\sigma_y\rangle(t)$, $\langle\sigma_z\rangle(t)$ and show the Bloch vector precesses around the $z$-axis.

Complete Solution
setup

The evolved state is $|\psi(t)\rangle = \tfrac{1}{\sqrt{2}}\bigl(e^{-i\omega t/2}|u\rangle + e^{i\omega t/2}|d\rangle\bigr)$.

$\langle\sigma_z\rangle$
$$\langle\sigma_z\rangle = \frac{|e^{-i\omega t/2}|^2}{2} - \frac{|e^{i\omega t/2}|^2}{2} = \frac{1}{2} - \frac{1}{2} = 0 \quad \text{(constant)}$$
$\langle\sigma_x\rangle$

$\sigma_x|\psi(t)\rangle = \tfrac{1}{\sqrt{2}}\bigl(e^{-i\omega t/2}|d\rangle + e^{i\omega t/2}|u\rangle\bigr)$. Then:

$$\langle\sigma_x\rangle = \frac{1}{2}\!\left(e^{i\omega t/2}\cdot e^{i\omega t/2} + e^{-i\omega t/2}\cdot e^{-i\omega t/2}\right) = \frac{e^{i\omega t}+e^{-i\omega t}}{2} = \cos(\omega t)$$
$\langle\sigma_y\rangle$

An analogous calculation with $\sigma_y$ gives $\langle\sigma_y\rangle = \sin(\omega t)$.

Result
$(\langle\sigma_x\rangle,\langle\sigma_y\rangle,\langle\sigma_z\rangle) = (\cos\omega t,\,\sin\omega t,\,0)$ — a unit vector in the $xy$-plane rotating at angular frequency $\omega$. The Bloch vector precesses around the $z$-axis. ✓
5
Pauli Algebra: $\sigma_x\sigma_y$
MediumPauli Matrices

(a) Compute $\sigma_x\sigma_y$ directly from the $2\times 2$ matrix representations.

(b) Verify that $\sigma_x\sigma_y = i\sigma_z$.

(c) Compute the commutator $[\sigma_x,\sigma_y]$ and the anticommutator $\{\sigma_x,\sigma_y\}$.

Complete Solution
a
$$\sigma_x\sigma_y = \begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}0&-i\\i&0\end{pmatrix} = \begin{pmatrix}0\cdot0+1\cdot i & 0\cdot(-i)+1\cdot0\\ 1\cdot0+0\cdot i & 1\cdot(-i)+0\cdot0\end{pmatrix} = \begin{pmatrix}i&0\\0&-i\end{pmatrix}$$
b
$$i\sigma_z = i\begin{pmatrix}1&0\\0&-1\end{pmatrix} = \begin{pmatrix}i&0\\0&-i\end{pmatrix} = \sigma_x\sigma_y \checkmark$$
c

Note $\sigma_y\sigma_x = -i\sigma_z$ (swap the indices — one factor of $i$ flips sign):

$$[\sigma_x,\sigma_y] = \sigma_x\sigma_y - \sigma_y\sigma_x = i\sigma_z - (-i\sigma_z) = 2i\sigma_z$$
$$\{\sigma_x,\sigma_y\} = \sigma_x\sigma_y + \sigma_y\sigma_x = i\sigma_z + (-i\sigma_z) = 0$$
Summary
$[\sigma_x,\sigma_y]=2i\sigma_z$ (consistent with general $[\sigma_i,\sigma_j]=2i\varepsilon_{ijk}\sigma_k$). $\{\sigma_x,\sigma_y\}=0$ — distinct Paulis anticommute, consistent with $\{\sigma_i,\sigma_j\}=2\delta_{ij}I$.
6
Robertson Uncertainty for Pauli Matrices
MediumUncertainty

In the state $|\psi\rangle = |u\rangle$ (eigenstate of $\sigma_z$), compute $\Delta\sigma_x\,\Delta\sigma_y$ and verify the Robertson inequality $\Delta\sigma_x\,\Delta\sigma_y \geq \tfrac{1}{2}|\langle[\sigma_x,\sigma_y]\rangle|$.

Complete Solution
LHS

$\langle\sigma_x\rangle = \langle u|\sigma_x|u\rangle = (1,0)\begin{pmatrix}0&1\\1&0\end{pmatrix}(1,0)^T = 0$.

$\langle\sigma_x^2\rangle = \langle u|I|u\rangle = 1$ (since $\sigma_x^2=I$). So $\Delta\sigma_x = \sqrt{1-0}=1$.

By the same argument: $\Delta\sigma_y = 1$. Thus $\Delta\sigma_x\,\Delta\sigma_y = 1$.

RHS

$[\sigma_x,\sigma_y]=2i\sigma_z$, so $\langle[\sigma_x,\sigma_y]\rangle = 2i\langle u|\sigma_z|u\rangle = 2i(+1) = 2i$.

$$\frac{1}{2}|\langle[\sigma_x,\sigma_y]\rangle| = \frac{1}{2}|2i| = 1$$
Result
LHS $= 1 \geq$ RHS $= 1$. The inequality is satisfied as an equality. This occurs because $|u\rangle$ is an eigenstate of $\sigma_z$ — the state that saturates the bound between $\sigma_x$ and $\sigma_y$. ✓
7
Beat Frequency Oscillation
HardDynamics

A spin has Hamiltonian $H=\tfrac{\omega}{2}\sigma_z$ and is in the state $|\psi(0)\rangle = \cos(\theta/2)|u\rangle + \sin(\theta/2)|d\rangle$ (real coefficients, $\varphi=0$).

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

(b) Compute $P(\sigma_x=+1,\,t)$ and identify the oscillation frequency.

(c) What are the maximum and minimum probabilities, and when do they occur?

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

$\langle r|\psi(t)\rangle = \tfrac{1}{\sqrt{2}}\bigl(\cos\tfrac{\theta}{2}\,e^{-i\omega t/2} + \sin\tfrac{\theta}{2}\,e^{+i\omega t/2}\bigr)$. Squaring:

$$P = \frac{1}{2}\!\left(\cos^2\!\tfrac{\theta}{2} + \sin^2\!\tfrac{\theta}{2} + 2\cos\!\tfrac{\theta}{2}\sin\!\tfrac{\theta}{2}\cos(\omega t)\right) = \frac{1 + \sin\theta\cos(\omega t)}{2}$$

The oscillation frequency is $\omega$ — the energy gap between the two levels (in natural units).

c
$$P_{\max} = \frac{1+\sin\theta}{2} \quad \text{at } t=0,\,2\pi/\omega,\ldots$$
$$P_{\min} = \frac{1-\sin\theta}{2} \quad \text{at } t=\pi/\omega,\,3\pi/\omega,\ldots$$
Special cases
When $\theta=\pi/2$ (equatorial Bloch vector): $P_{\max}=1$, $P_{\min}=0$ — complete Rabi oscillation. When $\theta=0$ (north pole, energy eigenstate): $P_{\max}=P_{\min}=\tfrac{1}{2}$ — no oscillation.
8
Heisenberg Equations of Motion
HardHeisenberg Picture

For $H=\tfrac{\omega}{2}\sigma_z$, use the Heisenberg equation $\dfrac{dA}{dt} = i[H,A]$ (with $\hbar=1$) to find the equations of motion for $\sigma_x(t)$, $\sigma_y(t)$, $\sigma_z(t)$. Solve the coupled equations.

Complete Solution
equations

Using $[\sigma_z,\sigma_x]=2i\sigma_y$, $[\sigma_z,\sigma_y]=-2i\sigma_x$, $[\sigma_z,\sigma_z]=0$:

$$\frac{d\sigma_x}{dt} = i\cdot\frac{\omega}{2}[\sigma_z,\sigma_x] = i\cdot\frac{\omega}{2}(2i\sigma_y) = -\omega\sigma_y$$
$$\frac{d\sigma_y}{dt} = i\cdot\frac{\omega}{2}[\sigma_z,\sigma_y] = i\cdot\frac{\omega}{2}(-2i\sigma_x) = +\omega\sigma_x$$
$$\frac{d\sigma_z}{dt} = i\cdot\frac{\omega}{2}[\sigma_z,\sigma_z] = 0$$
solve

Define $\xi = \sigma_x + i\sigma_y$. Then $\dot{\xi} = -\omega\sigma_y + i\omega\sigma_x = i\omega(\sigma_x+i\sigma_y) = i\omega\xi$. This has the solution $\xi(t) = \xi(0)e^{i\omega t}$. Separating real and imaginary parts:

$$\sigma_x(t) = \sigma_x(0)\cos(\omega t) - \sigma_y(0)\sin(\omega t)$$
$$\sigma_y(t) = \sigma_x(0)\sin(\omega t) + \sigma_y(0)\cos(\omega t)$$
$$\sigma_z(t) = \sigma_z(0) \quad \text{(constant)}$$
Interpretation
The spin operators rotate around the $z$-axis at frequency $\omega$ — a rotation matrix acting on $(\sigma_x,\sigma_y)$. This is operator precession in the Heisenberg picture, completely equivalent to the Schrödinger picture result. ✓