Lecture 01 · Foundation

The Language of Quantum Mechanics

Why visualization fails, classical vs quantum states, the qubit, Stern-Gerlach experiments, wave function collapse, and the mathematics of complex vector spaces.

Theory Exercises

Part I — Classical Mechanics: A Brief Recap

In classical mechanics, the state of a system is everything needed to predict its future. For a single particle in one dimension, the state is the pair $(x, p)$ — position and momentum. All possible states form phase space: a continuous plane where every point represents a fully specified condition.

The simplest classical system: a coin with two states — heads $H$ and tails $T$. The laws of motion are rules for updating the state at each step:

Law	Heads →	Tails →
"Nothing happens"	Heads	Tails
"Flip every step"	Tails	Heads

These are deterministic: knowing the state now determines the state at every future moment. Quantum mechanics shatters this principle.

Part II — Why Quantum Mechanics Cannot Be Visualized

You cannot visualize quantum mechanics. Stop trying.

This is not defeatism — it is the most practical advice a physicist can give. Classical mechanics maps onto everyday experience because it was built from everyday experience. Quantum mechanics describes atoms, electrons, and photons — objects so small that your neural architecture offers no reliable guide.

Every attempt to picture a quantum state — "the electron is here and there at once," "spin is a little top" — is a metaphor that eventually fails. The correct approach is abstraction: learn to manipulate symbols according to precise mathematical rules and accept that formal manipulation is understanding.

Part III — The Qubit

Just as Susskind started classical mechanics with the simplest system (the coin), he starts quantum mechanics with the qubit — a quantum system with two possible measurement outcomes, conventionally labeled:

|\uparrow\rangle \equiv +1 \qquad \text{and} \qquad |\downarrow\rangle \equiv -1

The physical realization: the spin of an electron, measured with a Stern-Gerlach device. Unlike a classical coin (definitely heads or tails), a qubit can exist in a superposition:

General Qubit State

$$|\psi\rangle = \alpha_1 |\uparrow\rangle + \alpha_2 |\downarrow\rangle, \quad \alpha_1, \alpha_2 \in \mathbb{C}$$

The probabilities of measuring each outcome are given by the Born rule:

P(+1) = |\alpha_1|^2, \qquad P(-1) = |\alpha_2|^2, \qquad |\alpha_1|^2 + |\alpha_2|^2 = 1

Part IV — Stern-Gerlach Experiments

Experiment 1: The Unprepared Beam

Send electrons through a vertical Stern-Gerlach apparatus. The beam splits into two: half deflect upward ($+1$), half downward ($-1$). No electron ever registers an intermediate value. This is already non-classical — a classical spinning top would show a continuous range.

Experiment 2: State Preparation

Select only the $+1$ (spin-up) electrons. Send them through a second identical vertical device. Result: every single electron registers $+1$. By selecting the outcome, we have prepared the state $|\uparrow\rangle$.

Experiment 3: Rotating the Detector 90°

Take prepared $|\uparrow\rangle_z$ electrons and pass them through a horizontal Stern-Gerlach device. Classical expectation: a vector pointing up has zero horizontal component, so it should read $0$.

What actually happens: every electron registers either $+1$ or $-1$ — never $0$ — randomly, with equal 50/50 probability. Why? Because:

|\uparrow\rangle_z = \frac{1}{\sqrt{2}}|+\rangle_x + \frac{1}{\sqrt{2}}|-\rangle_x

Each term has amplitude $\tfrac{1}{\sqrt{2}}$, so probability $\tfrac{1}{2}$. The average over many measurements is $0$ — agreeing with classical expectation — but individual measurements are never $0$.

Part V — Wave Function Collapse

After a measurement, the state collapses to the eigenstate corresponding to the observed outcome. The logical chain:

Prepare $|\uparrow\rangle_z$: measuring $\hat{\sigma}_z$ always gives $+1$. ✓
Measure $\hat{\sigma}_x$: get $+1$ (randomly). State collapses to $|+\rangle_x$.
Measure $\hat{\sigma}_x$ again: always get $+1$. ✓ (collapse confirmed)
Now measure $\hat{\sigma}_z$ again: result is random again — $\pm1$ each with probability $\tfrac{1}{2}$.

Step 4 is crucial: measuring $\hat{\sigma}_x$ destroyed the definite value of $\hat{\sigma}_z$. This is because $\hat{\sigma}_z$ and $\hat{\sigma}_x$ do not commute:

[\hat{\sigma}_z, \hat{\sigma}_x] = \hat{\sigma}_z\hat{\sigma}_x - \hat{\sigma}_x\hat{\sigma}_z \neq 0

Non-commuting observables cannot both have definite values simultaneously. This is a mathematical theorem, not a limitation of instruments.

Part VI — Complex Vector Spaces

A complex vector space $V$ over $\mathbb{C}$ supports addition and scalar multiplication with complex scalars. The state space of a qubit is $\mathbb{C}^2$ — pairs of complex numbers as column vectors:

|\psi\rangle = \begin{pmatrix} \alpha_1 \\ \alpha_2 \end{pmatrix}, \quad \alpha_1, \alpha_2 \in \mathbb{C}

The standard basis vectors correspond to the two spin eigenstates:

|\uparrow\rangle = \begin{pmatrix}1\\0\end{pmatrix}, \qquad |\downarrow\rangle = \begin{pmatrix}0\\1\end{pmatrix}

The Inner Product and Born Rule

The inner product of two states in $\mathbb{C}^2$:

\langle\phi|\psi\rangle = \phi_1^*\psi_1 + \phi_2^*\psi_2

where $^*$ denotes complex conjugation. Physical states satisfy $\langle\psi|\psi\rangle = 1$ (normalization). The Born rule — the fundamental link between mathematics and probability:

Born Rule

$$P(\phi\,|\,\psi) = |\langle\phi|\psi\rangle|^2$$

Why Complex Numbers?

The use of complex (rather than real) numbers is essential. Imaginary parts encode phase information — the relative phase between components — which has measurable consequences in quantum interference. Restricting to real numbers gives the wrong predictions.

Part VII — Classical vs Quantum: The Deep Contrast

Concept	Classical	Quantum
State	Point $(x,p)$ in phase space	Vector $\|\psi\rangle \in \mathbb{C}^2$
Evolution	Deterministic (Newton)	Deterministic (Schrödinger) between measurements
Measurement	Passive — reads pre-existing value	Active — collapses the state
Outcome	Continuous, predictable	Discrete $\pm1$, probabilistic
Math	Real numbers, ODEs	Complex vector spaces, linear algebra

General qubit state

$|\psi\rangle = \alpha_1|\uparrow\rangle + \alpha_2|\downarrow\rangle$

Normalization

$|\alpha_1|^2 + |\alpha_2|^2 = 1$

Inner product ($\mathbb{C}^2$)

$\langle\phi|\psi\rangle = \phi_1^*\psi_1 + \phi_2^*\psi_2$

Born rule

$P = |\langle\phi|\psi\rangle|^2$

Exercises — Lecture 1

Eight exercises covering all major topics. Click any exercise to reveal the full worked solution.

Normalization Condition

EasyBorn Rule

A qubit is in the state $|\psi\rangle = \dfrac{1}{\sqrt{3}}|\uparrow\rangle + \beta|\downarrow\rangle$ where $\beta$ is a real positive number.

(a) Find $\beta$ using the normalization condition.

(b) What is the probability of measuring spin-up? Spin-down?

Complete Solution

The normalization condition requires $|\alpha_1|^2 + |\alpha_2|^2 = 1$. Here $\alpha_1 = 1/\sqrt{3}$ and $\alpha_2 = \beta$:

$$\left|\frac{1}{\sqrt{3}}\right|^2 + \beta^2 = 1 \implies \frac{1}{3} + \beta^2 = 1 \implies \beta^2 = \frac{2}{3}$$

Since $\beta > 0$:

Answer

$$\beta = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{6}}{3}$$

By the Born rule, probabilities are the squared moduli of amplitudes:

$$P(+1) = |\alpha_1|^2 = \frac{1}{3}, \qquad P(-1) = \beta^2 = \frac{2}{3}$$

Check: $\frac{1}{3} + \frac{2}{3} = 1$ ✓

Inner Product Calculation

EasyVector Space

Let $|\phi\rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}$ and $|\psi\rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}$.

(a) Compute $\langle\phi|\psi\rangle$.

(b) Are these states orthogonal? What does that mean physically?

(c) If a system is in state $|\psi\rangle$, what is the probability of measuring outcome $|\phi\rangle$?

Complete Solution

The bra of $|\phi\rangle$ is the row vector of complex conjugates. Since all components are real here:

$$\langle\phi|\psi\rangle = \frac{1}{\sqrt{2}}(1,\; 1) \cdot \frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix} = \frac{1}{2}(1\cdot 1 + 1\cdot(-1)) = \frac{1}{2}(0) = 0$$

Yes — the inner product is zero, so these states are orthogonal. Physically, this means they are perfectly distinguishable: if the system is in state $|\psi\rangle$, there is zero probability of mistaking it for $|\phi\rangle$. These are exactly the $|r\rangle$ and $|l\rangle$ (right and left) spin eigenstates along the $x$-axis.

By the Born rule:

$$P = |\langle\phi|\psi\rangle|^2 = |0|^2 = 0$$

Answer

Probability = 0. Orthogonal states are mutually exclusive.

Born Rule: Measuring a Prepared State

MediumStern-Gerlach

The spin-up state along the $z$-axis is $|\uparrow_z\rangle = \begin{pmatrix}1\\0\end{pmatrix}$, and the spin-right state along the $x$-axis is $|r\rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}$.

(a) Compute $\langle r|\uparrow_z\rangle$.

(b) If a system is prepared in $|\uparrow_z\rangle$ and measured with a horizontal ($x$-axis) Stern-Gerlach device, what is $P(\sigma_x = +1)$? Does this match the experimental result described in the lecture?

(c) What is $P(\sigma_x = -1)$?

Complete Solution

$$\langle r|\uparrow_z\rangle = \frac{1}{\sqrt{2}}(1,\; 1)\begin{pmatrix}1\\0\end{pmatrix} = \frac{1}{\sqrt{2}}(1\cdot1 + 1\cdot0) = \frac{1}{\sqrt{2}}$$

$$P(\sigma_x = +1) = |\langle r|\uparrow_z\rangle|^2 = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}$$

Yes — this matches exactly: starting from spin-up along $z$, measuring along $x$ gives $+1$ with probability $\tfrac{1}{2}$. This is the 50/50 result of Experiment 3.

The left-state is $|l\rangle = \tfrac{1}{\sqrt{2}}(1,-1)^T$, so:

$$P(\sigma_x = -1) = |\langle l|\uparrow_z\rangle|^2 = \left|\frac{1}{\sqrt{2}}(1\cdot1 + (-1)\cdot0)\right|^2 = \frac{1}{2}$$

Summary

$P(\sigma_x=+1) = \frac{1}{2}$, $P(\sigma_x=-1) = \frac{1}{2}$. Both probabilities equal $\frac{1}{2}$ — 50/50, as observed.

Phase Freedom: Global vs Relative Phase

MediumConceptual

Consider two states:

$$|\psi\rangle = \frac{1}{\sqrt{2}}(|\uparrow\rangle + |\downarrow\rangle) \qquad \text{and} \qquad |\psi'\rangle = \frac{1}{\sqrt{2}}(|\uparrow\rangle - |\downarrow\rangle)$$

(a) Are these the same physical state? Compute $\langle\uparrow|\psi\rangle$, $\langle\uparrow|\psi'\rangle$ and compare probabilities.

(b) Now consider $|\chi\rangle = e^{i\pi/4}|\psi\rangle$. Are $|\chi\rangle$ and $|\psi\rangle$ the same physical state? Why?

Complete Solution

For $|\psi\rangle$: $\langle\uparrow|\psi\rangle = \frac{1}{\sqrt{2}}$, so $P(\uparrow) = \frac{1}{2}$.

For $|\psi'\rangle$: $\langle\uparrow|\psi'\rangle = \frac{1}{\sqrt{2}}$, so $P(\uparrow) = \frac{1}{2}$.

Both give $P(\uparrow) = P(\downarrow) = \frac{1}{2}$ when measured along $z$. However, these are different physical states! They differ in the relative phase between components. Measuring along the $y$-axis would distinguish them — $|\psi\rangle = |r\rangle$ (spin-right) while $|\psi'\rangle = |l\rangle$ (spin-left).

$$|\psi\rangle = \frac{1}{\sqrt{2}}(|\uparrow\rangle+|\downarrow\rangle) = |r\rangle, \quad |\psi'\rangle = \frac{1}{\sqrt{2}}(|\uparrow\rangle-|\downarrow\rangle) = |l\rangle$$

For any observable probability: $|\langle\phi|e^{i\pi/4}\psi\rangle|^2 = |e^{i\pi/4}|^2|\langle\phi|\psi\rangle|^2 = |\langle\phi|\psi\rangle|^2$.

The global phase $e^{i\pi/4}$ cancels in every measurement. So $|\chi\rangle$ and $|\psi\rangle$ are the same physical state — they differ only by a global phase, which has no observable consequences. States are really rays in Hilbert space, not individual vectors.

Complex Amplitudes and Phase

MediumComplex Numbers

A qubit is in state $|\psi\rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix}1\\i\end{pmatrix}$, which is the spin-"in" state $|i\rangle_y$ (eigenstate of $\sigma_y$ with eigenvalue $+1$).

(a) Verify this state is normalized.

(b) Compute $P(\sigma_z = +1)$ and $P(\sigma_z = -1)$.

(c) Is this the same state as $\tfrac{1}{\sqrt{2}}\begin{pmatrix}1\\-i\end{pmatrix}$? What observable distinguishes them?

Complete Solution

$$\langle\psi|\psi\rangle = \frac{1}{\sqrt{2}}(1^*, (-i)^*)\cdot\frac{1}{\sqrt{2}}\begin{pmatrix}1\\i\end{pmatrix} = \frac{1}{2}(1\cdot1 + i\cdot i) = \frac{1}{2}(1 + i^2)$$

Wait — the bra uses complex conjugates: $(i)^* = -i$. So:

$$\frac{1}{2}(1\cdot1 + (-i)\cdot i) = \frac{1}{2}(1 + (-i^2)) = \frac{1}{2}(1+1) = 1 \checkmark$$

The $z$-eigenstates are $|\uparrow\rangle = (1,0)^T$ and $|\downarrow\rangle = (0,1)^T$.

$$P(+1) = |\langle\uparrow|\psi\rangle|^2 = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}$$

$$P(-1) = |\langle\downarrow|\psi\rangle|^2 = \left|\frac{i}{\sqrt{2}}\right|^2 = \frac{|i|^2}{2} = \frac{1}{2}$$

The state $\tfrac{1}{\sqrt{2}}(1,-i)^T = |o\rangle_y$ is the spin-out eigenstate of $\sigma_y$ with eigenvalue $-1$. They give identical probabilities for $\sigma_z$ and $\sigma_x$ measurements, but $\sigma_y$ distinguishes them: $|i\rangle_y$ gives $P(\sigma_y=+1)=1$, while $|o\rangle_y$ gives $P(\sigma_y=-1)=1$.

Collapse Sequence

MediumCollapse

A spin is prepared in the state $|\uparrow\rangle_z$.

A measurement of $\sigma_x$ is performed and yields $+1$. What is the new state?
Immediately after, $\sigma_z$ is measured. What are the probabilities of each outcome?
If $\sigma_z = +1$ is obtained, what is the state now?
Now $\sigma_x$ is measured again. What are the probabilities?

What does this sequence demonstrate about non-commuting observables?

Complete Solution

Measuring $\sigma_x = +1$ collapses the state to the $x$-spin-up eigenstate: $|r\rangle = \tfrac{1}{\sqrt{2}}(1,1)^T$.

From $|r\rangle$, measuring $\sigma_z$:

$$P(\sigma_z=+1) = |\langle\uparrow|r\rangle|^2 = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}$$

$$P(\sigma_z=-1) = |\langle\downarrow|r\rangle|^2 = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}$$

Even though we started with definite $\sigma_z = +1$, the $\sigma_x$ measurement has destroyed that information — now $\sigma_z$ is random again.

Getting $\sigma_z = +1$ collapses the state back to $|\uparrow\rangle_z = (1,0)^T$.

Back in $|\uparrow\rangle_z$, measuring $\sigma_x$ again gives 50/50. We're back to step 1!

Key Insight

The measurements of $\sigma_z$ and $\sigma_x$ are irreversibly entangled in their collapse. Gaining information about one destroys the other. This is the essence of quantum non-commutativity: $[\sigma_z, \sigma_x] \neq 0$.

Functions as Vectors

MediumVector Space

The set of square-integrable functions $L^2(\mathbb{R})$ forms an infinite-dimensional complex vector space.

(a) Verify that $L^2(\mathbb{R})$ is closed under addition: if $\psi_1, \psi_2 \in L^2(\mathbb{R})$, is $\psi_1 + \psi_2 \in L^2(\mathbb{R})$? (Use the triangle inequality for integrals.)

(b) The inner product on $L^2(\mathbb{R})$ is $\langle\phi|\psi\rangle = \int_{-\infty}^{\infty}\phi^*(x)\psi(x)\,dx$. Explain why complex conjugation is required for this to be a valid inner product (hint: consider $\langle\psi|\psi\rangle$).

Complete Solution

We need to show $\int|(\psi_1+\psi_2)(x)|^2\,dx < \infty$ given both individual integrals are finite. Expanding:

$$|\psi_1+\psi_2|^2 \leq (|\psi_1|+|\psi_2|)^2 = |\psi_1|^2 + 2|\psi_1||\psi_2| + |\psi_2|^2$$

By the Cauchy-Schwarz inequality, $2|\psi_1||\psi_2| \leq |\psi_1|^2 + |\psi_2|^2$, so the sum is bounded by $2|\psi_1|^2 + 2|\psi_2|^2$, and the integral is finite. ✓

Without conjugation, we would have $\int \psi(x)\psi(x)\,dx = \int\psi^2(x)\,dx$. For a complex function $\psi = u + iv$, this gives $\int(u^2-v^2+2iuv)\,dx$, which is complex. A norm (squared length) must be real and non-negative.

With conjugation: $\langle\psi|\psi\rangle = \int\psi^*(x)\psi(x)\,dx = \int|\psi(x)|^2\,dx \geq 0$ — always real and non-negative.

Key Point

Complex conjugation in the inner product is the minimal modification needed to make a real, non-negative norm possible for complex-valued functions.

Average Spin Along Arbitrary Direction

HardIntegration

A spin is prepared along the $z$-axis in state $|\uparrow\rangle_z$. The apparatus is then rotated by angle $\theta$ from the $z$-axis in the $xz$-plane. The spin eigenstates along this direction are:

$$|+\rangle_\theta = \cos\frac{\theta}{2}|\uparrow\rangle + \sin\frac{\theta}{2}|\downarrow\rangle, \qquad |-\rangle_\theta = -\sin\frac{\theta}{2}|\uparrow\rangle + \cos\frac{\theta}{2}|\downarrow\rangle$$

(a) Compute $P(+1)$ and $P(-1)$ when measuring spin along the $\theta$-direction.

(b) Compute the average $\langle\sigma_\theta\rangle = P(+1)\cdot(+1) + P(-1)\cdot(-1)$.

(c) Show this equals $\hat{n}\cdot\hat{z} = \cos\theta$, confirming Susskind's correlation formula.

Complete Solution

$$P(+1) = |\langle +_\theta|\uparrow_z\rangle|^2 = \left|\cos\frac{\theta}{2}\langle\uparrow|\uparrow\rangle + \sin\frac{\theta}{2}\langle\downarrow|\uparrow\rangle\right|^2 = \cos^2\!\frac{\theta}{2}$$

$$P(-1) = |\langle -_\theta|\uparrow_z\rangle|^2 = \left|-\sin\frac{\theta}{2}\right|^2 = \sin^2\!\frac{\theta}{2}$$

$$\langle\sigma_\theta\rangle = (+1)\cos^2\!\frac{\theta}{2} + (-1)\sin^2\!\frac{\theta}{2} = \cos^2\!\frac{\theta}{2} - \sin^2\!\frac{\theta}{2}$$

Using the double-angle identity $\cos^2(A) - \sin^2(A) = \cos(2A)$:

$$\langle\sigma_\theta\rangle = \cos^2\!\frac{\theta}{2} - \sin^2\!\frac{\theta}{2} = \cos\theta$$

Result

$\langle\sigma_\theta\rangle = \cos\theta = \hat{n}_\theta\cdot\hat{z}$. The average spin equals the classical dot product — quantum mechanics recovers classical averages, even though individual outcomes are always $\pm 1$.

Next → Measurement & Quantum Logic