4. Measurement & Collapse
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Quantum measurement is fundamentally different from classical observation: the act of measurement changes the quantum state in a profound, irreversible way.
Video Lecture
Quantum Measurement Problem - Explained
Deep dive into the quantum measurement problem, wave function collapse, and the role of observation in quantum mechanics
💡 Tip: Watch at 1.25x or 1.5x speed for efficient learning. Use YouTube's subtitle feature if available.
Von Neumann's Measurement Theory
Mathematical Foundations of Quantum Mechanics
Von Neumann formalized the measurement process, distinguishing between two types of evolution: (1) continuous, deterministic evolution via the Schrödinger equation (unitary evolution), and (2) discontinuous, probabilistic collapse during measurement (projection postulate).
Why it matters: This 'measurement problem' - why and how does collapse occur? - remains one of the deepest puzzles in quantum foundations and drives modern interpretations of quantum mechanics.
The Measurement Problem
The paradox: Before measurement, a quantum system exists in a superposition of multiple states. Upon measurement, it "collapses" to a single definite state.
Before measurement: General superposition
where $|a_n\rangle$ are eigenstates of observable $\hat{A}$ with eigenvalues $a_n$
After measurement of outcome $a_k$: Instantaneous collapse
with probability $P(a_k) = |c_k|^2 = |\langle a_k|\psi\rangle|^2$
Key Questions:
- What constitutes a "measurement"?
- When exactly does collapse occur?
- Is collapse a physical process or an update of knowledge?
- What role does the observer play?
📝 Worked Example: Measurement Probabilities
Problem: A particle is in state $|\psi\rangle = \frac{3}{5}|E_1\rangle + \frac{4}{5}|E_2\rangle$ where $|E_1\rangle$ and $|E_2\rangle$ are energy eigenstates. What are the probabilities of measuring $E_1$ or $E_2$? What is the state after measuring $E_2$?
Step 1: Verify normalization
Step 2: Calculate measurement probabilities
Step 3: State after measuring $E_2$
The wave function collapses completely:
The state is now a pure eigenstate of energy with value $E_2$. A subsequent immediate measurement would yield $E_2$ with 100% certainty.
Projection Postulate (Born Rule)
The general collapse formula for projective measurements:
where $\hat{P}_k = |a_k\rangle\langle a_k|$ is the projector onto eigenstate $|a_k\rangle$
For degenerate eigenvalues:
If eigenvalue $a_k$ has degeneracy $g_k$ with orthonormal eigenstates $\{|a_k, i\rangle\}_{i=1}^{g_k}$:
The state collapses into the degenerate subspace but remains a superposition within it.
Properties of Projectors:
- Hermitian: $\hat{P}_k^\dagger = \hat{P}_k$
- Idempotent: $\hat{P}_k^2 = \hat{P}_k$ (measuring twice gives same result)
- Orthogonal: $\hat{P}_j\hat{P}_k = \delta_{jk}\hat{P}_k$
- Complete: $\sum_k \hat{P}_k = \hat{I}$
📝 Worked Example: Sequential Measurements
Problem: A spin-1/2 particle is in state $|\psi\rangle = \frac{1}{\sqrt{3}}|\uparrow_z\rangle + \sqrt{\frac{2}{3}}|\downarrow_z\rangle$. We measure $S_z$ and obtain $-\hbar/2$. What is the post-measurement state? If we immediately measure $S_x$, what are the possible outcomes and probabilities?
Step 1: After measuring $S_z = -\hbar/2$
The state has collapsed to the spin-down eigenstate.
Step 2: Express in $S_x$ basis
We know: $|\downarrow_z\rangle = \frac{1}{\sqrt{2}}(|\uparrow_x\rangle - |\downarrow_x\rangle)$
Step 3: Probabilities for $S_x$ measurement
Key insight: Even though we just "knew" the particle was in $|\downarrow_z\rangle$, measuring a different observable ($S_x$) gives probabilistic results. The measurements are incompatible!
🤔 Self-Check Question
True or False: If we measure an observable and get eigenvalue $a_k$, then immediately measure the same observable again, we are guaranteed to get $a_k$ again.
Show Answer
True. This is a consequence of wave function collapse. After the first measurement, the state is $|a_k\rangle$, which is an eigenstate of the observable with eigenvalue $a_k$. A subsequent measurement of the same observable yields $a_k$ with 100% probability (assuming no time has passed for the system to evolve).
Compatible Observables
Two observables $\hat{A}$ and $\hat{B}$ are compatible (or commute) if:
Fundamental Theorem:
Compatible observables share a complete set of simultaneous eigenstates:
Physical significance: Compatible observables can be measured simultaneously with definite values. Measuring one doesn't disturb the other.
Examples:
- Compatible: $[\hat{L}^2, \hat{L}_z] = 0$ (total and z-component of angular momentum)
- Compatible: $[\hat{H}, \hat{L}^2] = 0$ for central potentials
- Incompatible: $[\hat{x}, \hat{p}] = i\hbar \neq 0$ (position and momentum)
- Incompatible: $[\hat{L}_x, \hat{L}_y] = i\hbar\hat{L}_z \neq 0$ (different angular momentum components)
🔗 Related Topic: Uncertainty Principle - The incompatibility of observables (non-zero commutator) leads directly to the Heisenberg uncertainty principle.
Complete Set of Commuting Observables (CSCO)
A set of observables $\{\hat{A}, \hat{B}, \hat{C}, \ldots\}$ forms a CSCO if:
- All observables commute pairwise: $[\hat{A}, \hat{B}] = 0$, etc.
- Their simultaneous eigenvalues uniquely label all quantum states
Notation: States labeled by quantum numbers
where $a, b, c$ are eigenvalues of $\hat{A}, \hat{B}, \hat{C}$
Key Examples in Quantum Mechanics:
1. Hydrogen Atom:
$\{\hat{H}, \hat{L}^2, \hat{L}_z, \hat{S}_z\}$ → states $|n,\ell,m,m_s\rangle$
2. Particle in a box (1D):
$\{\hat{H}\}$ → states $|n\rangle$ (just one operator needed!)
3. 3D Harmonic Oscillator:
$\{\hat{H}, \hat{L}^2, \hat{L}_z\}$ → states $|n,\ell,m\rangle$
4. Free Particle:
$\{\hat{p}_x, \hat{p}_y, \hat{p}_z\}$ → states $|p_x,p_y,p_z\rangle$
📝 Worked Example: CSCO for 3D Harmonic Oscillator
Problem: Consider a 3D isotropic harmonic oscillator. Why is $\{\hat{H}, \hat{L}^2, \hat{L}_z\}$ a CSCO, but $\{\hat{H}, \hat{L}_z\}$ is not?
Check 1: Do all operators commute?
- $[\hat{H}, \hat{L}^2] = 0$ ✓ (spherical symmetry)
- $[\hat{H}, \hat{L}_z] = 0$ ✓
- $[\hat{L}^2, \hat{L}_z] = 0$ ✓
So $\{\hat{H}, \hat{L}^2, \hat{L}_z\}$ all commute.
Check 2: Do they uniquely label states?
For 3D harmonic oscillator, energy levels are:
Level $n$ has degeneracy $(n+1)(n+2)/2$ because multiple $(\ell, m)$ combinations give the same $n$.
- $\{\hat{H}\}$ alone: NOT complete (degenerate states)
- $\{\hat{H}, \hat{L}_z\}$: NOT complete (different $\ell$ can have same $m$)
- $\{\hat{H}, \hat{L}^2, \hat{L}_z\}$: COMPLETE! The triple $(n, \ell, m)$ uniquely specifies the state.
Answer: Only the full set $\{\hat{H}, \hat{L}^2, \hat{L}_z\}$ is a CSCO.
Types of Measurements
Strong (Projective) Measurements
The "standard" quantum measurement described by von Neumann:
- Complete collapse to eigenstate
- Maximal information gain
- Maximum disturbance to system
- Irreversible process
Example: Stern-Gerlach experiment measuring $S_z$
Weak Measurements
Minimal interaction with the system:
- Partial information extracted
- Minimal disturbance
- State remains nearly unchanged
- Can reveal "weak values" (sometimes outside eigenvalue spectrum!)
Weak value formula:
where $|\psi_i\rangle$ is pre-selected and $|\psi_f\rangle$ is post-selected
Applications: Precision metrology, amplification of small signals
Generalized Measurements (POVMs)
Positive Operator-Valued Measure: Most general quantum measurement framework
where $\hat{E}_m \geq 0$ and $\sum_m \hat{E}_m = \hat{I}$
POVMs generalize projective measurements and include indirect measurements, measurements with noise, etc.
💡 Application: Measurement-Based Quantum Computing
Measurement isn't just about extracting information—it's a computational resource! In one-way quantum computing, single-qubit measurements on a highly entangled 'cluster state' can perform universal quantum computation. The choice of measurement basis controls the computation flow.
Decoherence and the Classical Limit
The measurement problem revisited: Why don't we see macroscopic superpositions?
Decoherence mechanism:
Quantum systems interact with their environment. This entanglement causes rapid loss of coherence:
If environment states $|E_n\rangle$ are approximately orthogonal, the system behaves classically:
Density matrix is diagonal — no interference terms!
Decoherence Time Scales:
- Isolated atom: milliseconds to seconds
- Superconducting qubit: microseconds
- Dust particle: $\sim 10^{-40}$ seconds!
- Macroscopic object: essentially instantaneous
Decoherence explains why classical physics emerges from quantum mechanics without invoking "collapse."
🔗 Related Topic: Decoherence Theory - Advanced treatment of decoherence, density matrices, and the quantum-to-classical transition
💡 Application: Quantum Error Correction
To build useful quantum computers, we must fight decoherence! Quantum error correction codes protect quantum information by encoding logical qubits redundantly across multiple physical qubits. The breakthrough: we can measure error syndromes without measuring (and thus collapsing) the quantum state itself.
🤔 Self-Check Question: Qubit Measurement
A qubit is in state $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$. We perform a measurement in the computational basis.
(a) What are the possible outcomes?
(b) If we get outcome '0', what is the post-measurement state?
(c) Can we determine $\alpha$ and $\beta$ from a single measurement?
Show Answer
(a) Possible outcomes: 0 or 1 with probabilities $P(0) = |\alpha|^2$ and $P(1) = |\beta|^2$
(b) Post-measurement state after getting '0': $|\psi'\rangle = |0\rangle$ (complete collapse)
(c) No! A single measurement only gives one bit of information (0 or 1). To determine $\alpha$ and $\beta$ (which encode infinite classical information), we would need to prepare many identical copies and perform statistical measurements. This is related to the no-cloning theorem: we cannot copy an unknown quantum state.
Summary: The Measurement Postulate
- • Measurement of observable $\hat{A}$ yields eigenvalue $a_k$ with probability $P(a_k) = |\langle a_k|\psi\rangle|^2$
- • State immediately collapses: $|\psi\rangle \to |a_k\rangle$ (or into degenerate subspace)
- • Compatible observables $([\hat{A}, \hat{B}] = 0)$ can be measured simultaneously
- • A CSCO provides complete labeling of quantum states
- • Measurement types range from strong (projective) to weak (minimal disturbance)
- • Decoherence explains the emergence of classical behavior without explicit collapse
Video Lecture
Bell's Theorem and Measurement
How quantum measurement relates to Bell's inequalities and the nature of reality in quantum mechanics
💡 Tip: Watch at 1.25x or 1.5x speed for efficient learning. Use YouTube's subtitle feature if available.