1.2.1 Measurement Postulate#
Prompts
State the three parts of the measurement postulate. For each part, explain why it must be true for measurement to connect theory to experiment.
The Born rule says the probability of outcome \(\lambda_i\) is \(|\langle \psi_i \vert \psi \rangle|^2\). Why do we square the inner product instead of just taking the inner product itself or its absolute value?
After measuring observable \(\hat{O}\) and obtaining outcome \(\lambda_i\), the state collapses to the eigenstate \(\vert\psi_i\rangle\). Is this a physically real event that happens in nature, or is it just a bookkeeping device—a rule for updating our knowledge?
Suppose a qubit is in state \(\vert\psi\rangle = \frac{1}{\sqrt{2}}(\vert0\rangle + \vert1\rangle)\) and we measure \(\hat{Z}\). Compute the probabilities of each outcome and write down the state after measurement for each possible outcome.
Two experiments: (1) Prepare the state, measure once, record outcome. (2) Prepare the state, measure it, then immediately measure it again. Will the second measurement always give the same result as the first? Design an experiment to test this prediction.
Lecture Notes#
Overview#
Quantum measurement is fundamentally different from passive observation. A measurement is a quantum operation that:
Yields a definite outcome (an eigenvalue of the observable)
Collapses the state irreversibly to an eigenstate
Follows the Born rule: probability is the squared amplitude of overlap with the eigenstate
This section formalizes measurement through the measurement postulate (three axioms) and demonstrates its power by explaining the classic Stern-Gerlach experiments.
Stern-Gerlach Experiments as Motivation#
The Stern-Gerlach apparatus measures spin by splitting an atomic beam in an inhomogeneous magnetic field.
We will use the notation:
\(\vert\uparrow\rangle, \vert\downarrow\rangle\): eigenstates of \(\hat{\sigma}^z\) (z-basis)
\(\vert\rightarrow\rangle, \vert\leftarrow\rangle\): eigenstates of \(\hat{\sigma}^x\) (x-basis)
Fig. 2 Stern-Gerlach setup: an inhomogeneous magnetic field spatially separates beams by spin projection, enabling state preparation and measurement.#
Experiment 1: Z-Z-Z (Sequential Stern-Gerlach)
Assume we start with 1000 atoms in an unpolarized (maximally mixed) ensemble \(\rho_0 = \tfrac{1}{2}\vert\uparrow\rangle\langle\uparrow\vert + \tfrac{1}{2}\vert\downarrow\rangle\langle\downarrow\vert\).
Step |
Observable |
Incoming state |
Outcomes (Born rule) |
Selection |
Atoms (approx.) |
Outgoing state |
|---|---|---|---|---|---|---|
1 (prep) |
\(\hat{\sigma}^z\) |
depolarized ensemble |
\(P(m)=\begin{cases}\tfrac{1}{2}, & m=+1\\ \tfrac{1}{2}, & m=-1\end{cases}\) |
Keep \(+1\) branch |
500 |
\(\vert\uparrow\rangle\) |
2 |
\(\hat{\sigma}^z\) |
\(\vert\uparrow\rangle\) |
\(P(m)=\begin{cases}1, & m=+1\\ 0, & m=-1\end{cases}\) |
Keep \(+1\) branch |
500 |
\(\vert\uparrow\rangle\) |
3 |
\(\hat{\sigma}^z\) |
\(\vert\uparrow\rangle\) |
\(P(m)=\begin{cases}1, & m=+1\\ 0, & m=-1\end{cases}\) |
Keep \(+1\) branch |
500 |
\(\vert\uparrow\rangle\) |
What Experiment 1 teaches us:
The measurement outcome is binary: the beam always splits into exactly two components, never a continuum.
The first apparatus doubles as a state-preparation device: filtering one branch by post-selection prepares a definite pure state.
Repeatability: measuring the same observable twice in a row always confirms the first result — the outcome is reproducible, suggesting the existence of an underlying physical reality that persists between measurements.
Experiment 2: Z-X-Z (Incompatible Measurements)
Again start with 1000 atoms in the same depolarized ensemble.
Step |
Observable |
Incoming state |
Outcomes (Born rule) |
Selection |
Atoms (approx.) |
Outgoing state |
|---|---|---|---|---|---|---|
1 (prep) |
\(\hat{\sigma}^z\) |
depolarized ensemble |
\(P(m)=\begin{cases}\tfrac{1}{2}, & m=+1\\ \tfrac{1}{2}, & m=-1\end{cases}\) |
Keep \(+1\) branch |
500 |
\(\vert\uparrow\rangle\) |
2 |
\(\hat{\sigma}^x\) |
\(\vert\uparrow\rangle\) |
\(P(m)=\begin{cases}\tfrac{1}{2}, & m=+1\\ \tfrac{1}{2}, & m=-1\end{cases}\) |
Keep \(+1\) branch |
250 |
\(\vert\rightarrow\rangle\) |
3 |
\(\hat{\sigma}^z\) |
\(\vert\rightarrow\rangle\) |
\(P(m)=\begin{cases}\tfrac{1}{2}, & m=+1\\ \tfrac{1}{2}, & m=-1\end{cases}\) |
Keep \(-1\) branch |
125 |
\(\vert\downarrow\rangle\) |
What Experiment 2 teaches us:
The intermediate X measurement washes out the previously definite Z outcome — the final Z result is now random again.
Measurement is invasive: it is not a passive readout of a pre-existing value, but an active intervention that can alter the state.
Non-commuting measurements change reality — inserting \(\hat{\sigma}^x\) between two \(\hat{\sigma}^z\) measurements destroys the Z-definiteness that Experiment 1 established.
Building the Mathematical Model#
Goal: Find a formula \(P(m \vert \psi, \hat{O})\) for the probability of obtaining outcome \(m\) when measuring observable \(\hat{O}\) on state \(\vert\psi\rangle\).
Positivity constraint: Since \(P(m \vert \psi, \hat{O}) \geq 0\) for all states \(\vert\psi\rangle\), we need a mathematical object that is guaranteed non-negative for any input state. A natural choice is to model \(P\) as the expectation value of some operator \(\hat{P}_{O=m}\):
This is automatically non-negative if \(\hat{P}_{O=m}\) is positive semi-definite (PSD).
Positive Semi-Definite (PSD) Operator
An operator \(\hat{A}\) is positive semi-definite if \(\langle \psi \vert \hat{A} \vert \psi \rangle \geq 0\) for all states \(\vert\psi\rangle\). Equivalently, all eigenvalues of \(\hat{A}\) are \(\geq 0\).
Pinning down the operator: The SG experiments (Z-Z-Z) showed that repeatedly measuring the same observable on an eigenstate \(\vert O{=}m\rangle\) always returns \(m\). This means:
What PSD operator satisfies this? The projection operator \(\hat{P}_{O=m} = \vert O{=}m\rangle\langle O{=}m\vert\).
Verification: Projector Gives the Right Probabilities
For eigenstate \(\vert\psi\rangle = \vert O{=}m\rangle\):
For a different eigenstate \(\vert\psi\rangle = \vert O{=}m'\rangle\) with \(m' \neq m\):
(using orthogonality of eigenstates).
For a general superposition \(\vert\psi\rangle = c_1 \vert O{=}m_1\rangle + c_2 \vert O{=}m_2\rangle\) with \(\vert c_1\vert^2 + \vert c_2\vert^2 = 1\):
This gives us the Born rule: the probability of outcome \(m\) is the squared overlap with the corresponding eigenstate,
The Measurement Postulate#
The Measurement Postulate
Axiom 1 (Possible Outcomes): When measuring observable \(\hat{O}\), the only possible outcomes are its eigenvalues \(m\).
Axiom 2 (Born Rule): The probability of obtaining outcome \(m\) from state \(\vert\psi\rangle\) is
where \(|O=m\rangle\) is the (normalized) eigenstate corresponding to eigenvalue \(m\).
Axiom 3 (State Collapse): Immediately after obtaining outcome \(m\), the state collapses to the corresponding eigenstate:
Interpretation of collapse: After the measurement, the system is no longer in a superposition. All information about the pre-measurement state is lost (unless we knew the outcome beforehand). This is irreversible and fundamental to quantum mechanics.
Explaining the Stern-Gerlach Experiments#
The process tables above can be summarized algebraically:
\(\vert\uparrow\rangle\) is an eigenstate of \(\hat{\sigma}^z\), so repeated Z-measurement is deterministic.
Decomposed to \(\hat{\sigma}^x\)-basis:
\[ \vert\uparrow\rangle = \frac{1}{\sqrt{2}}\left(\vert\rightarrow\rangle + \vert\leftarrow\rangle\right)\]On Z-basis state \(\vert\uparrow\rangle\), X-measurement outcomes 50-50.
Symmetry argument: \(\vert\rightarrow\rangle\) and \(\vert\leftarrow\rangle\) looks symmetric from \(\vert\uparrow\rangle\) perspective, so they must have identical probability to be observed.
Select \(\vert\rightarrow\rangle\), decompose to \(\hat{\sigma}^z\)-basis again:
\[ \vert\rightarrow\rangle = \frac{1}{\sqrt{2}}\left(\vert\uparrow\rangle + \vert\downarrow\rangle\right)\]So selecting one X branch (e.g. \(\vert\rightarrow\rangle\)) makes the final Z outcomes 50-50. This is quantum complementarity: the intermediate X measurement erases prior Z-definiteness.
Discussion
Why can’t we prepare atoms to have definite spin-up in both z and x directions simultaneously?
The eigenstates of \(\hat{\sigma}^z\) are \(\{\vert\uparrow\rangle, \vert\downarrow\rangle\}\). The eigenstates of \(\hat{\sigma}^x\) are \(\{\vert\rightarrow\rangle, \vert\leftarrow\rangle\}\). These bases are orthogonal (perpendicular in Hilbert space). A state that is an eigenstate of one operator is a superposition in the other operator’s basis.
This is not a limitation of experiment—it is built into the structure of quantum mechanics. Non-commuting observables cannot be simultaneously definite.
Summary#
Measurement is fundamental: it yields a definite outcome (an eigenvalue) and irreversibly collapses the state.
The Born rule gives the probability as the squared overlap \(|\langle O=m\vert\psi\rangle|^2\).
The three axioms formalize the process: outcomes are eigenvalues, probabilities follow Born rule, state collapses to eigenstate.
The Stern-Gerlach experiments directly demonstrate collapse and complementarity.
Repeated measurement of the same observable yields the same outcome with probability 1 (because the state is now an eigenstate).
Complementary measurements (non-commuting observables) can wipe out prior measurement information—this is not a flaw but a feature.
See Also
1.1.2 State and Representation: Pure states as unit vectors in Hilbert space; the mathematical language used throughout this lesson
1.1.3 Hermitian Operators: Pauli operators and spectral decomposition; the observables whose eigenvalues are measurement outcomes
1.2.2 Uncertainty and Incompatibility: Why non-commuting observables cannot be simultaneously sharp; quantitative bound on \(\Delta A \cdot \Delta B\)
1.2.3 Measurement Operators: Projectors as the mathematical representation of “collapse”; Bayesian updating interpretation
Homework#
1. A qubit is in the state \(\vert\psi\rangle = \frac{2\vert0\rangle + \sqrt{5}\vert1\rangle}{3}\). Measure \(\hat{Z}\). (a) Find the probability of each outcome using the Born rule. (b) Write down the state immediately after obtaining outcome \(-1\).
2. Start with the state \(\vert+\rangle = \frac{1}{\sqrt{2}}(\vert0\rangle + \vert1\rangle)\). (a) Measure \(\hat{Z}\): find the probabilities and the post-measurement state for each outcome. (b) Immediately after measuring \(\hat{Z}\) and obtaining outcome \(+1\), measure \(\hat{X}\). What are the outcome probabilities now? (c) Compare to measuring only \(\hat{X}\) on the original state \(\vert+\rangle\). Explain the difference physically.
3. Show that if measuring observable \(\hat{O}\) yields outcome \(m\) (with eigenstate \(\vert O=m\rangle\)), an immediately repeated measurement of \(\hat{O}\) gives \(m\) again with probability 1. (Hint: Apply the collapse postulate, then the Born rule.)
4. The Stern-Gerlach experiment: A qubit prepared in state \(\vert\psi\rangle = \frac{1}{\sqrt{3}}\vert0\rangle + \sqrt{\frac{2}{3}}\vert1\rangle\) is measured in sequence: first \(\hat{Z}\), then \(\hat{X}\), then \(\hat{Z}\) again. (a) Compute the probability of each outcome at each stage, assuming all measurements succeed. (b) Write down the final state after all three measurements.
5. For the state \(\vert0\rangle\), show that measuring \(\hat{X}\) gives outcomes \(+1\) and \(-1\) with equal probability \(1/2\). (Hint: Express \(\vert0\rangle\) in the \(\hat{X}\) eigenbasis \(\{\vert+\rangle, \vert-\rangle\}\).)
6. Two states: \(\vert\psi_1\rangle = \frac{1}{\sqrt{2}}(\vert0\rangle + \vert1\rangle)\) and \(\vert\psi_2\rangle = \frac{1}{\sqrt{2}}(\vert0\rangle + \mathrm{e}^{\mathrm{i}\varphi}\vert1\rangle)\) (real \(\varphi\)). (a) Show that both states give identical measurement probabilities for \(\hat{Z}\). (b) Compute the probability of outcome \(+1\) when measuring \(\hat{X}\) on \(\vert\psi_2\rangle\) as a function of \(\varphi\). (c) Explain why the relative phase is invisible in \(\hat{Z}\) measurements but affects \(\hat{X}\) measurements.
7. Consider the three-apparatus Stern-Gerlach experiment: \(\hat{Z}\) \(\to\) (filter \(+1\)) \(\to\) \(\hat{X}\) \(\to\) (filter \(-1\)) \(\to\) \(\hat{Z}\) again. Starting with \(\vert0\rangle\), what fraction of atoms make it through all three stages? What is the final state for those that succeed?
8. Suppose we measure \(\hat{Z}\) on a superposition state and obtain outcome \(-1\). (a) Write down the state after collapse. (b) If we immediately measure \(\hat{Z}\) again, what is the probability of obtaining \(-1\)? (c) If instead we measure \(\hat{X}\) immediately after, what are the outcome probabilities?