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Eigenvalues as the only possible results, $|\langle f_n|\Psi\rangle|^2$ as their probabilities, collapse after a measurement, and averages over ensembles.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to predict the possible results of a measurement, their probabilities and their average, and the state the system is left in.
You know Born's rule for position, $|\Psi|^2$ as a probability density, and from the last lesson that observables are Hermitian operators whose eigenvectors form an orthonormal basis. You can expand a state in stationary states and compute $\langle Q\rangle$. This lesson ties those together into the rule for what any measurement does.
| Term | What it means |
|---|---|
| Generalized statistical interpretation | A measurement of $Q$ gives an eigenvalue $q_n$ with probability $\vert \langle f_n\vert \Psi\rangle\vert ^2$. |
| Expansion coefficient | $c_n = \langle f_n\vert \Psi\rangle$, the component of the state along eigenvector $f_n$. |
| Collapse | The change of the state, on measurement, to the eigenvector of the value found. |
| Ensemble | Many identically prepared systems, over which averages are taken. |
| Expectation value | $\langle Q\rangle = \sum q_n\vert c_n\vert ^2$, the ensemble average. |
| Standard deviation | $\sigma_Q = \sqrt{\langle Q^2\rangle - \langle Q\rangle^2}$, the spread of the results. |
| Degenerate eigenvalue | An eigenvalue shared by more than one independent eigenvector; its probability sums over them. |
Let $\hat{Q}$ be the operator of an observable, with orthonormal eigenvectors $f_n$ and eigenvalues $q_n$. Any state can be expanded as
$$\Psi = \sum_n c_nf_n, \qquad c_n = \langle f_n|\Psi\rangle.$$
The generalized statistical interpretation makes three claims. First, a measurement of $Q$ always returns one of the eigenvalues $q_n$ — nothing in between. Second, the probability of $q_n$ is $|c_n|^2$; since the state is normalized, $\sum|c_n|^2 = 1$. Third, immediately afterward the state is $f_n$, the eigenvector of the value found: the wave function collapses, so an immediate repeat is certain to give $q_n$ again.
The expectation value follows at once,
$$\langle Q\rangle = \sum_n q_n|c_n|^2 = \langle\Psi|\hat{Q}\Psi\rangle,$$
and it is an average over an ensemble of identically prepared systems, not the result of any one measurement. Born's rule for position is the special case where the eigenfunctions are delta functions and the coefficients are $\Psi(x)$ itself. For a continuous spectrum, such as momentum, $|c(p)|^2dp$ is the probability of a result in $[p, p + dp]$.
Another way: picture
Picture the state as an arrow and the observable's eigenvectors as a set of perpendicular axes. A measurement asks, in effect, which axis the arrow points along, and forces an answer: it snaps the arrow onto one axis, choosing each with probability equal to the square of the arrow's shadow on it. Once snapped, the arrow lies on that axis, so asking again gives the same answer. A different observable has different axes, and asking about those scrambles the answer again.
Another way: steps
Checks. The probabilities must be between zero and one and add to one. The average must lie between the smallest and largest eigenvalue that has nonzero probability. The spread is zero only if the state is an eigenstate. And an expectation value need not be a possible result: a coin that gives $0$ or $1$ averages $\tfrac{1}{2}$.
The states $\tfrac{3}{5}f_1 + \tfrac{4}{5}f_2$ and $\tfrac{3}{5}f_1 - \tfrac{4}{5}f_2$ give identical probabilities for $Q$: $0.36$ and $0.64$. They are nevertheless different states. Measure a different observable, one whose eigenvectors are $(f_1 \pm f_2)/\sqrt{2}$, and the first state gives $|\tfrac{3}{5} + \tfrac{4}{5}|^2/2 = 0.98$ for the plus eigenvector while the second gives $|\tfrac{3}{5} - \tfrac{4}{5}|^2/2 = 0.02$.
So relative phases between components are physical: they show up as interference when the state is measured in a different basis. Only an overall phase multiplying the whole state, $e^{i\alpha}\Psi$, is unobservable. This is why time evolution matters even for probabilities: the phases $e^{-iE_nt/\hbar}$ of stationary states leave energy probabilities unchanged but make the position distribution slosh back and forth.
An immediate repeat of the same measurement always agrees with the first. Waiting between them changes things: after an energy measurement, the state is a stationary state and stays one, so later energy measurements still agree; after a position measurement, the collapsed state is sharply localized and immediately begins to spread, so a later position measurement can differ.
Measuring a different observable between two measurements of $Q$ also changes things, unless the two observables share eigenvectors. Measure a spin component along $z$, then along $x$, then along $z$ again: the middle measurement collapses the state to an $x$ eigenvector, which is an equal mixture of the $z$ eigenvectors, and the final result is up or down at random, whatever the first gave. Observables that can share eigenvectors are called compatible, and the next lesson shows that they are exactly those whose operators commute.
When several independent eigenvectors share an eigenvalue, the probability of that value is the sum of the squared coefficients over all of them. In the two-dimensional square well, for example, the states $(n_x, n_y) = (1, 2)$ and $(2, 1)$ have the same energy, and a state with coefficients $0.6$ and $0.8$ on them has probability $0.36 + 0.64 = 1$ of that energy.
After the measurement, the state collapses not to a single eigenvector but to the projection of the old state onto the whole degenerate subspace, renormalized. This is the rule that lets a measurement of energy leave angular momentum untouched in the hydrogen atom, where many states share each energy. Degeneracy always signals a symmetry, and it is the starting point of degenerate perturbation theory later in the course.
The rules above are not in dispute; every experiment agrees with them. What they mean is. The Copenhagen view, associated with Niels Bohr, treats the wave function as a tool for predicting measurement results and collapse as an update of that tool. The many-worlds view of Hugh Everett keeps the Schrödinger equation always and says every outcome occurs in a separate branch. Pilot-wave theory, due to Louis de Broglie and David Bohm, gives particles definite positions guided by the wave.
All three predict the same probabilities for every experiment done so far. What has been settled experimentally, by the Bell tests discussed at the end of this course, is that no theory in which measured values exist in advance and signals travel no faster than light can reproduce them. The practical content of this lesson — eigenvalues, squared coefficients, collapse — is what every physicist, whatever their interpretation, uses to predict results.
A probability cannot be checked on one system. If a state gives $0.36$ for one result and $0.64$ for the other, a single measurement gives one of them and says almost nothing about the numbers. To test the prediction, an experimenter prepares the same state again and again, measures each copy once, and counts. With a thousand copies the fraction of the first result should come out near $0.36$, typically within about $0.015$; with a million, within about $0.0005$.
This is why the expectation value is called an ensemble average. It is not the average over repeated measurements of one system, because the first measurement collapses the state and every later result simply repeats it. It is the average over many systems, each prepared identically and each measured once. Laboratory practice follows this closely: an atomic clock interrogates millions of atoms, a quantum computer repeats each circuit thousands of times, and a particle-physics experiment collects billions of collisions, all to turn single random outcomes into measured probabilities that can be compared with theory.
In the BB84 protocol, devised by Charles Bennett and Gilles Brassard in 1984, a sender encodes bits in single photons polarized along one of two bases, horizontal–vertical or diagonal, and a receiver measures each photon in one of the two bases, chosen at random. When the bases match, the result is certain; when they differ by $45°$, it is random, with probability $\cos^2 45° = \tfrac{1}{2}$ each way. Afterward the two compare bases publicly, keep only the matching ones, and share a secret key.
The security comes straight from this lesson. An eavesdropper who measures a photon must pick a basis, and a wrong guess collapses the photon into the wrong basis, introducing errors that the legitimate parties detect by comparing a sample of their key: an intercept-and-resend attack produces a $25$ percent error rate. Commercial systems now run over fiber links in several countries, and in 2017 the Chinese satellite Micius distributed keys between ground stations $1200$ km apart.
Every quantum computation ends with a measurement. A qubit left in $\alpha|0\rangle + \beta|1\rangle$ reads out $0$ with probability $|\alpha|^2$ and $1$ with probability $|\beta|^2$, so a single run gives one random bit string. Programmers therefore run each circuit many times — typically thousands of shots — and build up a histogram whose frequencies estimate the probabilities.
The number of shots follows from statistics. Estimating a probability of about $0.5$ to within $\pm 0.01$ needs roughly $2500$ runs, because the standard error of a frequency falls as $1/\sqrt{N}$. Algorithms are designed so that the answer appears with high probability, making a few runs enough: Grover's search, for instance, amplifies the amplitude of the right answer until its squared magnitude is close to one. Collapse also makes measurement a one-way step, which is why quantum error correction must detect errors by measuring carefully chosen joint properties that reveal the error without collapsing the encoded data.
It is tempting to think a particle in $\tfrac{3}{5}f_1 + \tfrac{4}{5}f_2$ secretly has value $q_1$ or $q_2$, and the measurement simply finds out which. That picture fails: the relative phase between the components produces interference in other measurements that no mixture of definite values could produce, and Bell's theorem rules out any local version of hidden values. Before the measurement the state is genuinely a superposition; the measurement produces a definite value.
Two smaller errors are common. One is to expect a single measurement to give the expectation value; it always gives an eigenvalue, and the expectation value is only the average. The other is to forget collapse: after a measurement the old coefficients no longer apply, and an immediate repeat gives the same value.
An observable has eigenvalues $2$ and $7$ with eigenvectors $f_1, f_2$. The state is $\Psi = \tfrac{3}{5}f_1 - \tfrac{4i}{5}f_2$. Check the normalization.
$\left(\tfrac{3}{5}\right)^2 + \left|\tfrac{4i}{5}\right|^2 = \tfrac{9}{25} + \tfrac{16}{25} = 1$
The coefficients' squared magnitudes must add to one.
Find the probability of the value $2$.
$P(2) = \tfrac{9}{25} = 0.36$
The first squared coefficient.
Find the probability of the value $7$.
$P(7) = \left|-\tfrac{4i}{5}\right|^2 = 0.64$
The phase $-i$ drops out of the magnitude.
Find the expectation value.
$\langle Q\rangle = 0.36 \times 2 + 0.64 \times 7 = 0.72 + 4.48 = 5.20$
The average over many identical systems.
Find the standard deviation.
$\langle Q^2\rangle = 0.36 \times 4 + 0.64 \times 49 = 32.8; \quad \sigma_Q = \sqrt{32.8 - 27.04} = 2.40$
The spread of results about the average.
A spin is in the state $\chi = \tfrac{1}{5}\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ in the $z$ basis. Write the $x$ eigenvectors.
$\chi_\pm^{(x)} = \tfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ \pm 1 \end{pmatrix}$
The eigenvectors of $S_x$, with values $\pm\hbar/2$.
Project onto the plus eigenvector.
$c_+ = \langle\chi_+^{(x)}|\chi\rangle = \tfrac{1}{\sqrt{2}} \cdot \tfrac{1}{5}(3 + 4) = \tfrac{7}{5\sqrt{2}}$
The inner product gives the coefficient.
Project onto the minus eigenvector.
$c_- = \tfrac{1}{\sqrt{2}} \cdot \tfrac{1}{5}(3 - 4) = -\tfrac{1}{5\sqrt{2}}$
The same for the other eigenvector.
Square both coefficients.
$P_+ = \tfrac{49}{50} = 0.98, \qquad P_- = \tfrac{1}{50} = 0.02$
They add to one, as they must.
Find the average spin along $x$.
$\langle S_x\rangle = \tfrac{\hbar}{2}(0.98 - 0.02) = 0.48\hbar$
Nearly all the results are up along $x$.
Compare with the $z$ probabilities.
$P_\uparrow^{(z)} = 0.36, \quad P_\downarrow^{(z)} = 0.64$
The same state gives very different probabilities in different bases.
An electron in a well with $E_1 = 0.376$ eV is in $\Psi = A(\psi_1 + 2\psi_2 + \psi_3)$. Find the normalization.
$|A|^2(1 + 4 + 1) = 1 \quad\Rightarrow\quad |A|^2 = \tfrac{1}{6}$
Orthonormal states: add the squared coefficients.
List the probabilities of the three energies.
$P_1 = \tfrac{1}{6}, \quad P_2 = \tfrac{4}{6}, \quad P_3 = \tfrac{1}{6}$
Squared coefficients times $|A|^2$.
List the energies.
$E_1 = 0.376, \quad E_2 = 1.504, \quad E_3 = 3.384\ \text{eV}$
$E_n = n^2E_1$.
Compute the expectation value.
$\langle E\rangle = \tfrac{1}{6}(0.376 + 4 \times 1.504 + 3.384) = \tfrac{9.776}{6} = 1.629\ \text{eV}$
Weight each energy by its probability.
Find the probability of a result above $1$ eV.
$P(E > 1\ \text{eV}) = P_2 + P_3 = \tfrac{5}{6}$
Add the probabilities of the qualifying eigenvalues.
Suppose a measurement gives $1.504$ eV. Write the new state.
$\Psi \to \psi_2$
The state collapses to the matching eigenfunction.
Find the expectation value after the measurement.
$\langle E\rangle = E_2 = 1.504\ \text{eV}, \quad \sigma_E = 0$
Now an eigenstate: every further measurement gives $E_2$.
Identify the coefficient of $\psi_2$.
$c_2 = \tfrac{i}{\sqrt{2}}$
The component along the second eigenstate.
Square its magnitude.
$|c_2|^2 = \tfrac{|i|^2}{2}$
Multiply by the complex conjugate.
Evaluate the probability.
A particle is in a superposition of several energy eigenstates. Its energy is measured and found to be $E_{4}$. The energy is measured again immediately. What is the result?
Complete the worked solution: a particle is in the state $\Psi = A(\psi_1 + 7\psi_2)$. Normalize it and find the probability of each energy.
Add the squared coefficients.
$1^2 + 7^2 =$ n
Orthonormal states: the squared norm is this sum times $|A|^2$.
Divide the first squared coefficient by the norm.
$P_1 = \dfrac{1}{1 + 7^2} =$ p
With $|A|^2$ fixed by normalization.
Divide the second squared coefficient by the norm.
$P_2 = \dfrac{7^2}{1 + 7^2} =$ q
The two probabilities add to one.
A state $\Psi$ is expanded in the orthonormal eigenfunctions $f_n$ of an observable $\hat{Q}$, with eigenvalues $q_n$. Match each quantity to its expression.
| $\langle f_{5}|\Psi\rangle$ | $|c_{5}|^2$ | $\sum_n q_n|c_n|^2$ | $f_{5}$ | |
|---|---|---|---|---|
| the coefficient $c_{5}$ | ||||
| the probability of $q_{5}$ | ||||
| the average $\langle Q\rangle$ | ||||
| the state after finding $q_{5}$ |
An observable has eigenvalues $q_1 = 7$ and $q_2 = 9$ with eigenfunctions $f_1$ and $f_2$. A system is in the state $\Psi = \tfrac{3}{5}f_1 + \tfrac{4i}{5}f_2$. Fill in the probabilities and averages.
| value | |
|---|---|
| $P(q_1)$ | |
| $P(q_2)$ | |
| $\langle Q\rangle$ | |
| $\langle Q^2\rangle$ |
A harmonic oscillator with $\hbar\omega = 66$ meV is in the state $\Psi = \tfrac{3}{5}\psi_0 + \tfrac{4}{5}\psi_1$. What is the average of many energy measurements on identically prepared oscillators, in meV?
Answer: meV
An electron in an infinite square well $1$ nm wide, where $E_1 = 0.376$ eV, is in the state $\Psi = A(\psi_1 + 1\psi_2 + 1\psi_3)$. What is the expectation value of its energy, in eV?
Answer: eV
In a test of a quantum key distribution link, $1444$ single photons are sent, each polarized along one axis, and each is measured by a polarizer turned $60°$ from that axis. About how many photons pass?
Answer: photons
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
An observable has eigenvalues $q_1 = 2$ and $q_2 = 6$ with eigenfunctions $f_1$ and $f_2$. A system is in the state $\Psi = \tfrac{3}{5}f_1 + \tfrac{4i}{5}f_2$. Fill in the probabilities and averages.
| value | |
|---|---|
| $P(q_1)$ | |
| $P(q_2)$ | |
| $\langle Q\rangle$ | |
| $\langle Q^2\rangle$ |
You can apply the generalized statistical interpretation. Explain to someone why an immediate repeat of a measurement always gives the same result.
17. Your turn: a particle is in $\Psi = \tfrac{1}{\sqrt{2}}(\psi_1 + i\psi_2)$. What is the probability of finding $E_2$?, step 3
$P(E_2) = \tfrac{1}{2}$
The factor of $i$ does not change it.