Measuring in any basis, and Stern–Gerlach

1 / 12 · Learn

Before you start

Lesson 2 measured in the ± basis without saying how, and Lesson 1 left a Stern–Gerlach puzzle unanswered. With $H$ from Lesson 4 you can now settle both.

By the end of this lesson you can:

rewrite a state in any orthonormal basis and apply the Born rule there;
measure in the ± basis by applying $H$ , then measuring in the 0/1 basis;
say which basis a Stern–Gerlach analyzer measures, from the way it points;
predict the fraction of atoms leaving each exit of a cascade;
explain why a later $z$ analyzer shows no trace of an earlier $z$ result.

Check yourself

A warm-up: make a guess before reading on. The lesson comes back to it.

A beam of atoms that all left the $+$ exit of a $z$ analyzer enters a second analyzer, turned to point along $x$ . What happens at the second analyzer?

ANo atom is deflected either way: the spin points along $z$ , at right angles to the $x$ analyzer, so it has nothing to act on.
BEvery atom leaves by the $+$ exit: their spin was already selected.
CThe atoms spread out evenly over a range of positions between the two exits.
DAbout half the atoms leave by each exit.

Show the answer

Answer: D. About half the atoms leave by each exit.

An $x$ analyzer measures in the ± basis, and $\ket{+Z} = \ket0 = (\ket+ + \ket-)/\sqrt2$ , so each exit gets $\tfrac12$ .

Why not the others

A. The $z$ and $x$ directions are at right angles in the room, but the states $\ket{+Z}$ and $\ket{+X}$ are not orthogonal: their inner product is $1/\sqrt2$ , so each exit gets half.
B. Selecting at the first analyzer fixes the $z$ result only. In the $x$ basis the state is $\ket{+Z} = (\ket{+X} + \ket{-X})/\sqrt2$ , so each exit gets half.
C. Every atom leaves by one exit or the other; the experiment shows two separate beams, never a smear.

2 / 12 · Learn

Any orthonormal basis is a measurement

So far "measure" has mostly meant the 0/1 basis, and in Lesson 2 you also met the ± basis. Any pair of orthonormal states $\{\ket a, \ket b\}$ defines a measurement, and the recipe is always the same:

Write the state in that basis: $\ket\psi = \alpha'\ket a + \beta'\ket b$ .
The outcomes are $a$ and $b$ , with $P(a) = \abs{\alpha'}^2$ and $P(b) = \abs{\beta'}^2$ .
Afterwards the qubit is in $\ket a$ or $\ket b$ , whichever was observed.

For the ± basis, substituting $\ket0 = (\ket+ + \ket-)/\sqrt2$ and $\ket1 = (\ket+ - \ket-)/\sqrt2$ gives the rewrite you know:

\alpha\ket0 + \beta\ket1 = \frac{\alpha+\beta}{\sqrt2}\ket+ + \frac{\alpha-\beta}{\sqrt2}\ket-

Check yourself

Write $\ket0$ in the ± basis: give its coefficients on $\ket+$ and on $\ket-$ .

Hints

What are $\alpha$ and $\beta$ for the state $\ket0$ ?
The coefficients are $(\alpha+\beta)/\sqrt2$ on $\ket+$ and $(\alpha-\beta)/\sqrt2$ on $\ket-$ .
With $\alpha = 1$ and $\beta = 0$ , both are $1/\sqrt2$ .

Show the answer

Answer: $\ket{+} = \frac{1}{\sqrt{2}},\quad \ket{-} = \frac{1}{\sqrt{2}}$

$\ket0 = (\ket+ + \ket-)/\sqrt2$ . Measured in the ± basis it gives each outcome with probability $\abs{1/\sqrt2}^2 = \tfrac12$ .

Refresher: Inner products and bases Bras, complex conjugates and inner products, if the shortcut below is new to you.

A shortcut saves the rewriting: the coefficient of $\ket a$ is the inner product $\alpha' = \braket{a}{\psi}$ . Take the entries of $\ket a$ , complex-conjugate them, multiply each by the matching entry of $\ket\psi$ , and add. Here is an orthonormal basis that is neither 0/1 nor ±; you will meet it again on a tilted magnet:

\begin{aligned}\ket a &= \frac{\sqrt3}{2}\ket0 + \frac12\ket1,\\ \ket b &= \frac12\ket0 - \frac{\sqrt3}{2}\ket1\end{aligned}

Check yourself

A qubit in $\ket0$ is measured in the basis $\{\ket a, \ket b\}$ above. What is $P(a)$ ?

$P(a) = \;?$

Hints

Which entry of $\ket0$ is nonzero, and which entry of $\ket a$ does it pick out?
The coefficient of $\ket a$ is $\braket{a}{0}$ ; the Born rule squares its size.
$\braket{a}{0} = \tfrac{\sqrt3}{2}\cdot 1 + \tfrac12\cdot 0$ . Square it.

Show the answer

Answer: $P(a) = \frac{3}{4}$

$\braket{a}{0} = \tfrac{\sqrt3}{2}$ , so $P(a) = \tfrac34$ , and $P(b) = \abs{\braket{b}{0}}^2 = \tfrac14$ . They add to 1, as they must.

Common slips

$\frac{\sqrt{3}}{2}$ : That is $\braket{a}{0}$ , the amplitude. The Born rule squares its size: $(\sqrt3/2)^2$ .
$1$ : $\ket0$ is certain only in the 0/1 basis. In the basis $\{\ket a, \ket b\}$ it is a superposition, $\ket0 = \tfrac{\sqrt3}{2}\ket a + \tfrac12\ket b$ , so either outcome can occur.
$\frac{1}{4}$ : That is $P(b) = \abs{\braket{b}{0}}^2 = (1/2)^2$ . For $P(a)$ use the entries of $\ket a$ .
$\frac{1}{2}$ : Not every basis splits $\ket0$ evenly; only the ± basis does. Compute $\braket{a}{0}$ and square its size.

3 / 12 · Learn

Certain in one basis, random in another

Lesson 2 showed that whether a result is certain depends on the basis, not on the state alone. $\ket0$ gives 0 every time in the 0/1 basis, yet $\ket0 = (\ket+ + \ket-)/\sqrt2$ , so in the ± basis it gives $+$ or $-$ at random, half each. The general rule: every state is certain in exactly one basis, the one that contains it.

The Bloch sphere of Lesson 3 shows why. A measurement basis is an axis, its two outcomes are the two ends, and the end at angle $\gamma$ from the state has probability $\cos^2(\gamma/2)$ : the rule $P(0) = \cos^2(\theta/2)$ , turned to face another axis. So only a state on the axis gives a certain result.

Check yourself

A qubit gives 0 every time it is measured in the 0/1 basis. You measure it in the ± basis instead. What do you get?

ANothing meaningful: only the 0/1 basis is a real measurement; ± is just another way of writing the state.
B $+$ or $-$ at random, half the time each.
C $+$ every time: the qubit has a definite value, so every measurement is predictable.
DIt depends on the unknown global phase of the state.

Hints

Which state gives 0 every time in the 0/1 basis?
Write that state in the ± basis, $\ket0 = (\ket+ + \ket-)/\sqrt2$ , and square the size of each coefficient.

Show the answer

Answer: B. $+$ or $-$ at random, half the time each.

The qubit is $\ket0 = (\ket+ + \ket-)/\sqrt2$ (up to an unmeasurable global phase), so each ± outcome has probability $\tfrac12$ . On the sphere $\ket0$ is $90°$ from both ends of the $x$ axis, and $\cos^2 45° = \tfrac12$ .

Why not the others

A. Any orthonormal basis gives a real measurement with real outcomes. On hardware you do it with $H$ followed by a 0/1 meter (next segment), and a Stern–Gerlach analyzer turned along $x$ does it directly. Writing a state in a basis is bookkeeping; measuring in it is an experiment.
C. A common idea is that certain in one basis means certain in all. The rewrite says otherwise: the qubit is $\ket0 = (\ket+ + \ket-)/\sqrt2$ , so $P(+) = P(-) = \tfrac12$ . Certainty belongs to a state and a basis. "Definite" sounds like a property of the qubit alone, which is what makes this tempting.
D. A global phase $e^{i\gamma}$ multiplies both ± coefficients by the same factor, and $\abs{e^{i\gamma}} = 1$ , so no probability in any basis changes. Whatever its global phase, this qubit gives $P(\pm) = \tfrac12$ .

Check yourself

Take $\ket\psi = \cos\tfrac{\pi}{8}\ket0 + \sin\tfrac{\pi}{8}\ket1$ . In how many different measurement bases does one of the outcomes happen with certainty? (Write $\ket{\psi^\perp}$ for the state orthogonal to $\ket\psi$ .)

ANone: it is a superposition, so every measurement is random.
BTwo: $\{\ket\psi, \ket{\psi^\perp}\}$ and $\{\ket{\psi^\perp}, \ket\psi\}$ .
CInfinitely many, because there are infinitely many states.
DExactly one: the basis $\{\ket\psi, \ket{\psi^\perp}\}$ .

Hints

On the Bloch sphere a measurement basis is an axis. When is a result along an axis certain?
Only a state sitting at one end of the axis gives a certain result. How many axes pass through the point of $\ket\psi$ ?

Show the answer

Answer: D. Exactly one: the basis $\{\ket\psi, \ket{\psi^\perp}\}$ .

On the Bloch sphere $\ket\psi$ sits at $\theta = \pi/4$ . Only the axis through that point, with $\ket{\psi^\perp}$ at the opposite end, gives a certain result. Every other axis leaves both outcomes possible.

Why not the others

A. "Superposition" is always relative to a basis. $\ket\psi$ is a superposition of $\ket0$ and $\ket1$ , but it is itself one member of the basis $\{\ket\psi, \ket{\psi^\perp}\}$ , where $\ket{\psi^\perp}$ is the state orthogonal to it. Measured in that basis it gives the outcome $\psi$ every time.
B. A basis is a set of states. Listing the same two states in the other order is the same measurement, with the same two outcomes; on the sphere both describe one axis.
C. There are infinitely many bases, but only one axis passes through the point of $\ket\psi$ . Any other axis makes an angle $\gamma$ with $0° < \gamma < 180°$ to the state, and $\cos^2(\gamma/2)$ is then strictly between 0 and 1.

4 / 12 · Learn

The ± measurement: H, then a 0/1 meter

Many quantum computers measure only in the 0/1 basis. To measure in the ± basis, apply $H$ first, then measure in the 0/1 basis, and read 0 as $+$ and 1 as $-$ . This works because $H\ket+ = \ket0$ and $H\ket- = \ket1$ (Lesson 4). For a general state, $H$ moves the ± coefficients onto $\ket0$ and $\ket1$ :

H(\alpha\ket0 + \beta\ket1) = \frac{\alpha+\beta}{\sqrt2}\ket0 + \frac{\alpha-\beta}{\sqrt2}\ket1

So the meter reads 0 with probability $P(+)$ and 1 with probability $P(-)$ .

Check yourself

Your device can only measure in the 0/1 basis. Which procedure measures a qubit in the ± basis?

AApply $H$ , then measure in the 0/1 basis; read 0 as $+$ and 1 as $-$ .
BApply $X$ , then measure in the 0/1 basis.
CMeasure in the 0/1 basis many times: $\ket+$ and $\ket-$ give different histograms.
DMeasure in the 0/1 basis, then apply $H$ .

Hints

Which gate turns $\ket+$ and $\ket-$ into $\ket0$ and $\ket1$ ?
That gate has to act before the meter fires: once it has fired, its result answers the 0/1 question and nothing else.

Show the answer

Answer: A. Apply $H$ , then measure in the 0/1 basis; read 0 as $+$ and 1 as $-$ .

$H$ sends $\ket+ \to \ket0$ and $\ket- \to \ket1$ , so after it the 0/1 meter’s 0 and 1 mean $+$ and $-$ . The probabilities are $P(+)$ and $P(-)$ of the state you started with.

Why not the others

B. $X$ swaps $\ket0$ and $\ket1$ , so the meter reports the 0/1 result upside down. $X\ket+ = \ket+$ and $X\ket- = -\ket-$ , so both still read 0 or 1 at random and can’t be told apart.
C. In the 0/1 basis $\ket+$ and $\ket-$ both give 50/50, since $\abs{\pm1/\sqrt2}^2 = \tfrac12$ . Their difference is a relative sign, which 0/1 probabilities can’t see. A gate before the meter has to turn that sign into a difference in probabilities, and $H$ does.
D. Order matters. Once the 0/1 meter has fired, the result is recorded and the qubit is $\ket0$ or $\ket1$ ; an $H$ afterwards changes the qubit but not the result, which still answers the 0/1 question. The $H$ must come first, to move the ± coefficients onto $\ket0$ and $\ket1$ .

Check yourself

A qubit in $\tfrac{\sqrt3}{2}\ket0 + \tfrac12\ket1$ goes through $H$ and then a 0/1 meter. What is the probability that the meter reads 0? (Exact, or 3 decimals.)

$P(0) = \;?$

Hints

After the $H$ , what is the amplitude on $\ket0$ ? Look at the formula above.
It is $(\alpha+\beta)/\sqrt2$ , and the Born rule squares its size.
$\alpha + \beta = (\sqrt3 + 1)/2$ . Square it, then divide by 2.

Show the answer

Answer: $P(0) = 0.933$

The amplitude on $\ket0$ after $H$ is $(\sqrt3+1)/(2\sqrt2)$ ; its square is $(4+2\sqrt3)/8 = (2+\sqrt3)/4 \approx 0.933$ . That is $P(+)$ for the original state, which is the point: reading 0 means $+$ . (Lesson 4 used the same formula for $H$ on a general state.)

Common slips

$\frac{3}{4}$ : $\tfrac34$ is $P(0)$ for the state before the $H$ . After $H$ the amplitude on $\ket0$ is $(\alpha+\beta)/\sqrt2$ .
$0.067$ : That is the probability of reading 1, which uses $\alpha - \beta$ . Reading 0 uses $\alpha + \beta$ .
$\frac{1}{2}$ : A common idea is that $H$ just randomizes a qubit. It doesn’t: the two contributions to each output amplitude add or partly cancel. Here they add on $\ket0$ , giving $(\alpha+\beta)/\sqrt2$ .
$0.966$ : That is the amplitude $(\alpha+\beta)/\sqrt2$ . The probability is its square.

5 / 12 · Learn

Superposition or coin? Ask in another basis

In Lesson 2 you told these two registers apart by measuring in the ± basis. Now you can build that measurement from a gate and a 0/1 meter. Register A holds 1000 qubits, each in $\ket+$ . Register B holds 1000 qubits, each prepared as $\ket0$ or $\ket1$ by a coin toss you didn’t see. Measured in the 0/1 basis, the two look the same: about half 0s and half 1s.

Now apply $H$ to every qubit and measure in the 0/1 basis, which is a ± measurement. Every qubit of A becomes $\ket0$ and reads 0. Each qubit of B becomes $\ket+$ or $\ket-$ , and those still read 0 or 1 at random. Run both tests.

Check yourself

Register C holds 1000 copies of $\ket-$ . Register D holds 1000 qubits, each $\ket0$ or $\ket1$ by a coin toss. Which statement is true?

AThey are physically the same: both give 0 or 1 with probability $\tfrac12$ .
BApply $H$ to every qubit, then measure in the 0/1 basis: C gives all 1s, D gives about half 0s and half 1s.
CThey differ in principle, but no experiment can show it.
DMeasure in the 0/1 basis: C gives exactly 500 of each result, D only about 500.

Hints

What does $H$ do to $\ket-$ ? What does it do to $\ket0$ and to $\ket1$ ?
A state that becomes a basis state under $H$ gives one result every time; one that becomes $\ket+$ or $\ket-$ still reads 0 or 1 at random.

Show the answer

Answer: B. Apply $H$ to every qubit, then measure in the 0/1 basis: C gives all 1s, D gives about half 0s and half 1s.

$H\ket- = \ket1$ , so C reads 1 every time. D’s qubits become $\ket+$ or $\ket-$ , each 50/50 in the 0/1 basis. Only a measurement in another basis, such as ±, separates them.

Why not the others

A. A very common idea is that a superposition is a coin you haven’t looked at. Then no test could separate C from D. But $H\ket- = \ket1$ , so every qubit of C reads 1 after $H$ , while D stays 50/50. The 50/50 in the 0/1 basis is a fact about that basis, not about missing knowledge.
C. One gate is enough. $H$ followed by a 0/1 meter gives 1 every time for C and about half 1s for D, exactly as the lab showed for A and B.
D. Probabilities predict frequencies, not exact counts. In the 0/1 basis C scatters around 500 just as D does, as Register A did in the lab. You need a measurement in another basis.

6 / 12 · Learn

A Stern–Gerlach analyzer is a measurement

In the Stern–Gerlach experiment, atoms fly through a magnet that deflects each one along the magnet’s axis. Every atom lands in one of two separate beams, never in between. Treat the magnet as a black box, an analyzer with a $+$ exit and a $-$ exit. The qubit model of an analyzer has three rules:

a $z$ analyzer measures in the basis $\{\ket{+Z}, \ket{-Z}\} = \{\ket0, \ket1\}$ ;
an $x$ analyzer measures in $\{\ket{+X}, \ket{-X}\} = \{\ket+, \ket-\}$ ;
each atom leaves in the state of the exit it took.

So an analyzer is a measurement whose basis you choose by turning it.

Check yourself

Which pair of states is orthogonal (inner product 0)?

A $\ket{-Z}$ and $\ket{-X}$
B $\ket{+Z}$ and $\ket{+X}$
C $\ket{+X}$ and $\ket{-X}$

Hints

Write each state as a column vector, then take the inner product of each pair.
$\ket{\pm Z} = \ket0, \ket1$ and $\ket{\pm X} = (\ket0 \pm \ket1)/\sqrt2$ . Multiply matching entries and add; orthogonal means the sum is 0.

Show the answer

Answer: C. $\ket{+X}$ and $\ket{-X}$

$\braket{-X}{{+X}} = \tfrac12(1\cdot 1 + (-1)\cdot 1) = 0$ . The two exits of one analyzer always give orthogonal states, at opposite ends of the Bloch sphere. States along axes $90°$ apart have an inner product of size $1/\sqrt2$ .

Why not the others

A. $\braket{-X}{{-Z}} = \tfrac{1}{\sqrt2}(1\cdot 0 + (-1)\cdot 1) = -\tfrac{1}{\sqrt2}$ , not 0. States $90°$ apart on the Bloch sphere have an inner product of size $1/\sqrt2$ , a 50/50 split; orthogonal states sit $180°$ apart.
B. The $z$ and $x$ directions are at right angles in the room, but the states are not orthogonal: $\braket{+X}{{+Z}} = 1/\sqrt2$ . That is why a $+Z$ atom splits 50/50 at an $x$ analyzer. Orthogonal spin states belong to opposite directions.

Turn an analyzer by an angle $\theta$ from $z$ toward $x$ and it measures along that direction of the Bloch sphere. Its exits leave atoms in

\begin{aligned}\ket{{+}_\theta} &= \cos\tfrac\theta2\ket0 + \sin\tfrac\theta2\ket1,\\ \ket{{-}_\theta} &= \sin\tfrac\theta2\ket0 - \cos\tfrac\theta2\ket1\end{aligned}

At $\theta = 90°$ these are $\ket+$ and $\ket-$ ; at $\theta = 60°$ they are the basis $\ket a$ , $\ket b$ from earlier in this lesson. Notice the halving: a $90°$ turn in the room is a $45°$ angle between the kets, which is why $\ket{+Z}$ and $\ket{+X}$ have inner product $\cos 45° = 1/\sqrt2$ .

Check yourself

A beam of $\ket{-Z}$ atoms enters an analyzer tilted $60°$ from $z$ toward $x$ , whose exits are $\ket a$ ( $+$ ) and $\ket b$ ( $-$ ). What fraction of the atoms leaves by the $+$ exit?

Hints

Which basis state is $\ket{-Z}$ , and which entry of $\ket a$ does it pick out?
The share through the $+$ exit is $\abs{\braket{a}{1}}^2$ : the Born rule in the analyzer’s basis.
$\braket{a}{1} = \tfrac12$ . Square it.

Show the answer

Answer: $\frac{1}{4}$

$\ket{-Z} = \ket1$ and $\braket{a}{1} = \tfrac12$ , so the $+$ exit gets $\tfrac14$ and the $-$ exit $\tfrac34$ . On the Bloch sphere the state is $120°$ from the analyzer’s $+$ end, and $\cos^2 60° = \tfrac14$ .

Common slips

$\frac{3}{4}$ : $\tfrac34$ is the share of the $-$ exit here (or of the $+$ exit for $\ket{+Z}$ atoms). For $\ket{-Z} = \ket1$ the coefficient of $\ket a$ is $\braket{a}{1} = \tfrac12$ .
$\frac{1}{2}$ : $\tfrac12$ is the amplitude $\braket{a}{1}$ , not its square. (It would also be the answer for a $90°$ tilt, not $60°$ .)
$0$ : The atoms point along $-z$ and the analyzer’s $+$ end is $120°$ away, not $180°$ . Only the exact opposite direction gives probability 0.

7 / 12 · Explore

Predict: the z, x, z cascade

This is the cascade from Lesson 1. You now have the model; predict before you run it.

Predict, then see

Atoms from the oven pass a $z$ analyzer (its $+$ exit feeds the next one), then an $x$ analyzer (its $+$ exit feeds the next one), then a second $z$ analyzer. Of the atoms that reach the last analyzer, what fraction leaves by its $-$ exit?

ANone: the first analyzer selected them as $+Z$ , and they still are.
BAbout $\tfrac18$ .
CAll of them: keeping the $+X$ exit flips the spin to $-Z$ .
DAbout half.

In a sentence or two: every atom left the first analyzer by $+$ . Why does the last analyzer still split them?

Show what happens

What happens: D. About half.

Each analyzer is a measurement along its own axis, and every atom leaves in the state of the exit it took. Atoms leaving the $x$ analyzer’s $+$ exit are in $\ket{+X} = (\ket{+Z} + \ket{-Z})/\sqrt2$ whatever happened before, so the last $z$ analyzer sends half each way. The first $+Z$ result is not stored in that state. Turn the middle analyzer to $z$ and the split at the end disappears.

Why not the others

A. A very common prediction: the atoms "remember" being $+Z$ . The run says otherwise: the last analyzer splits evenly. The $x$ analyzer is a measurement too, and it leaves every atom it passes in $\ket{+X}$ , which keeps no trace of the $+Z$ result.
B. $\tfrac18$ is the share of all the oven’s atoms that leave by that exit. Of the atoms that reach the last analyzer, half leave by each exit.
C. The $x$ analyzer doesn’t flip anything along $z$ . It leaves each atom in $\ket{+X} = (\ket{+Z} + \ket{-Z})/\sqrt2$ , which gives $+Z$ and $-Z$ equally often.

Why does the first analyzer split the oven’s atoms 50/50?

The oven’s atoms have random spin directions. An analyzer along $\hat n$ sends an atom whose spin points along $\hat r$ out of its $+$ exit with probability $(1 + \hat n\cdot\hat r)/2$ , which is the Bloch-sphere rule $\cos^2(\gamma/2)$ written with a dot product. Averaged over all directions, $\hat n\cdot\hat r$ is 0, so the average is $\tfrac12$ along any axis. A beam like this is a mixture, not a superposition. Describing mixtures takes a tool called the density operator, which is beyond this course.

Check yourself

Now make the middle analyzer a $z$ analyzer (still passing its $+$ exit), so the atoms meet $z$ , $z$ , $z$ . What fraction of the atoms that reach the last analyzer leaves by its $-$ exit? You can test it in the widget above.

$P(-Z) = \;?$

Hints

In which state do the atoms leave the second analyzer?
An analyzer gives a certain result when the arriving state is one of its own two states.
They arrive in $\ket{+Z}$ . What is $\abs{\braket{-Z}{{+Z}}}^2$ ?

Show the answer

Answer: $P(-Z) = 0$

The atoms leave the second analyzer in $\ket{+Z}$ , a state of the $z$ basis, so the last $z$ analyzer passes all of them through $+$ and none through $-$ . Repeating a measurement in the same basis repeats its result, as in Lesson 1.

Common slips

$\frac{1}{2}$ : A repeat of the same measurement gives the same result. The atoms leave the second $z$ analyzer in $\ket{+Z}$ , one of the $z$ basis states, so the last $z$ analyzer sends every one through $+$ .
$\frac{1}{4}$ : That treats each later $z$ analyzer as halving the beam again, $\tfrac12 \cdot \tfrac12$ . An analyzer splits the beam only when the arriving state is not one of its own two states. Atoms arriving in $\ket{+Z}$ all leave a $z$ analyzer by $+$ , so none leaves by $-$ .

8 / 12 · Worked example

Cascade fractions, step by step

Worked example

z, then x, then z, from the oven

Lesson 1 asked about the atoms that reach the last analyzer: half of them leave by $-$ . Now count against everything the oven sends out. What fraction of the oven’s atoms leaves the last analyzer’s $-$ exit?

Analyzer 1: z, keep +
The oven’s spins are random, so a $z$ analyzer sends half of them out of each exit. The kept half leave in $\ket{+Z} = \ket0$ .
Share of the oven’s atoms still in the beam: $\tfrac12$ .
Analyzer 2: x, keep +
An $x$ analyzer measures in the ± basis. Rewrite the arriving state there: $\ket0 = (\ket+ + \ket-)/\sqrt2$ , so $P(+X) = \abs{1/\sqrt2}^2 = \tfrac12$ . The kept atoms leave in $\ket{+X} = \ket+$ .
Share still in the beam: $\tfrac12 \cdot \tfrac12 = \tfrac14$ .
Analyzer 3: z
Your turn
Atoms arrive at the last analyzer in $\ket{+X} = \ket+$ . What is the probability that one of them leaves by the $-$ exit?
$P(-Z) = \;?$
Hints
1. Write $\ket+$ in the $z$ basis.
2. The $-Z$ coefficient of $\ket+ = (\ket0 + \ket1)/\sqrt2$ is $1/\sqrt2$ . Square it.
Show the answer
Answer: $P(-Z) = \frac{1}{2}$
$\ket+ = (\ket0 + \ket1)/\sqrt2$ , so $P(-Z) = \abs{1/\sqrt2}^2 = \tfrac12$ .
Common slips
- $0$ : That assumes the atoms still carry their first $+Z$ result. The $x$ analyzer replaced their state with $\ket+ = (\ket0 + \ket1)/\sqrt2$ , so $-Z$ has probability $\abs{1/\sqrt2}^2$ .
Put it together
Your turn
So what fraction of the oven’s atoms leaves the last analyzer’s $-$ exit?
Hint
1. Multiply the share that reaches the last analyzer by the probability of the $-$ exit there.
Show the answer
Answer: $\frac{1}{8}$
$\tfrac12 \cdot \tfrac12 \cdot \tfrac12 = \tfrac18$ . Another $\tfrac18$ leaves by $+$ : the last analyzer receives $\tfrac14$ of the oven’s atoms and splits them evenly.
Common slips
- $\frac{1}{2}$ : $\tfrac12$ is the share of the atoms that reach the last analyzer. Multiply by the $\tfrac14$ of the oven’s atoms that get that far.
- $\frac{1}{4}$ : $\tfrac14$ is the share that reaches the last analyzer. Half of those leave by $-$ .
- $0$ : That treats the atoms as if they remembered the first $+Z$ result. After the $x$ analyzer their state is $\ket+$ , which splits evenly at a $z$ analyzer.
The rule
Follow the kept beam. At each analyzer, rewrite the arriving state in that analyzer’s basis, take $\abs{\text{coefficient}}^2$ for the exit you follow, and multiply. The state passed on is always the state of that exit, so nothing earlier enters the next step.

Here is one more cascade, with a repeated analyzer. What fraction of the oven’s atoms leaves by the plus exit of the last analyzer?

Another worked example

x, x, z, from the oven

Analyzer 1: x, keep −
The oven’s spins are random: half leave by each exit. The kept atoms are in $\ket{-X} = \ket-$ . Share: $\tfrac12$ .
Analyzer 2: x, keep −
Atoms arrive in $\ket-$ , one of the $x$ analyzer’s own two states, so every one of them leaves by $-$ : probability 1. Share: $\tfrac12 \cdot 1 = \tfrac12$ .
Analyzer 3: z, the + exit
$\ket- = (\ket0 - \ket1)/\sqrt2$ , so $P(+Z) = \abs{1/\sqrt2}^2 = \tfrac12$ . The minus sign has no effect on 0/1 probabilities.
Total
$\tfrac12 \cdot 1 \cdot \tfrac12 = \tfrac14$ of the oven’s atoms leave by the last $+$ exit. A repeated analyzer multiplies by 1, or by 0 if you follow its other exit.

Check yourself

Atoms from the oven pass through three analyzers in a row:

a $z$ analyzer, whose $-$ exit feeds the next one;
a $z$ analyzer, whose $-$ exit feeds the next one;
an $x$ analyzer.

What fraction of the atoms that leave the oven leaves the last analyzer’s $+$ exit?

Hints

In which state does each kept exit leave the atoms, and in which basis does the next analyzer measure?
Each analyzer is a measurement in its own basis. Multiply the probabilities of the exits you follow; each analyzer passes on the state of its exit.
Analyzer 1 ( $z$ , keep $-$ ): the oven’s spins are random, so half leave by each exit: $\tfrac12$ . The atoms passed on are in $\ket{-Z}$ .

Show the answer

Answer: $\frac{1}{4}$

Multiply the shares of the kept exits: $\tfrac{1}{2} \cdot 1 \cdot \tfrac{1}{2} = \tfrac{1}{4}$ .

Follow the kept beam and multiply.

Analyzer 1 ( $z$ , keep $-$ ): the oven’s spins are random, so half leave by each exit: $\tfrac12$ . The atoms passed on are in $\ket{-Z}$ .
Analyzer 2 ( $z$ , keep $-$ ): atoms arrive in $\ket{-Z}$ , one of this analyzer’s own states, so every one leaves by $-$ : $1$ . The atoms passed on are in $\ket{-Z}$ .
Analyzer 3 ( $x$ , the $+$ exit): atoms arrive in $\ket{-Z} = (\ket{+X} - \ket{-X})/\sqrt2$ , so half leave by each exit: $\tfrac12$ .

Total: $\tfrac{1}{2} \cdot 1 \cdot \tfrac{1}{2} = \tfrac{1}{4}$ of the atoms that leave the oven.

Common slips

$\frac{1}{2}$ : $\tfrac{1}{2}$ is the fraction of the atoms arriving at the last analyzer that leave by its $+$ exit. The question asks about all the atoms that leave the oven, so multiply by the shares kept at the earlier analyzers.
$\frac{1}{8}$ : This halves the beam at every analyzer. An analyzer only splits the beam when the arriving state is not one of its own states; a repeat of the same analyzer passes every atom through the same exit (or none through the other).

9 / 12 · Practice

Practice

Check yourself

A qubit in $\tfrac{1}{\sqrt2}\left(\ket0 + e^{i\pi/3}\ket1\right)$ is measured in the ± basis. What is $P(-)$ ?

$P(-) = \;?$

Hints

In the ± basis, which combination of $\alpha$ and $\beta$ multiplies $\ket-$ ?
Born rule in the new basis: $P(-) = \abs{\alpha - \beta}^2/2$ .
Here $\alpha - \beta = \tfrac{1-e^{i\pi/3}}{\sqrt2}$ . Square its size and halve it. Use $\abs{1 \pm e^{i\varphi}}^2 = 2 \pm 2\cos\varphi$ .

Show the answer

Answer: $P(-) = \frac{1}{4}$

The coefficient of $\ket-$ is $(\alpha - \beta)/\sqrt2$ with $\alpha - \beta = \tfrac{1-e^{i\pi/3}}{\sqrt2}$ , so $P(-) = \tfrac12\big/2 = \tfrac{1}{4}$ .

Rewrite in the ± basis: $\alpha\ket0 + \beta\ket1 = \tfrac{\alpha+\beta}{\sqrt2}\ket+ + \tfrac{\alpha-\beta}{\sqrt2}\ket-$ .

Here $\alpha = 1/\sqrt2$ and $\beta = e^{i\pi/3}/\sqrt2$ , so $\alpha - \beta = \tfrac{1-e^{i\pi/3}}{\sqrt2}$ and $\abs{\alpha - \beta}^2 = \tfrac12$ . Use $\abs{1 \pm e^{i\varphi}}^2 = 2 \pm 2\cos\varphi$ .

So $P(-) = \tfrac12\big/2 = \tfrac{1}{4}$ . Check: $P(+) + P(-) = \tfrac{3}{4} + \tfrac{1}{4} = 1$ .

Common slips

$\frac{1}{2}$ : $\tfrac{1}{2}$ is $P(1)$ , the probability of reading 1 in the 0/1 basis. In the ± basis the question is different: the coefficient of $\ket-$ is $(\alpha - \beta)/\sqrt2$ , not $\beta$ .
$\frac{3}{4}$ : That is $P(+)$ . You used $\alpha + \beta$ ; the coefficient of $\ket-$ has $\alpha - \beta$ .

Check yourself

A beam of $\ket{+Z}$ atoms enters an analyzer tilted 135° from $z$ toward $x$ . What fraction of the atoms leaves by its $+$ exit? (Give an exact value or 3 decimals.)

Hints

On the Bloch sphere, where does $\ket{+Z}$ point, and where does the analyzer’s $+$ end point? What is the angle between them?
Born rule along any axis: the share is $\cos^2(\gamma/2) = (1 + \cos\gamma)/2$ , where $\gamma$ is the angle on the sphere between the state and that exit’s end.
Here $\gamma = 135°$ and $\cos 135° = -\tfrac{\sqrt2}{2}$ . Compute $(1 + \cos 135°)/2$ .

Show the answer

Answer: $0.146$

$\braket{{+}_\theta}{0} = \cos 67.5°$ , so the share is $\cos^2 67.5° = (1 + \cos 135°)/2 = \tfrac{2-\sqrt{2}}{4} \approx 0.146$ .

The analyzer’s $+$ exit state is $\ket{{+}_\theta} = \cos\tfrac\theta2\ket0 + \sin\tfrac\theta2\ket1$ with $\theta = 135°$ . The atoms arrive in $\ket{+Z} = \ket0$ , so the coefficient is $\braket{{+}_\theta}{0} = \cos 67.5°$ .

Share through $+$ : $\cos^2 67.5° = \tfrac{1 + \cos 135°}{2} = \tfrac{1 - \tfrac{\sqrt2}{2}}{2} = \tfrac{2-\sqrt{2}}{4} \approx 0.146$ . The $-$ exit gets the rest, $\tfrac{2+\sqrt{2}}{4} \approx 0.854$ .

On the Bloch sphere the state is 135° from the analyzer’s $+$ end, and $\cos^2(\gamma/2)$ with $\gamma = 135°$ gives the same number.

Common slips

$\frac{1}{2}$ : That uses the full angle: $\cos^2 135° = \tfrac{1}{2}$ . The amplitude inside the ket uses half of it, $\cos(67.5°)$ , so the share is $\cos^2 67.5° = (1 + \cos 135°)/2$ .
$0.383$ : That is $\cos 67.5°$ , the amplitude $\braket{{+}_\theta}{0}$ . The Born rule squares it.
$0.854$ : That is the share of the $-$ exit, $\sin^2 67.5°$ . The $+$ exit gets $\cos^2 67.5°$ .

Check yourself

A source makes many copies of one of five things: $\ket0$ , $\ket1$ , $\ket+$ , $\ket-$ , or a coin source that sends $\ket0$ or $\ket1$ at random. You measure 20 copies in the 0/1 basis and 20 more in the ± basis.

0/1 basis: 0 came up 0 times, 1 came up 20 times.
± basis: $+$ came up 11 times, $-$ came up 9 times.

Which is the source?

AThe coin source
B $\ket-$
C $\ket+$
D $\ket0$
E $\ket1$

Hints

Which candidates give the same result every time in the 0/1 basis? Which do so in the ± basis?
A state is certain only in the basis that contains it. A coin source is random in both bases.
Your 0/1 data are all one result. Which candidates can do that?

Show the answer

Answer: E. $\ket1$

All 20 copies read 1 in the 0/1 basis, which only $\ket1$ does. Its ± results are mixed, as they should be: $\ket1 = (\ket+ - \ket-)/\sqrt2$ .

All 20 copies read 1 in the 0/1 basis, which only $\ket1$ does. Its ± results are mixed, as they should be: $\ket1 = (\ket+ - \ket-)/\sqrt2$ . Every other candidate is ruled out by one of the two data sets.

Why not the others

A. A coin source sends $\ket0$ or $\ket1$ , so it gives a mix in the 0/1 basis, and a mix in the ± basis too, since each of those states is 50/50 there. It would give the same result 20 times in a row only about once in a million tries, yet the 0/1 data came out 1 all 20 times.
B. $\ket-$ gives $-$ every time in the ± basis, but 11 of your copies gave $+$ .
C. $\ket+$ gives $+$ every time in the ± basis, but 9 of your copies gave $-$ .
D. $\ket0$ reads 0 every time in the 0/1 basis, but 20 of your copies read 1.

10 / 12 · Explain

Explain it

After a $z$ analyzer (keep $+$ ) and an $x$ analyzer (keep $+$ ), why does a final $z$ analyzer split the beam 50/50, instead of sending every atom out of $+$ ?

Show a model answer

Each analyzer is a measurement in its own basis, and it leaves every atom in the state of the exit it took. The atoms kept at the $x$ analyzer are in $\ket{+X} = (\ket0 + \ket1)/\sqrt2$ , whatever their earlier $z$ result was; that result is not stored in the new state. Measured in the $z$ basis, $\ket{+X}$ gives $+Z$ and $-Z$ with probability $\tfrac12$ each. The $x$ and $z$ directions are $90°$ apart in space, but $\ket{+Z}$ and $\ket{+X}$ are not orthogonal: their inner product is $1/\sqrt2$ .

A good answer covers

An analyzer is a measurement: it leaves each atom in the state of the exit it took, replacing the earlier state.
The atoms leaving the $+X$ exit are in $\ket{+X} = (\ket0 + \ket1)/\sqrt2$ , an equal superposition of $+Z$ and $-Z$ , so $P(\pm Z) = \tfrac12$ .
Notes that $x$ and $z$ are $90°$ apart in space, while the states $\ket{+Z}$ and $\ket{+X}$ have inner product $1/\sqrt2$ (not 0).

11 / 12 · Mastery check

Mastery check

Check yourself

Atoms prepared in $\ket{+Z}$ pass through three analyzers in a row:

an $x$ analyzer, whose $+$ exit feeds the next one;
a $z$ analyzer, whose $-$ exit feeds the next one;
an $x$ analyzer.

What fraction of the atoms in the beam leaves the last analyzer’s $+$ exit?

Show the answer

Answer: $\frac{1}{8}$

Multiply the shares of the kept exits: $\tfrac{1}{2} \cdot \tfrac{1}{2} \cdot \tfrac{1}{2} = \tfrac{1}{8}$ .

Follow the kept beam and multiply.

Analyzer 1 ( $x$ , keep $+$ ): atoms arrive in $\ket{+Z} = (\ket{+X} + \ket{-X})/\sqrt2$ , so half leave by each exit: $\tfrac12$ . The atoms passed on are in $\ket{+X}$ .
Analyzer 2 ( $z$ , keep $-$ ): atoms arrive in $\ket{+X} = (\ket{+Z} + \ket{-Z})/\sqrt2$ , so half leave by each exit: $\tfrac12$ . The atoms passed on are in $\ket{-Z}$ .
Analyzer 3 ( $x$ , the $+$ exit): atoms arrive in $\ket{-Z} = (\ket{+X} - \ket{-X})/\sqrt2$ , so half leave by each exit: $\tfrac12$ .

Total: $\tfrac{1}{2} \cdot \tfrac{1}{2} \cdot \tfrac{1}{2} = \tfrac{1}{8}$ of the atoms in the beam.

Common slips

$0$ : This treats the atoms as if they remembered earlier results. Each analyzer leaves an atom in the state of its exit, so once an analyzer along the other axis has acted, the earlier result is gone: the next analyzer splits the beam again.
$\frac{1}{2}$ : $\tfrac{1}{2}$ is the fraction of the atoms arriving at the last analyzer that leave by its $+$ exit. The question asks about all the atoms in the beam, so multiply by the shares kept at the earlier analyzers.
$\frac{1}{4}$ : $\tfrac{1}{4}$ is the share that reaches the last analyzer. You still need the last analyzer’s own factor for the $+$ exit.

Check yourself

A beam of atoms that all left the $+$ exit of an $x$ analyzer enters a second analyzer, turned to point along $z$ . What happens at the second analyzer?

AAbout half the atoms leave by each exit.
BEvery atom leaves by the $+$ exit: their spin was already selected.
CThe atoms spread out evenly over a range of positions between the two exits.
DNo atom is deflected either way: the spin points along $x$ , at right angles to the $z$ analyzer, so it has nothing to act on.

Show the answer

Answer: A. About half the atoms leave by each exit.

A $z$ analyzer measures in the $\{\ket{+Z}, \ket{-Z}\}$ basis, and $\ket{+X} = (\ket{+Z} + \ket{-Z})/\sqrt2$ . Each exit gets $\abs{1/\sqrt2}^2 = \tfrac12$ .

Why not the others

B. Selecting at the first analyzer fixes the $x$ result only. In the $z$ basis the state is $\ket{+X} = (\ket{+Z} + \ket{-Z})/\sqrt2$ , so each exit gets half. Being certain in one basis does not make a state certain in every basis.
C. Every atom leaves by one exit or the other: the experiment shows two separate beams, never a continuous smear. That two-valued result is why the spin is modelled as a qubit.
D. The $x$ and $z$ directions are at right angles in the room, but the states are not orthogonal: their inner product is $1/\sqrt2$ . So half the atoms leave by each exit. Orthogonal spin states belong to opposite directions, such as $+X$ and $-X$ .

Check yourself

An analyzer tilted 120° from $z$ toward $x$ measures in the basis $\ket a = \tfrac12\ket0 + \tfrac{\sqrt3}{2}\ket1$ , $\ket b = \tfrac{\sqrt3}{2}\ket0 - \tfrac12\ket1$ . A qubit in $\ket-$ is measured in this basis. What is $P(a)$ ?

$P(a) = \;?$

Show the answer

Answer: $P(a) = 0.067$

$\braket{a}{-} = \tfrac{1}{\sqrt2}\left(\tfrac12 - \tfrac{\sqrt3}{2}\right)$ , so $P(a) = \tfrac{2-\sqrt{3}}{4} \approx 0.067$ .

The coefficient of $\ket a$ is the inner product $\braket{a}{\psi}$ : multiply matching entries and add (all entries are real, so conjugating changes nothing). With $\ket\psi = \ket-$ this is $\tfrac{1}{\sqrt2}\left(\tfrac12 - \tfrac{\sqrt3}{2}\right)$ .

So $P(a) = \tfrac{2-\sqrt{3}}{4} \approx 0.067$ , and $P(b) = 1 - P(a)$ .

Common slips

$0.259$ : That is $\abs{\braket{a}{\psi}}$ , the size of the amplitude. The Born rule squares it.
$0.933$ : That is $P(b)$ , the other outcome. $P(a) = \abs{\braket{a}{\psi}}^2$ uses the entries of $\ket a$ .
$\frac{1}{2}$ : That is the probability of either 0/1 outcome. In this basis the amplitudes are different; compute $\braket{a}{\psi}$ .

Check yourself

A qubit is prepared in $\tfrac35\ket0 + \tfrac45\ket1$ . You apply $H$ and then measure in the 0/1 basis. What is the probability of reading 1?

$P(1) = \;?$

Show the answer

Answer: $P(1) = \frac{1}{50}$

After $H$ the amplitude on $\ket1$ is $(\alpha - \beta)/\sqrt2$ with $\alpha - \beta = -\tfrac15$ , so $P(1) = \tfrac{1}{25}\big/2 = \tfrac{1}{50}$ .

Reading 1 after $H$ is the outcome $-$ of a ± measurement. Rewrite in the ± basis: $\alpha\ket0 + \beta\ket1 = \tfrac{\alpha+\beta}{\sqrt2}\ket+ + \tfrac{\alpha-\beta}{\sqrt2}\ket-$ .

Here $\alpha = 3/5$ and $\beta = 4/5$ , so $\alpha - \beta = -\tfrac15$ and $\abs{\alpha - \beta}^2 = \tfrac{1}{25}$ .

So $P(-) = \tfrac{1}{25}\big/2 = \tfrac{1}{50}$ . Check: $P(+) + P(-) = \tfrac{49}{50} + \tfrac{1}{50} = 1$ .

Common slips

$\frac{16}{25}$ : $\tfrac{16}{25}$ is $P(1)$ , the probability of reading 1 without the $H$ . In the ± basis the question is different: the coefficient of $\ket-$ is $(\alpha - \beta)/\sqrt2$ , not $\beta$ .
$\frac{49}{50}$ : That is $P(+)$ . You used $\alpha + \beta$ ; the coefficient of $\ket-$ has $\alpha - \beta$ .
$\frac{1}{5\sqrt{2}}$ : That is $\abs{\alpha - \beta}/\sqrt2$ , the size of the amplitude. The Born rule squares it: $P = \abs{\alpha - \beta}^2/2$ .
$\frac{1}{25}$ : That is $\abs{\alpha - \beta}^2$ . The coefficient carries a $1/\sqrt2$ , so divide by 2.

Check yourself

For a qubit in $\ket-$ , which measurement has an outcome that is certain?

AA measurement in the 0/1 basis
BAny measurement: the state is definite, so every result is predictable
CA measurement in the ± basis
DNone: quantum measurement results are always random

Show the answer

Answer: C. A measurement in the ± basis

$\ket-$ belongs to the ± basis, so a measurement there gives $-$ every time. In the 0/1 basis, $\ket- = (\ket0 - \ket1)/\sqrt2$ splits 50/50.

Why not the others

A. In the 0/1 basis, $\ket- = (\ket0 - \ket1)/\sqrt2$ , so the two outcomes each have probability $\tfrac12$ . It is certain only in the ± basis, which contains it.
B. A common idea is that a definite state is definite in every basis. But $\ket- = (\ket0 - \ket1)/\sqrt2$ , so in the 0/1 basis each outcome has probability $\tfrac12$ . A state is certain in exactly one basis, the one that contains it.
D. $\ket-$ is one of the two states of the ± basis, so measuring in that basis gives $-$ every time. Randomness comes from measuring in a basis that does not contain the state.

12 / 12 · Recap

Recap

From memory

Before reading the summary, write down what you remember. Pulling it back from memory is what makes it last.

How do you find the outcome probabilities for a measurement in a basis $\{\ket a, \ket b\}$ other than 0/1?
Show the key points
- Write $\ket\psi = \alpha'\ket a + \beta'\ket b$ (the coefficient of $\ket a$ is $\braket{a}{\psi}$ , with conjugated entries of $\ket a$ )
- $P(a) = \abs{\alpha'}^2$ and $P(b) = \abs{\beta'}^2$
- Afterwards the qubit is in $\ket a$ or $\ket b$
- A state is certain only in the basis that contains it
How do you measure in the ± basis with a 0/1 meter, and why does it work?
Show the key points
- Apply $H$ , then measure in the 0/1 basis; 0 means $+$ , 1 means $-$
- $H\ket+ = \ket0$ and $H\ket- = \ket1$ , so $H$ moves the ± coefficients onto $\ket0$ and $\ket1$
- This test separates $\ket+$ from a coin of $\ket0$ / $\ket1$ , which no 0/1 measurement can
Explain the result of the $z$ (keep $+$ ), $x$ (keep $+$ ), $z$ cascade with the qubit model.
Show the key points
- An analyzer is a measurement along its axis: $z \leftrightarrow \{\ket0, \ket1\}$ , $x \leftrightarrow \{\ket+, \ket-\}$
- It leaves each atom in its exit’s state, so the $x$ analyzer’s $+$ beam is in $\ket+$ whatever came before
- $\ket+$ splits 50/50 at $z$ ; overall $\tfrac12\cdot\tfrac12\cdot\tfrac12 = \tfrac18$ at each final exit
- Directions $90°$ apart in space are not orthogonal states (inner product $1/\sqrt2$ )

Key ideas

To measure in a basis $\{\ket a, \ket b\}$ , write $\ket\psi = \alpha'\ket a + \beta'\ket b$ ; then $P(a) = \abs{\alpha'}^2$ , with $\alpha' = \braket{a}{\psi}$ . Afterwards the qubit is in $\ket a$ or $\ket b$ .
Certainty depends on the basis. A state is certain in exactly one basis, the one that contains it: $\ket0$ is certain in 0/1 and 50/50 in ±.
A ± measurement is $H$ followed by a 0/1 meter, read as 0 → $+$ and 1 → $-$ . It tells $\ket+$ apart from a coin that gives $\ket0$ or $\ket1$ .
A Stern–Gerlach analyzer measures along its own axis of the Bloch sphere. A state along one axis splits 50/50 at an analyzer turned $90°$ from it; only opposite directions give orthogonal states.
In a cascade, multiply the probabilities of the exits you follow. Each analyzer leaves the atom in its exit’s state, so $z$ , $x$ , $z$ sends $\tfrac18$ of the oven’s atoms to each final exit.

Go further

Free resources for when you want more than this lesson covers.

Superposition states and mixed statesQuVis, University of St Andrews · Interactive
Run a Stern–Gerlach apparatus yourself and try to tell a superposition from a 50/50 mixture, written in spin notation.
Lecture 1: Introduction to superpositionAllan Adams, MIT OpenCourseWare · Video
A university physics lecture whose color and hardness boxes repeat the z, x, z cascade, then build superposition from it.
Change of basis3Blue1Brown · Video
How to translate vectors and matrices between two coordinate systems, the general form of rewriting a state in another basis.
Stern–Gerlach experimentWikipedia · Article
The 1922 silver-atom experiment, its history, and the sequential set-ups behind the cascades in this lesson.

Math refreshers

The toolkit topics this lesson leans on.

Inner products and bases⟨φ|ψ⟩, orthogonality, and |⟨φ|ψ⟩|² as a probability