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Stern–Gerlach experiments

Learning objectives

After this chapter, you should be able to:

The final two objectives are developed in optional laboratory and inquiry extensions after the chapter’s core conceptual sequence.

1.1 An experiment before a formalism

Quantum mechanics is easiest to trust when it grows out of a real experiment, rather than being handed to you as a list of abstract postulates. So this book begins with one of the simplest experiments in physics: the one performed by Otto Stern and Walther Gerlach in 1922. Their question was simple. Does an atom’s magnetism point in only a few special directions, or can it point anywhere at all? The answer they found was strange enough to force physicists to rethink what it even means for a particle to have a direction.

Picture the apparatus in three stages. First, an oven heats a sample of silver until atoms boil off as a vapor. Second, that vapor escapes through a narrow opening, and a pair of slits downstream accepts only atoms traveling within a narrow range of directions; what emerges is a collimated beam. Collimation does not, by itself, select one exact speed: an oven beam retains a thermal spread of speeds unless a separate velocity selector is added. In the idealized diagrams below we suppress both spreads so that the spin-dependent splitting is easy to see. Third, the beam passes between the shaped poles of a magnet and then strikes a glass screen, where each atom leaves a visible mark.

One feature of the magnet matters more than any other: its poles are shaped so that the field is much stronger near one pole than the other. Without this asymmetry the experiment would show nothing interesting, as the next few paragraphs will make clear.

To see why the asymmetry matters, picture each atom as a tiny bar magnet, described by a magnetic moment vector μ\boldsymbol\mu—the same quantity that tells a compass needle which way to twist. Just like a compass needle, the atom’s energy depends on how μ\boldsymbol\mu is oriented relative to the external field B\mathbf B: lined up with the field is low energy, opposed to the field is high energy, and every angle in between interpolates smoothly. In symbols,

U=−μ⋅B.U=-\boldsymbol\mu\cdot\mathbf B.

A uniform field, however, cannot push the atom anywhere—it can only twist it. The force pulling one “pole” of the tiny magnet forward is exactly canceled by the force pushing the other pole backward, so the atom feels a torque but no net force. A push appears only when the field itself is stronger on one side of the atom than the other, that is, only when the field has a gradient. For a gradient along the zz direction, the net force on the atom works out to

Fz=−∂U∂z≈μz∂Bz∂z.F_z=-\frac{\partial U}{\partial z} \approx \mu_z\frac{\partial B_z}{\partial z}.

Equation (2) is the whole operating principle of the device. It says that each atom is pushed by an amount proportional to μz\mu_z, the component of its magnetic moment along the gradient direction. A larger μz\mu_z means a larger deflection; a reversed μz\mu_z means deflection the other way. In short, the magnet turns an otherwise invisible internal property of the atom—its magnetic orientation—into something we can see directly: a spot’s position on the screen.

Now we can ask what classical physics predicts for the result. Nothing in classical physics picks out a preferred direction for μ\boldsymbol\mu, so the atoms emerging from the oven should have their magnetic moments pointing every which way. That means μz=μcos⁡θ\mu_z=\mu\cos\theta should take every value between −μ-\mu and +μ+\mu, depending on each atom’s orientation angle θ\theta. Combined with equation (2), this predicts a continuous smear of deflections on the screen. For fixed-magnitude moments pointing isotropically, cos⁡θ\cos\theta is uniformly distributed, so the idealized classical pattern is an approximately uniform band between the two extreme deflections. Thermal speeds, finite slits, and detector resolution can broaden or reshape that band, but they do not turn it into two discrete spots.

That is not what Stern and Gerlach saw. Instead of the predicted continuous band, their deposited silver pattern separated into two traces. In the ideal two-outcome model used throughout this chapter, we represent those traces as two spots labeled z+z+ and z−z-.

A silver-atom oven, collimating slits, and a field-gradient magnet send atoms to a detector. The classical prediction is a continuous smear, while the observed pattern has two distinct spots labeled z plus and z minus.

Figure 1:The Stern–Gerlach apparatus converts the spin component along the field gradient into a visible deflection. The decisive observation is not merely a deflected beam, but two discrete spots in place of the classical smear.

The two-spot pattern tells us something remarkable: the component of the atom’s magnetic moment that couples to the apparatus takes two discrete values rather than a continuum. Historically, Stern and Gerlach observed this space quantization before electron spin had been identified. In the modern description, a ground-state silver atom has closed inner shells and one unpaired outer electron with zero orbital angular momentum, so its relevant two-valued angular momentum is spin. No matter which axis we build the magnet along, an analyzer of this effective spin-12\tfrac12 system produces only two outcomes—never a value in between. The measured zz component can only be

Sz=+ℏ2orSz=−ℏ2.S_z=+\frac{\hbar}{2}\quad\text{or}\quad S_z=-\frac{\hbar}{2}.

This restriction—only two outcomes, no matter how the analyzer is oriented—is the first genuinely quantum fact in this book. Everything that follows is an attempt to build a mathematical theory around it.

From now on we abbreviate an analyzer oriented along zz as SGz\mathrm{SG}_z. Something useful happens if we place a physical block in front of one of its two exits. Say we block the z−z- exit and let only the z+z+ beam through. Every atom that makes it past the block is now guaranteed to have been recorded as z+z+—so the same device that just measured spin has become a device that prepares it. This dual role, one physical apparatus used first as a measurement and then as a preparation, will come up again and again in this chapter, so it is worth fixing firmly in mind now.

Reading the experiment from left to right

To talk about these experiments precisely, it helps to separate every one of them into three stages:

  1. A preparation specifies how the incoming ensemble was made. A source by itself may be unpolarized; a source followed by a filter prepares a selected spin state.

  2. A transformation changes the state without yet recording a spin result. Examples include a uniform magnetic field or a coherent path recombiner.

  3. A measurement ends in macroscopically distinguishable records, such as counts at the two exits of an analyzer.

This vocabulary heads off a common confusion. The symbol z+z+ gets used for three different things: an outcome recorded at a detector, the physical exit path leading to that detector, and the state that gets prepared when we select that path. Context usually makes clear which meaning is intended, but keep in mind that these are three related, not identical, ideas.

In the laboratory, what we actually observe are raw counts: some atoms land in the z+z+ spot, some in the z−z- spot. If N+N_+ and N−N_- are the numbers of atoms detected at the two exits, the measured relative frequencies are

f+=N+N++N−,f−=N−N++N−.f_+=\frac{N_+}{N_++N_-},\qquad f_-=\frac{N_-}{N_++N_-}.

Quantum theory predicts the probabilities that these frequencies approach after many repetitions of the same experiment. It does not, in general, predict which exit any one particular atom will take.

1.2 Repeatability

Here is the simplest possible follow-up experiment: put a second SGz\mathrm{SG}_z analyzer right after the first one, oriented along the same zz axis, and keep only the atoms that came out of the first analyzer’s z+z+ exit. What happens at the second analyzer?

The answer is unambiguous: every atom that enters the second analyzer leaves through its z+z+ exit. None comes out z−z-.

This tells us that the first analyzer did more than sort atoms into two piles. Selecting the z+z+ output prepared something new: a beam in which every atom is now definite for a repeated SzS_z measurement, meaning that measuring it again along zz can no longer produce a surprise.

Repeatability also gives us an operational definition of the label ∣+z⟩|+z\rangle, which will be introduced formally in Chapter 2: it means the preparation that produces z+z+ with certainty when immediately analyzed along zz. We do not need to imagine what the atom is “really doing” between devices in order to define the state reproducibly.

Example 1.1: counting atoms

A source sends 12,00012{,}000 atoms into SGz\mathrm{SG}_z. The source has no preferred orientation, so half emerge from each exit on average. The z−z- exit is blocked, and the z+z+ beam enters a second SGz\mathrm{SG}_z.

About 6,0006{,}000 atoms reach the second analyzer, and all 6,0006{,}000 leave its z+z+ exit. Statistical fluctuations affect the first split, but an ideal repeat measurement does not create a second 50–50 split.

Screenshot of the Stern Gerlach simulation

Figure 2:Assemble your own chain of ovens, analyzers, and counters, fire spin-12\tfrac12 or spin-1 atoms through it, and watch Monte Carlo counts accumulate next to the analytic prediction. Rebuild the repeated-analyzer setup from Example 1.1 before reading on, then try chaining analyzers at other angles.

Interactive simulation: Stern Gerlach

1.3 Incompatible questions

Repeating a measurement along the same axis gave a simple, if important, result: nothing changes. The next natural question is what happens if we measure along a different axis instead. Prepare a z+z+ beam as before, but this time send it into SGx\mathrm{SG}_x, an analyzer whose field gradient points along xx rather than zz. The output divides equally:

P(x+∣z+)=P(x−∣z+)=12.P(x+\mid z+)=P(x-\mid z+)=\frac12.

So a z+z+ preparation is not definite for SxS_x: knowing that an atom is z+z+ tells us nothing about which way it will go in SGx\mathrm{SG}_x. Here is the more surprising part. Keep only the x+x+ output of that measurement, and send those atoms into a third analyzer, this one measuring SzS_z again. If the atom really still had a definite zz value all along, we should recover z+z+ with certainty, exactly as in Section 1.2. Instead, both zz outcomes now occur with equal probability, as if the intervening xx measurement had erased what we thought we already knew:

z+⟶x+⟶{z+with probability 1/2,z−with probability 1/2.z+\longrightarrow x+\longrightarrow \begin{cases} z+ & \text{with probability }1/2,\\ z- & \text{with probability }1/2. \end{cases}

This is the key point: measuring SxS_x changed what we could predict about SzS_z, even though nothing directly measured SzS_z in between. It is tempting to imagine that the atom secretly carried permanent values for both SzS_z and SxS_x, and that each analyzer simply reads the relevant value without changing anything else. The experiment rules out that specific, noninvasive fixed-label model: if the xx analyzer merely revealed an unchanged xx label, the final SzS_z measurement should still return z+z+ with certainty—but it does not. The sequence does not, by itself, rule out every possible hidden-variable account; a more elaborate model could allow the apparatus to alter hidden variables contextually. Bell’s test in Chapter 5 will impose additional locality assumptions and reach a stronger conclusion.

Two Stern–Gerlach analyzer chains. Repeating a z measurement after preparing z plus gives z plus with certainty. Selecting x plus between the preparation and final z measurement produces equal z plus and z minus probabilities.

Figure 3:Aligned analyzers repeat a definite result, whereas an intervening analyzer along an incompatible axis replaces that certainty with a new preparation. The diagrams are read from left to right, just like the laboratory sequence.

Concept check 1.1

A z+z+ beam enters SGx\mathrm{SG}_x. The x−x- exit is blocked, and the transmitted x+x+ beam enters another SGx\mathrm{SG}_x. What fraction of the original z+z+ beam reaches the final x+x+ detector?

Answer

One half. The first xx measurement transmits half the atoms into x+x+. Repeating the SxS_x measurement then returns x+x+ with certainty.

What “disturbance” does and does not explain

One might instead say that the middle analyzer simply adds uncontrolled mechanical noise. Repeatability shows why that description is inadequate: if the device only randomized everything, its own SxS_x result would not be reliable on repetition. Yet an x+x+ selection is perfectly repeatable. The analyzer performs a reproducible transformation—it replaces a preparation definite for SzS_z with one definite for SxS_x—even though the later SzS_z result becomes uncertain. This does not prove that no underlying disturbance occurs; it shows that “mere random jostling” lacks the structure needed to explain the observations.

Order matters here in a way that has no classical analog. Compare

z+⟶z+⟶xz+\longrightarrow z+\longrightarrow x

with

z+⟶x+⟶z.z+\longrightarrow x+\longrightarrow z.

In the first sequence, repeating the zz measurement changes nothing, so the final xx analyzer still sees a 50–50 split, exactly as it would from a fresh z+z+ beam. In the second sequence, the intervening x+x+ selection changes the outcome seen by the final zz analyzer, turning certainty into a 50–50 split. Chapter 3 will express this order dependence precisely, as a statement about products of noncommuting matrices.

Analyzers at a general angle

So far we have only compared axes at 0∘0^\circ and 90∘90^\circ. What happens at angles in between? Nothing stops us from building an analyzer tilted at any angle relative to the preparation axis, and it seems reasonable to guess that the probability of recovering the original outcome should fall off smoothly as the analyzer tilts away from the preparation axis. If spin behaved like an ordinary arrow, we might guess that this probability varies as cos⁡α\cos\alpha, matching the projection of one vector onto another. Experiment gives a different, and at first surprising, answer. For a beam prepared spin-up along a\mathbf a and analyzed along b\mathbf b, separated by angle α\alpha,

P(+b∣+a)=cos⁡2α2,P(−b∣+a)=sin⁡2α2.P(+\mathbf b\mid+\mathbf a)=\cos^2\frac{\alpha}{2},\qquad P(-\mathbf b\mid+\mathbf a)=\sin^2\frac{\alpha}{2}.

Notice the half-angle, α/2\alpha/2, rather than α\alpha itself. This is not a typo or a minor detail—it is an early hint that spin does not transform like an ordinary arrow in space. As Chapter 4 will show, a representative spinor acquires an overall minus sign under a 360∘360^\circ rotation and returns to the identical ket only after 720∘720^\circ. The minus sign does not change the physical state of an isolated spin, because overall phase is unobservable; it becomes detectable only relative to another coherent amplitude. The half-angle in equation (9) is an early fingerprint of this spinor transformation law.

It is worth checking that this formula reproduces everything found so far. At α=0∘\alpha=0^\circ (aligned analyzers), cos⁡2(0)=1\cos^2(0)=1: certainty of recovering the same outcome, matching Section 1.2. At α=180∘\alpha=180^\circ (oppositely aligned analyzers), cos⁡2(90∘)=0\cos^2(90^\circ)=0: certainty of the opposite outcome. At α=90∘\alpha=90^\circ (perpendicular analyzers, like xx versus zz), cos⁡2(45∘)=1/2\cos^2(45^\circ)=1/2: an equal split, matching Section 1.3. Equation (9) is therefore not a separate new rule; it is the general law that contains every case already found as a special case. For now we treat it as an empirical summary of the data. Chapter 2 will derive it from the mathematics of state vectors.

Two analyzer axes separated by angle alpha beside a graph. The probability for the plus outcome decreases from one to zero as alpha goes from zero to 180 degrees, while the minus probability increases from zero to one; both are one half at 90 degrees.

Figure 4:The half-angle rule smoothly joins repeatability at 0∘0^\circ, equal splitting at 90∘90^\circ, and certain reversal at 180∘180^\circ. At every angle the two exclusive probabilities add to one.

Example 1.2: three analyzer directions

A z+z+ beam passes through an analyzer whose axis n\mathbf n is 60∘60^\circ from +z+z, and only the +n+\mathbf n output is retained. That beam is then measured along zz. This means applying equation (9) twice in a row, using the selected output of the first analyzer as the input to the second.

At the middle analyzer, equation (9) applies with α=60∘\alpha=60^\circ, so the half-angle appearing in the formula is α/2=30∘\alpha/2=30^\circ:

P(+n∣z+)=cos⁡230∘=(32)2=34.P(+\mathbf n\mid z+)=\cos^2 30^\circ =\left(\frac{\sqrt3}{2}\right)^2=\frac34.

The selected middle output is a new +n+\mathbf n preparation. Since the angle back to +z+z is again 60∘60^\circ,

P(z+∣+n)=34,P(z−∣+n)=14.P(z+\mid+\mathbf n)=\frac34,\qquad P(z-\mid+\mathbf n)=\frac14.

Thus 9/169/16 of the original atoms reach the final z+z+ detector and 3/163/16 reach the final z−z- detector; the remaining quarter was blocked at the middle analyzer. Conditional probabilities must be multiplied along a selected path.

Concept check 1.2

A z+z+ beam is analyzed along an axis n\mathbf n only 10∘10^\circ from +z+z. One student predicts exactly z+z+ because the axes are nearly aligned; another predicts a 50–50 split because the axes are different. What does the experiment predict?

Answer

Neither. The probability of +n+\mathbf n is cos⁡2(5∘)≈0.9924\cos^2(5^\circ)\approx0.9924, while the probability of −n-\mathbf n is about 0.0076. Quantum outcomes are discrete, but their probabilities vary continuously with analyzer orientation.

1.4 Recombining alternatives

Every experiment so far has ended with a measurement: an atom lands in one detector or another, and that is the end of the story. What happens if, instead of recording which path an atom took, we let two paths recombine before anything is measured?

Imagine a carefully built device that splits a z+z+ beam into two separate paths, one associated with x+x+ and one with x−x-, and then brings the two paths back together—coherently, without ever revealing which path a given atom traveled. Remarkably, the recombined beam turns out to be back in the original z+z+ state, exactly as if the device had never touched it. But if we block either path, so that only atoms from the other path get through, the later SzS_z statistics change completely.

This kills a natural but wrong guess: that the two paths behave like two ordinary, independent sub-beams, so that whatever happens to the recombined beam should just be the sum of what happens along each path separately. Quantum theory instead assigns a complex number—a probability amplitude—to each indistinguishable alternative. When alternatives are indistinguishable, we add the amplitudes first, and only afterward take the squared magnitude to get a probability.

This distinction will later become the difference between a coherent superposition and an incoherent mixture.

The word coherently carries real experimental content. The apparatus must preserve a stable relation between the phases accumulated on the two paths. If vibrations, stray fields, or a path detector randomize that relation, the interference washes out even when nobody reads the detector. What matters is whether a durable physical record could distinguish the alternatives, not whether a person happens to look.

One can summarize the alternatives with two limiting arrangements:

Two path experiments. With no path record, x plus and x minus amplitudes recombine coherently and recover z plus with certainty. With path records, interference is lost and the final z plus probability is one half.

Figure 5:Recombination restores the original state only while the two alternatives remain coherent. A durable path record changes the calculation from adding amplitudes to adding probabilities, even if nobody reads the record.

This is why “half the atoms went through each arm” is not a complete description of the coherent experiment. The path amplitudes can cancel at one output and reinforce at another, even though each open arm alone sends atoms to both outputs.

Concept check 1.3

Why is a 50–50 collection of separately prepared x+x+ and x−x- atoms not automatically equivalent to a coherent recombination of the two paths?

Answer

The collection records mutually exclusive preparations and is an incoherent mixture. A coherent recombination preserves the relative phase between the two amplitudes, so later probabilities can contain interference terms.

1.5 A minimal model

We now have three empirical rules—repeatability, incompatibility, and interference—plus the half-angle probability law. Before building a full mathematical theory, it helps to state in plain language exactly what that theory needs to accomplish:

For spin 12\tfrac12, two independent states suffice to describe every possible pure spin state. We will represent them by vectors in a two-dimensional complex vector space. The word “two-dimensional” refers to state space, not ordinary physical space.

If “complex vector space” does not yet mean much to you, that is completely expected at this point—nothing so far has required it. Chapter 2 builds this idea from the ground up, starting with a review of ordinary arrows in the plane and of complex-number arithmetic itself, before ever writing down a spin state. The two-outcome rule, the incompatibility of SzS_z and SxS_x, and the interference rule found above are the experimental reasons that vector space is the right tool; the vocabulary comes second.

From laboratory rules to mathematical demands

Let’s translate the experimental facts directly into mathematical requirements. Whatever structure we build, it needs:

An ordinary list of two probabilities, say (1/2,1/2)(1/2,1/2), cannot do this job: it records the weight of each alternative but throws away the phase that controls interference. A pair of complex numbers can do both at once, since a complex number naturally carries both a magnitude and a phase. This is exactly why Chapter 2 introduces state vectors with complex-number coefficients—called kets—before ever writing down a wavefunction.

1.6 What spin is—and is not

The name “spin” invites a picture: a tiny sphere spinning on its axis, like a planet. That picture is a useful crutch for the imagination, but it is not literally correct, and taking it too literally leads to wrong predictions. Spin does carry angular momentum, and often a magnetic moment, which is exactly why it interacts with the field in a Stern–Gerlach apparatus in the first place. But a rigid, rotating sphere would have a magnetic-moment component that varies continuously as its axis tilts—precisely the smooth spread that classical physics predicted and that the apparatus did not show. The name “spin” is a historical label; it is the experiments, not the name, that tell us what spin actually is.

There is a second way the arrow picture fails, beyond just the two discrete outcomes. A classical arrow pointing at a known angle α\alpha to the analyzer would have a projected component proportional to cos⁡α\cos\alpha, and that component would vary continuously as α\alpha varies. Real spin does not do this: the outcome is always ±ℏ/2\pm\hbar/2, and the angle only controls how often each outcome occurs, through equation (9).

One more warning about pictures. Chapter 2 introduces the Bloch sphere, which looks like a sphere covered in little arrows. Resist the urge to picture each arrow as a literal miniature compass needle sitting inside the particle. Each arrow is shorthand for an entire preparation recipe and the statistics it produces—not a claim that the particle contains a tiny spinning pointer.

1.7 A systematic way to predict analyzer chains

Until the vector formalism is available, a short operational recipe is enough for ideal sequential measurements:

  1. Identify the state entering the first analyzer. If the source is unpolarized, the first analyzer produces equal counts.

  2. At each analyzer, measure the angle from the most recently selected preparation axis, not necessarily from the original axis.

  3. Use equation (9) for each branch.

  4. Multiply conditional probabilities along one branch; add probabilities only when mutually exclusive detected branches are pooled.

  5. If paths are recombined coherently, stop using this probability-tree recipe. Amplitudes and their phases are then required.

Example 1.3: keep all branches straight

An unpolarized source sends 16,00016{,}000 atoms into SGz\mathrm{SG}_z. The z+z+ exit is selected, then the x−x- exit of an SGx\mathrm{SG}_x analyzer is selected, and finally both exits of SGz\mathrm{SG}_z are counted.

The expected bookkeeping is

16,000→z+8,000→x−4,000→z{2,000z+,2,000z−.16{,}000\xrightarrow{z+}8{,}000 \xrightarrow{x-}4{,}000 \xrightarrow{z} \begin{cases} 2{,}000&z+,\\ 2{,}000&z-. \end{cases}

The final counts are not 4,0004{,}000 and zero. The state entering the last analyzer is x−x-, not the earlier z+z+ preparation. Writing the state beneath each segment of an apparatus sketch is a simple way to avoid this mistake.

1.8 Optional laboratory extension: from probabilities to data

The conceptual route to the state-vector formalism is complete at this point. This optional section connects ideal probabilities to the fluctuations and systematic effects encountered in laboratory data.

Equations such as (9) predict probabilities, whereas a real experiment produces integer counts. The link between the two is statistical.

If the preparation and analyzer are held fixed and NN independent atoms are recorded, the number N+N_+ at the positive exit fluctuates around Np+Np_+, where

p+=cos⁡2α2.p_+=\cos^2\frac{\alpha}{2}.

For ideal independent trials, the count follows a binomial distribution. Its standard deviation is

σN+=Np+(1−p+),\sigma_{N_+}=\sqrt{Np_+(1-p_+)},

and the measured frequency f+=N+/Nf_+=N_+/N has standard deviation

σf+=p+(1−p+)N.\sigma_{f_+}=\sqrt{\frac{p_+(1-p_+)}{N}}.

The 1/N1/\sqrt N dependence matters. Recording four times as many atoms reduces the typical frequency fluctuation by only a factor of two. Quantum theory does not predict that every batch divides in exactly the theoretical ratio; it predicts the distribution of results from which those batches are drawn.

Example 1.4: is a split consistent with the angle rule?

A z+z+ source is analyzed along an axis 60∘60^\circ from +z+z. Among N=1,000N=1{,}000 detected atoms, N+=735N_+=735 leave the positive exit. The ideal prediction is p+=3/4p_+=3/4, so the expected count and its standard deviation are

Np+=750,σN+=1000(34)(14)≈13.7.Np_+=750, \qquad \sigma_{N_+}=\sqrt{1000\left(\frac34\right)\left(\frac14\right)} \approx13.7.

The observed count is 15 below the mean, about 1.1 standard deviations. Such a difference is ordinary statistical variation, not persuasive evidence against the angle rule. By contrast, a persistent discrepancy that grows in statistical significance as more data are collected suggests imperfect preparation, a misaligned analyzer, unequal detector efficiencies, or a failure of the model.

Concept check 1.4

A run with 100 atoms gives 54 counts at x+x+ and 46 at x−x-. Has the expected 50–50 split failed?

Answer

No. For N=100N=100 and p=1/2p=1/2, the expected standard deviation of the positive count is 100(1/2)(1/2)=5\sqrt{100(1/2)(1/2)}=5. A count of 54 is less than one standard deviation above the mean of 50 and is entirely typical.

Estimating an unknown analyzer angle

The same experiment can be used in reverse. For a known z+z+ input and an analyzer known to lie between 0 and π\pi from +z+z, the observed positive frequency estimates the angle:

αest=2cos⁡−1f+.\alpha_{\mathrm{est}}=2\cos^{-1}\sqrt{f_+}.

If 81%81\% of a large sample exits through +n+\mathbf n, then αest=2cos⁡−1(0.9)≈51.7∘\alpha_{\mathrm{est}}=2\cos^{-1}(0.9)\approx51.7^\circ. Counts can therefore characterize an apparatus as well as test a prepared state. In later chapters, this reversal of viewpoint becomes quantum-state tomography: known analyzers are used to infer an unknown preparation.

1.9 Optional inquiry extension: separating competing explanations

This optional section turns the chapter’s observations into a model-testing strategy: choose arrangements for which plausible explanations make different predictions.

A useful experiment does more than exhibit an effect; it distinguishes between models that would otherwise make the same prediction. A single zz analyzer, on its own, cannot distinguish a genuine quantum state from a classical population of atoms carrying fixed zz labels—both would give the same repeatable z+z+ result. Sequential analyzers are powerful precisely because they force competing stories to disagree.

Consider three hypotheses for a beam selected from the z+z+ exit:

  1. Fixed-label hypothesis: every atom carries permanent values for all analyzer directions, and an analyzer only reads the relevant value.

  2. Classical-arrow hypothesis: every atom carries an ordinary magnetic arrow, and the analyzer reports its component along the chosen axis.

  3. Quantum preparation hypothesis: selecting an outcome prepares a state definite for that analyzer, while other components generally remain indefinite.

The sequence z+→zz+\rightarrow z is compatible with the first and third hypotheses, so repeatability alone does not decide between them. The sequence z+→x+→zz+\rightarrow x+\rightarrow z, however, restores a 50–50 zz distribution. That result conflicts with permanent, merely revealed zz labels. The fact that every analyzer still has only two exits conflicts with a classical arrow whose projected component can vary continuously. Coherent recombination then adds a third test: interference depends on a controllable relative phase that neither classical population model retains.

This pattern is a general method for reasoning in quantum mechanics:

Example 1.5: measurement or mere beam splitting?

An ideal device separates a z+z+ input into x+x+ and x−x- paths. Compare three arrangements.

The last comparison shows why spatial overlap alone is not sufficient for interference. The relevant condition is whether the alternatives remain coherent—whether any physical degree of freedom carries usable which-path information. Chapter 2 will represent this distinction by retaining or discarding the cross terms between amplitudes.

Concept check 1.5

Why can the sequence z+→x+→zz+\rightarrow x+\rightarrow z test more than the sequence z+→zz+\rightarrow z?

Answer

Both a fixed-label model and quantum theory can imitate immediate repeatability. They differ after an incompatible intermediate measurement: quantum theory predicts that selecting x+x+ prepares a new state and makes the final zz result 50–50, whereas a device that merely reads a permanent zz label should not erase it.

Summary

Exercises

Exercises 1–6 rehearse the core analyzer logic, 7–11 consolidate angle rules and coherent alternatives, and 12–16 extend the chapter into data analysis and experimental design. Complete the first group before choosing from the later two.

  1. An unpolarized beam of 20,00020{,}000 atoms passes through SGz\mathrm{SG}_z; the z+z+ output then passes through SGx\mathrm{SG}_x. Estimate the counts in the two final outputs.

  2. Predict the possible final outcomes and their probabilities for z+→x−→zz+\rightarrow x-\rightarrow z.

  3. Insert a second SGx\mathrm{SG}_x after the x−x- selection in Exercise 2. Explain why its result differs qualitatively from the final SzS_z result.

  4. A student says, “The first SzS_z analyzer tells us which atoms were already pointing up.” Identify two observations in this chapter that this classical explanation fails to capture.

  5. Explain operationally what must be changed to turn a spin analyzer into a state-preparation apparatus.

  6. Design a sequence of ideal analyzers that begins with z+z+ and has a final z−z- detection probability of 1/21/2. Give at least two sequences and explain them.

  7. A z+z+ beam is sent to an analyzer tilted 120∘120^\circ from +z+z. Find both output probabilities using equation (9).

  8. An unpolarized beam of 40,00040{,}000 atoms passes through three analyzers. The selected outputs are first z+z+ and then +n+\mathbf n, where n\mathbf n is 60∘60^\circ from +z+z. Both outputs of the final zz analyzer are recorded. Predict all final counts and account for atoms blocked earlier.

  9. Draw probability trees for z+→x+→zz+\rightarrow x+\rightarrow z and z+→x−→zz+\rightarrow x-\rightarrow z. If the two selected xx beams are later pooled incoherently, what are the final zz probabilities?

  10. A two-path apparatus gives deterministic z+z+ when both paths are open and recombined. With either path blocked, the final zz analyzer gives both results. Explain why adding the two single-path probability distributions cannot reproduce the both-open result.

  11. Classify each element as preparation, transformation, or measurement: an oven source, a z+z+ filter, a uniform magnetic field, an unobserved path marker, a coherent recombiner, and a two-channel detector. Some elements may play more than one role; explain your choices operationally.

  12. A z+z+ beam is analyzed 90∘90^\circ from its preparation axis. In a run of 2,5002{,}500 atoms, estimate the mean and standard deviation of the ++ count. Is a count of 1,2901{,}290 surprising on this scale?

  13. A large sample prepared in z+z+ gives f+=0.64f_+=0.64 at an analyzer whose axis is known to lie between 0∘0^\circ and 180∘180^\circ from +z+z. Estimate the angle.

  14. Derive equation (15) from equation (14). Explain why the absolute uncertainty in a count grows with NN while the uncertainty in a relative frequency shrinks.

  15. Design a three-configuration experiment that distinguishes coherent recombination, an incoherent mixture of the two paths, and accidental blocking of one path. State the qualitative final zz statistics expected in each configuration.

  16. A source drifts so that its intensity falls by 10%10\% during a run. Explain why raw output counts could then give a misleading comparison between two analyzer settings. Propose a frequency or normalization that reduces the problem.

Selected exercise guidance

Use these answers only after making a complete attempt.

Exercise 2

The x−x- selection occurs with probability 1/21/2. Conditional on that selection, the final z+z+ and z−z- probabilities are each 1/21/2, so each complete detected route contains 1/41/4 of the original beam.

Exercise 7

P(+n∣z+)=cos⁡2(60∘)=1/4P(+\mathbf n|z+)=\cos^2(60^\circ)=1/4 and P(−n∣z+)=3/4P(-\mathbf n|z+)=3/4.

Exercise 12

The mean is 1,2501{,}250 and the standard deviation is 2500(1/2)(1/2)=25\sqrt{2500(1/2)(1/2)}=25. A count of 1,2901{,}290 is 1.6 standard deviations above the mean, so it is not especially surprising.

Exercise 13
α=2cos⁡−10.64=2cos⁡−1(0.8)≈73.7∘.\alpha=2\cos^{-1}\sqrt{0.64}=2\cos^{-1}(0.8)\approx73.7^\circ.