This article reviews the predictions of quantum mechanics (QM) for one- and two-particle Stern–Gerlach experiments and then frames Bell's results, which rule out hidden-variable alternatives to QM, as attempts by a skeptical Eve to fool Alice and Bob with (first example) classical probability mixtures of non-entangled quantum states and (second example) a classical hidden-variable theory. With hidden variables, Eve can succeed when Alice and Bob limit themselves to two Stern–Gerlach directions, but always fails for some choices of three directions. For three random directions, hidden-variable theories are impossible (we show) exactly 2/3 of the time. All the calculations in this article are available in Mathematica and Python files in the supplementary material, allowing readers to experiment with their own variants.

This article gives an elementary exploration of Bell's demonstration that no classical theory can reproduce the results of quantum mechanics. Our treatment is “contemporary” in the sense of making contact with basic ideas from quantum computing. Bell's results1 sharpen and accentuate the supposed paradox raised by Einstein, Podolsky, and Rosen (EPR)2 that the measurement of one particle of a two-particle state seems to propagate some kind of information “instantaneously” to the other particle. Historically, doubters of the then-new quantum mechanics (QM) tried to explain such results by postulating “hidden variables” supposedly carried by the particles as additional information. Bell demonstrated (and we will see below) that no such hidden variables can possibly give consistency with the predictions of QM, which are by now experimentally confirmed to high accuracy.

The treatment here of hidden variables (Sec. VII) closely follows that of Mermin,3 in turn based on a version of Bell due to Bohm.4 In this Bohm–Mermin formulation, Bell's famous “inequalities” on correlations among separated observations are equivalently replaced by explicit “programs” for supposed hidden variable theories.

A single qubit is represented abstractly by a complex unimodular two-vector. We can define basis vectors as

|0(10),|1(01).
(1)

The most general state of a single qubit can be written in terms of three Euler angles, 0aπ and 0b,c2π as

|ψ(a,b,c)eic(cos(12a)eibsin(12a)).
(2)

Its corresponding orthogonal state is (up to an arbitrary phase)

|ψ(a,b,c)=|ψ(πa,π+b,c)=eic(sin(12a)eibcos(12a)).
(3)

The most general linear operator on such a vector preserving unimodularity is a 2 × 2 complex unitary matrix. In its most general form, we can write this as parametrized by an angle θ and a unit 3-vector n,

SU2(θ,nx,ny,nz)=[cos(12θ)inzsin(12θ)(nyinx)sin(12θ)(nyinx)sin(12θ)cos(12θ)+inzsin(12θ)].
(4)

It is easy to verify that the general state Eq. (2) can be reached by the composition of three rotations (“Euler angles”) acting on |0, specifically,

|a,b,c=SU2(b,0,0,1)·SU2(a,0,1,0)·SU2(b2c,0,0,1)·|0.
(5)

The parametrization in Eqs. (2) and (4), with notation seemingly referencing rotations in three-dimensional space, is valid even when qubits are implemented as a physical system with no rotational symmetries (e.g., as superconducting circuits). However, the parametrization takes on an additional physical meaning when the qubits in Eq. (1) are identified with the two physical spin states of a spin-1/2 particle such as an electron, taking the basis states to be up and down along the z-axis. In that case, the possibility arises of simply making measurements in a (three-dimensional) coordinate system that is rotated from the original one. How do spin states like Eqs. (1) and (2) transform then?

The most general three-dimensional rotation of an angle θ around an axis with unit vector n can be written as the real, orthogonal rotation matrix,

SO3(θ,nx,ny,nz)=[cosθ+nx2(1cosθ)nxny(1cosθ)nzsinθnxnz(1cosθ)+nysinθnynx(1cosθ)+nzsinθcosθ+ny2(1cosθ)nynz(1cosθ)nxsinθnznx(1cosθ)nysinθnzny(1cosθ)+nxsinθcosθ+nz2(1cosθ)].
(6)

This is simply geometry, not QM. QM asserts that, however, a spatial rotation (θ,n) in Eq. (6) induces a spin transformation in Eq. (4) with the same(θ,n). This is a deep property of spin-1/2 particles and is possible only because the structures of the continuous groups SU(2) and SO(3) are the same. In fact, as is well known, SU(2) is exactly a double covering of SO(3). An accessible proof of this is given in  Appendix A.

“Stern–Gerlach” refers to experiments where the state of one or more electrons is prepared in one coordinate system, but is then measured in the coordinate system of a physically rotated apparatus.

This is elementary QM. Suppose the electron is prepared to be in the z-axis state |0 and measured in a frame rotated by an angle θ, say (without loss of generality) around the x axis. Let | and | denote the up and down directions in that frame. Then, the measurement probabilities can be calculated using Eq. (4),

Prob()=||0|2=|0|SU2(θ,1,0,0)|0|2=cos2(12θ)Prob()=||0|2=1Prob()=sin2(12θ).
(7)

An important special case is θ=π/2, yielding equal probabilities 1/2 for seeing the | and | states.

Two-particle states for electrons A and B lie in the product space with basis vectors [cf. Eq. (1)],

|0A|0B|00,|0A|1B|01,|1A|0B|10,|1A|1B|11.
(8)

It is customary to “flatten” the coefficients in this basis to a vector of length four in the above (lexical) order. For typographical convenience, we write the coefficients as a row vector, even when a column vector is more logical,

(a,b,c,d)a|00+b|01+c|10+d|11.
(9)

If the individual particle states |A=(p,q) and |B=(r,s) are separately prepared (e.g., by “classically separated” apparatuses), then the two-particle state is just the outer product, |A|B|AB=(pr,ps,qr,qs), but this is not the general case |AB=(a,b,c,d). It is easy to tell from the numerical components whether a two-particle state is a pure outer product: just take the determinant of the un-flattened 2 × 2 matrix of coefficients and see if it is zero, for example,

det[(pr,ps,qr,qs)]=|prpsqrqs|=(pr)(qs)(qr)(ps)=0.
(10)

Two-particle measurements of two particle states obey the usual projection rules: the probability of measuring Ψ=(a,b,c,d) to be in some measurement apparatus state Φ=(e,f,g,h) is the modulus-squared of the Hermitian dot product |Ψ|Φ|2. There is now the additional possibility of making a measurement on just one of the two particles. One would conventionally introduce the formalism of density matrices at this point; but for the limited purposes of this article, we can do without it. As one simple example, the measurement of |ψA, for Ψ=(a,b,c,d) in the basis frame of Eq. (8), leaves particle B in the state

|ψB=1|a|2+|b|2abwithprobability|a|2+|b|21|c|2+|d|2cdwithprobability|c|2+|d|2.
(11)

We will be interested in scenarios where Eve prepares a two-particle state |Ψ with the particles having opposite momentum, so that they fly off in opposite directions, A towards Alice and B towards Bob. Here, Eve is not playing her usual role as an eavesdropper, but is rather a skeptic of whether QM must truly be nonclassical. Alice will perform a measurement on A and Bob on B. Their measurement apparatuses are classically separated, and they may or may not be aligned with each other or with Eve's basis states. There are four possible outcomes, namely,

|A|B,|A|B,|A|B,|A|B,
(12)

giving four probabilities summing to unity. A typical calculation, among the four, is

Prob(AB)=|Ψ|(|A|B)|2,
(13)

which we can compute in any basis by calculating the product state |AB as a vector of length four in that basis. We can denote any particular two-particle Stern–Gerlach setup by a mapping “Stern,” giving the four probabilities, thus, calculated

Stern(|Ψ,|A,|B)Probs(AB,AB,AB,AB).
(14)

Example: Eve prepares the state |01. Both Alice and Bob measure in the basis of the different state |ψ(a,b,c) (Eq. (2)), aligned with each other, but not with Eve. Then, a calculation gives

Stern(|01,|ψ,|ψ)={14sin2(a),cos4(12a),sin4(12a),14sin2(a)}.
(15)

Three things to notice: (1) when a =0, so that Alice, Bob, and Eve are all aligned, we get probabilities (0,1,0,0) as expected. (2) When a=π/2, so that Alice and Bob are at right angles to Eve, we get (14,14,14,14), also as expected. (3) Although Eve prepared the system with the particles in opposite spin states, Alice and Bob sometimes (for a0) measure them as having the same spin state (first and fourth probabilities). Physically, this is because the state |01 is not the state of zero angular momentum—next to be discussed.

If Eve prepares the state |10 instead of |01, we, of course, correspondingly get

Stern(|10,|ψ,|ψ)={14sin2(a),sin4(12a),cos4(12a),14sin2(a)}.
(16)

When we consider rotations in three-space, the basis vectors in Eq. (8) are not the most natural ones. The more natural basis is a triplet state with angular momentum 1,

Ψ1,1|00=(1000),Ψ1,0|01+|102=(012120),Ψ1,1|11=(0001),
(17)

whose components transform among themselves under rotations, and a singlet state of angular momentum zero,

Ψ0|01|102=(012120),
(18)

which is invariant under rotations. (This singlet state is elsewhere often termed a “Bell state” and denoted Ψ.) We can verify these assertions by (i) substituting the components of the infinitesimal version of Eq. (4), SU2(δθ,nx,ny,nz) [see  Appendix A, Eq. (A1)] into a 4 × 4 matrix in two ways, corresponding to their action on particles A and B separately, (ii) multiplying these two matrices to get their joint action in a two-particle state, (iii) checking that the result is still unitary (it has to be), and (iv) verifying that its action is closed in the states in (17) and (separately) the Bell state (18). These calculations are included in the Mathematica file in the supplementary material of this article.6 

An even easier check that the state Ψ0 is rotation-invariant (that is, spherically symmetrical) is to calculate it using basis vectors that point in a different direction, e.g., from Eqs. (2) and (3),

Ψ0=12(|ψ(a,b,c)|ψ(a,b,c)|ψ(a,b,c)|ψ(a,b,c))=ei(b+2c)(012120),
(19)

which differs from Eq. (18) for all the values of (a, b, c) only by an unmeasurable phase.

In the most general case, Eve constructs a Bell singlet state |Ψ0 (we now know that she can do this in any coordinate system) that is then measured by Alice and Bob in arbitrary different coordinate systems. By symmetry, the only dependence can be on the angle θ between Alice and Bob's “up” direction, and so, without loss of generality, we can take

|A=|ψ(12θ,0,0),|B=|ψ(12θ,0,0).
(20)

With the machinery already defined, we calculate the four measurement probabilities and get

Stern(|Ψ0,|A,|B)={12sin2(12θ),12cos2(12θ),12cos2(12θ),12sin2(12θ)}.
(21)

What can we notice about Eq. (21)?

First, no matter how they choose their respective coordinate systems, Alice and Bob each separately get a random 50% mixture of up and down measurements [sum of the first and second vs third and fourth; or first and third vs second and fourth probabilities in (21)].

Second, if θ=π/2, so that Alice and Bob measure orthogonal directions, then all four outcomes have equal probability 1/4. So, in particular, there is no correlation between Alice's and Bob's measurements in this case.

Third, in the other extreme, when Alice and Bob are co-aligned, with θ = 0, then the probabilities are {0,12,12,0}, and so they always get opposite results, either up-down or down-up, no matter what coordinate direction they mutually choose. They don't have to tell Eve about their choice or even make a choice until after Eve's two particles have been dispatched in opposite directions.

Fourth, for intermediate θ, Alice and Bob get opposite measurements with probability cos2(θ/2) and the same measurements with probability sin2(θ/2). But these are exactly the single-particle probabilities in Eq. (7). It is as if Alice's measurement—happening first, say—establishes the coordinate system with respect to which Bob's measurement occurs, with the results that are (in that frame) not isotropic. This is Einstein's “spooky action at a distance” (“spukhafte Fernwirkung”). Whether you consider it the paradox that Einstein did depends on how “quantum friendly” you are. Causality is not violated: Alice and Bob each obtain random measurement results. It is only when they get together and compare the statistics of previous results that the paradox emerges as a correlation. That correlation is due to the fact that they were each making partial measurements of an intrinsically two-particle state. (Full disclosure: This author has never been able to see this as a paradox and so has a hard time making it seem so in this paragraph; but he is trying his best.)

We can easily imagine a world in which QM is the correct theory, but entanglement is not practically achievable. Indeed, that is almost our world, where entanglement is measurable only under very special conditions and at cryogenic temperatures. So the imagined world is simply one with higher temperatures and more consequent thermal noise.

Atomic singlet states with zero angular momentum and spherical symmetry still exist in this world. What do we measure for such states if, prior to the measurement, entanglement is lost to noise? One might guess from Eq. (18) that we always see a classical mixture of |01 and |10, each with 50% probability. That guess is wrong. If we add Eqs. (15) and (16) each with weight 1/2, we get a vector of probabilities

{14sin2(a),14[1+cos2(a)],14[1+cos2(a)],14sin2(a)},
(22)

which depends on the angle a, an artifact of the z-oriented basis in which |01 and |10 were defined. This violates the spherically symmetry of the singlet state! This is in part why we chose to avoid density matrices, above. The commonly taught prescription for breaking entanglement by zeroing the off-diagonal components of the density matrix is coordinate dependent. In different coordinate systems, this prescription gives observably different classical mixtures of product states. Decoherence by thermal noise cannot, of course, depend on the coordinate system, and so it does not correspond to zeroing the off-diagonal components.

The more sensible calculation is instead to take an isotropic average of Eq. (15) over all directions

120π{14sin2(a),cos4(12a),sin4(12a),14sin2(a)}sin(a)da={16,13,13,16}.
(23)

We see that, with co-aligned coordinate systems, Alice and Bob get opposite measurements 2/3 of the time and get the same measurements 1/3 of the time. This isotropic, classical mixture of up-down QM product states does not reproduce the result of an entangled singlet, where opposite measurements are obtained with a probability of 1. The measurement probabilities of a singlet state are not reproduced by an isotropic classical mixture of product states.

In hidden-variable theories—proposed alternatives to standard QM—particles are allowed to carry with them any amount of ancillary “hidden” classical information. Eve's goal is to provide enough hidden information so that the predictions of standard QM can be duplicated without the nonlocal effects of “spooky action at a distance.”

Alice and Bob agree that they will each set their apparatuses in one of two pre-determined directions and call them 1 and 2. Eve knows in advance the two directions, but not the choice between them, which may be the same or be different for Alice and Bob on each test.

QM predicts that when Alice and Bob choose the same direction, they always measure opposite spins; if they choose different directions, they measure opposite spins with probability cos2(12θ), where θ is the angle between 1 and 2, cf. Eq. (21).

Eve's most robust classical (i.e., hidden-variable) option is to send with each particle a complete deterministic program for whether it should record as spin-up (↑) or spin-down (↓) for each angle setting that it later encounters. When the particle arrives at Alice's (or Bob's) apparatus, it looks at the angle, consults its program, and records its spin accordingly. Eve can, thus, choose among exactly four programs: ,,,, where the first arrow is the outcome for measurement direction 1 and the second for direction 2. If we adopt the purely notational convention that Bob's program interchanges the meaning of up and down, then Eve must assign the same program number to Alice's and Bob's particle on each test because there would, otherwise, be a same-angle case that did not produce opposite spins.

Suppose that Eve sends particles for the four programs with proportions p0,p1,p2,p3, respectively. These must satisfy the following relations:

p0+p1+p2+p3=1,[normalization]p0+p1=p2+p3,[AandBboth1andequally]p0+p3=p2+p4,[AandBboth2andequally]sin2(12θ)(p0+p3)=cos2(12θ)(p1+p2)[12or21].
(24)

The last equation enforces QM's ratio of probabilities, Eq. (21), when Alice and Bob choose different angles. The solution of Eq. (24) is

(p0,p1,p2,p3)={12cos2(12θ),12sin2(12θ),12sin2(12θ),12cos2(12θ)}.
(25)

Success! If Eve programs her particle pairs with the above classical probabilities, her hidden variables will exactly reproduce QM's predictions for this experiment.

Flushed with success, Eve accepts a more difficult challenge: Bob and Alice pre-determine three directions on the sphere, 1,2,3, with pairwise angular separations θ12,θ13,θ23. Eve now chooses among the eight programs ,,,, ,,, in respective proportions p0,p1,p2,p3,p4,p5,p6,p7, satisfying the following equations:

p0+p1+p2+p3+p4+p5+p6+p7=1,[normalization]p0+p1+p2+p3=p4+p5+p6+p7,[AandBboth1andequally]p0+p1+p4+p5=p2+p3+p6+p7,[AandBboth2andequally]p0+p2+p4+p6=p1+p3+p5+p7,[AandBboth3andequally]sin2(12θ12)(p0+p1+p6+p7)=cos2(12θ12)(p2+p3+p4+p5)[12or21]sin2(12θ13)(p0+p2+p5+p7)=cos2(12θ13)(p1+p3+p4+p6)[13or31]sin2(12θ23)(p0+p3+p4+p7)=cos2(12θ23)(p1+p2+p5+p6)[23or32].
(26)

Eve notes with satisfaction that these are seven equations for eight unknowns and so should have not just one solution, but a one-dimensional family of them. Indeed so, if we take p0, 0p01, as arbitrary, the solutions are

p0=p0p1=12cos2(12θ12)p0p2=12cos2(12θ13)p0p4=12cos2(12θ23)p0p3=12[1cos2(12θ12)cos2(12θ13)]+p0p5=12[1cos2(12θ12)cos2(12θ23)]+p0p6=12[1cos2(12θ13)cos2(12θ23)]+p0p7=12[cos2(12θ12)+cos2(12θ13)+cos2(12θ22)1]p0.
(27)

Eve points out that any of these solutions will deterministically give exactly the QM results. Not so fast, Eve! It is easy for Alice and Bob to agree on three directions that put at least one of the values p1,,p7 outside of the allowed range [0,1] for all p0 in [0,1]. For example, if 1,2, and 3 are spaced 120° apart around a great circle, then p7=(18+p0), which is always negative.3 

In fact, if Alice and Bob use the same three random directions, a straightforward numerical simulation (included as a Python script in the supplementary material of this article) shows that p7 has no allowed (i.e., positive) value 0.166 of the time and that 0p0,,p71 is violated in that or some other way 0.666 of the time. That is to say, Eve can succeed in mimicking QM in three random directions only for 0.333 of such random choices. It is not coincidence that this fraction looks to be 1/3, see the Acknowledgments.

As is well known, John Bell set out to show that QM allowed the loophole of local, hidden-variable alternatives, but showed to his own surprise that no such alternative can reproduce QM's predictions. Stern–Gerlach experiments on two spin-1/2 particles in the (Bell) singlet state are the simplest such example. A contradiction with hidden variables first emerges when we demand consistency with QM probabilities at three or more angles. Bell's more general inequalities, applying to states more general than the singlet state, are known to require four angles for a contradiction; there exist known results on how often classical theory is violated in those cases.5 

See the supplementary material6 for the code in Mathematica for verifying the above equations (or trying variants) and in Python for verifying the simulation mentioned in Sec. VII B.

Thanks to Scott Aaronson for discussions and to two anonymous reviewers for suggestions that improved this paper. The author must make special acknowledgement to his friend of many decades, Freeman Dyson. Some weeks before Dyson's death, the author communicated to him Eq. (27) and mentioned that Monte Carlo simulation yielded a value of 0.333 for random angles (see above). Was this exactly 1/3, the author wondered? Freeman wrote back that it was indeed, that it was a simple calculation, and that he would send the author the details later. Alas, his death on February 28, 2020, intervened. The author eventually did laboriously prove this result ( Appendix B), but will always wonder what was Dyson's simple calculation, perhaps here the last published instance of his mathematical ability, undimmed at age 96.

To show that SU(2) and SO(3) have the same structure, we must show that compositions of rotations in one space (Eq. (4)) give the same result as compositions of the same rotations in the other space (Eq. (6)). Let us check this: The Baker–Campbell–Hausdoff formula (see Wikipedia) tells us that we only have to test whether the infinitesimal generators of the two groups have the same commutators, because then they will exponentiate up to compositions of finite rotations identically. For SU(2), we calculate

SU2(θ,nx,ny,nz)θ|θ=0=12(inzinxnyinx+nyinz),
(A1)

from which we can read off the generators,

Tx=(0i2i20),Ty=(012120),Tz=(i200i2),
(A2)

and verify that their commutators satisfy

TxTyTyTx=Tz,TyTzTzTy=Tx,TzTxTxTz=Ty.
(A3)

For SO(3), we calculate

SO3(θ,nx,ny,nz)θ|θ=0=(0nznynz0nxnynx0),
(A4)

giving the generators

Jx=(000001010),Jy=(001000100),Jz=(010100000),
(A5)

whose commutators can be verified to be

JxJyJyJx=Jz,JyJzJzJy=Jx,JzJxJxJz=Jy.
(A6)

Aha, Eqs. (A3) and (A6) are the same. To see that it is a double covering, note that under θθ+2π, Eqs. (2) and (4) go to minus themselves, a distinct value, while Eq. (6) is unchanged.

The starting point is Eq. (27) in the main text and the conditions 0pi1 for i=0,,7. Define a, b, and c by

acosθ13,bcosθ12,ccosθ23.
(B1)

Using the identity 12cos(12x)2=14(1+cosx), we get these equivalences (algebra shown in the Mathematica Notebook in the supplementary material)

0p1,2,44p01+{a,b,c}(threeconditionsonp0)p1,2,41{a,b,c}3+4p0(automaticsincea,b,c1)0p3,5,6{a+b,b+c,c+a}4p0(threeconditionsonp0)p3,5,614p04+{a+b,b+c,c+a}(dominatedbyfirstline)0p74p01+a+b+c(oneconditiononp0)p71a+b+c3+4p0(automaticsincea,b,c1).
(B2)

Equation (B2), the condition for which there is a classical hidden-variable solution, can be summarized in one line as

max(0,a+b,b+c,c+a)4p01+min(a,b,c,a+b+c).
(B3)

How often is Eq. (B3) violated if we choose three random directions? We saw in the main text that numerical simulation suggests an exact answer 1/3. Can we prove this result analytically? Without loss of generality, let direction 1 be along the z-axis; direction 3 be at an angle λ to it in the (y, z) plane; and direction 2 be at an angle η from direction 1 and an angle φ out of the (y, z) plane, as shown in Fig. 1. The cosines a, b, and c, of the angles between these are readily calculated from the figure as

a=cosλ,b=cosη,c=sinλsinηcosφ+cosλcosη.
(B4)
Fig. 1.

Three arbitrary directions in three-space are shown in a convenient coordinate system. The pairwise cosines of the angles between them are the pairwise dot products of the Cartesian coordinates shown.

Fig. 1.

Three arbitrary directions in three-space are shown in a convenient coordinate system. The pairwise cosines of the angles between them are the pairwise dot products of the Cartesian coordinates shown.

Close modal

Because the area on the two-sphere is proportional to dcosλ or dcosη, random choices for directions 2 and 3 are uniform choices in a and b in the interval (1,1); so we must integrate (something)dadb. The “something” is the fraction of the range 0φ2π, for every a and b, which produces a value c that satisfies Eq. (B3).

Figure 2 shows the domain of the desired integral for two typical values of a=cosλ, as determined numerically. Each side of the four-sided boundary corresponds to a different case of Eq. (B3). Noting that, on both the left and right sides, a and b merely change their sign between the upper and lower branches, we can write the left and right boundaries for all the cases as

cleft=1|ab|,cright=|a+b|1.
(B5)
Fig. 2.

The domain satisfying the inequalities of Eq. (B3) is shown as a function of cosη (=b) and cosφ for two values of cosλ (=a). The boundary is shown in green. Red and blue contour lines show where the inequalities are satisfied (solid) or violated (dashed) by constant amounts.

Fig. 2.

The domain satisfying the inequalities of Eq. (B3) is shown as a function of cosη (=b) and cosφ for two values of cosλ (=a). The boundary is shown in green. Red and blue contour lines show where the inequalities are satisfied (solid) or violated (dashed) by constant amounts.

Close modal

Using the last equation in (B4) to get φleft and φright from cleft and cright, we can write the desired integral as

φleft=arccos1|ab|ab1a21b2,φright=arccos|a+b|1ab1a21b2,(fractionhiddenvariableOK)=141111(φrightφleft)πdadb=13.
(B6)

The prefactor 1/4 reflects division by the area integrated over, namely, 1111dadb=4. I find it astonishing that the integral in Eq. (B6) can be done analytically, both by Freeman Dyson (see the Acknowledgements) and by Mathematica (with some coaching, see the supplementary materials for the code). In fact, Mathematica version 9 gets the integral wrong, while version 12 gets it right. The numerical results leave no doubt that both Dyson and Mathematica 12 are correct and that the answer is 1/3.

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See the supplementary material at https://doi.org/10.1119/10.0001189 for computer files.

Supplementary Material