Saturday, January 07, 2017

Only probabilities, not observables, are spherically symmetric around singlet states

Physicist and spy Klaus Fuchs has expressed the opinion that Born's rule (squared complex amplitudes are interpreted as probabilities or probability densities) could be derived from something deeper. I think that this wishful thinking is demonstrably impossible. Why? We just don't have any method or theorem in mathematics or physics that could allow us not to assume any statement of the sort
the probability is \(f(\theta)\)
and deduce the conclusion of the form
the probability is \(f(\theta)\)
For example, think about an electron whose spin is prepared to be aligned "up" with respect to an axis, and then measure the projection of the spin \(j_z\) with respect to the \(z\)-axis. The angle between the two axes is \(\theta\), the amplitude is \(\cos(\theta/2)\), up to a phase, and the probability to get "up" again is therefore \(\cos^2(\theta/2)\).

How could you possibly derive that from something "deeper"? We don't have anything "deeper" than probabilities that probabilities could be constructed from. At most, we may define probabilities as \(N/N_{\rm total}\), the frequentist formula by which we measure it – which would give us rational numbers if \(N_{\rm total}\) were some "fundamentally real" options. And we may deduce that the probability is \(p=1/N\) if \(N\) options are related by a symmetry. Or we may say that each state on a "shell of the phase space" – quantum mechanically, a subspace of the Hilbert space – has the probability \(p=1/N\) to be realized during a random evolution as envisioned by the ergodic theorem.

None of those Ansätze can produce the statement "the probability is \(\cos^2(\theta/2)\)" and there are no other candidates of the "methods" in mathematics and physics. So I find it rather clear that unless someone finds a totally new mathematics that finds completely new definitions or laws for probabilities, and e.g. calculates probabilities from Bessel's function of the number of Jesus' disciples (which seems like a quantity of a different type than probability, and that's the main reason why this example should sound ludicrous), it is clearly impossible to derive statements like "the probability of 'up' is \(\cos^2(\theta/2)\)" from something that says nothing about the values of probabilities.

The people saying "Born's rule smells like it's derived" never respond to the argument above – which I consider a proof of a sort. I think that if one carefully looks at the task, he will agree that the only way to deduce that the probability is a continuous function of some variables is to make at least some assumptions that the probability is a continuous function of some variables. Quantum mechanics including Born's rule is making statements about Nature of the form the probability is a continuous function of some variables. But if you have nothing like that as a fundamental law of physics, you just can't possibly derive any conclusion like that.

Quantum mechanics and its statistical character can't be "emergent". The statements about the values of probabilities have to appear somewhere in our derivations for the first time. So the only way how a physical theory may make predictions of probabilities at all is that it contains an axiom with the formula telling us what the probabilities are, namely (in the case of quantum mechanics) Born's rule. Such a rule can't be born out of nothing or out of something unrelated to probabilities, it's that simple.

Klaus Fuchs also said another thing that implicitly hides a misconception. He says that when a spherically symmetric composite particle decays, the decay products randomly choose some directions. This symmetry breaking is ugly and the ugliness suggests that something dirty is going on, so there must be some missing physics.

Except that the correct experimental analysis of the experiments shows that the random direction is being chosen, indeed, and a correct theoretical analysis unambiguously shows that the result is correctly predicted by quantum mechanics and couldn't be predicted by any fundamentally different theory. Moreover, the claim "the singlet composite state is spherically symmetric" is being implicitly misinterpreted – to say something that is demonstrably false – by the "realists". (Just to be sure, Fuchs hasn't explicitly made this mistake but others, more hardcore "realists", have.)

What do I mean?

To be specific, let's consider the initial state \(\ket\psi\) with a composite particle, a bound state of the electron and the positron known as positronium. The relative spins of the two leptons aren't determined. As a result, due to the simple rules for the addition of two spins \(1/2+1/2\), the composite particle has either \(S=0\) or \(S=1\) which (mostly) decay to two photons and three photons, respectively, and they are called para-positronium and ortho-positronium, respectively. In Greek, "ortho-" is straight or erect and is therefore used to denote "the same direction" (but not "erect by the same sex", which wouldn't be straight), while "para-" means against and is used all over science to denote the opposite directions – in this case the directions of the spin. (The exception is Paraguay which comes from the Guarani language, not Greek, and means "born from water" or "water-born", "Para-guay".)

The \(S=0\) para-positronium state has the standard "Bell's" singlet combination of the spins\[

\ket\psi = \frac{ \ket\uparrow \ket\downarrow - \ket\downarrow \ket\uparrow }{\sqrt{2}}

\] Needless to say, this "Bell's state" was invented and heavily used some 40 years before Bell wrote it down – when Pauli and others started to play with the spins in the mid 1920s – and the "Bell's terminology" is utterly idiotic both historically and physically.

OK, the \(S=0\) condition means that the wave function is spherically symmetric. If you had a problem with the components' being odd under rotations by 360 degrees (like all fermions), you could think about bound states of two bosons and/or their orbital angular momentum etc.

People who don't quite understand quantum mechanics can never get rid of the idea that the wave function describes "what the system really is", some objectively real i.e. observation-independent or observer-independent i.e. classical degrees of freedom. Because everything seems to be predictable from the wave function, they incorrectly think, everything that can be measured is spherically symmetric as well.

However, this reasoning is completely wrong.

All that Penny gave to Sheldon was the napkin. Analogously, it is only the wave function (of the para-positronium) that is invariant under the \(SU(2)\sim SO(3)\) rotations of the three-dimensional space. (You have to appreciate The Big Bang Theory for its ability to explain the word "only".) And just like the napkin doesn't contain Leonard Nimoy, the wave function doesn't contain any observables whatsoever. The wave function only contains the probability amplitudes – the square roots of the probability distributions, along with some quantum phases (which affect the probabilities of outcomes for other, non-commuting observables) – and probability amplitudes are completely different things than observables. In quantum mechanics, observables – everything whose value may be shown on a measurement apparatus once you only do the measurement once – must be represented by Hermitian linear operators, not by state vectors.

The probabilities of various outcomes in directions related by the rotational symmetry are the same. But the observables in different directions themselves are not symmetric.

Let's be very explicit and slow to see what is true and what is not true.

Take the state \(\ket\psi\) for the para-positronium. Apply a rotation by the angle \(\alpha\) around some axis \(\hat n\). You will see that\[

R_{\alpha, \hat n} \ket \psi = \ket \psi.

\] The wave function i.e. state vector doesn't change under the rotation. You may write\[

R = \exp(i\alpha \vec J \cdot \hat n)

\] if you wish. You may also send \(\alpha\to 0\). For a small angle \(\alpha\), it's useful to Taylor-expand the exponential in \(R\) above. The term \(1\) will cancel against the left hand side and right hand side of \(R\ket\psi = \ket\psi\) while the following term proportional to \(\alpha\) (the others are negligible) will give us\[

\vec J \ket \psi = 0.

\] That's why the state \(\ket\psi\) carries no angular momentum. The angular momentum is the generator of the \(SU(2)\sim SO(3)\) rotations so the vanishing of the angular momentum is the same thing as the spherical symmetry – of the wave function.

However, does it mean that the measurements done in two directions away from the positronium will be the same? Not at all. If we measure the direction of the outgoing photons, for example, we will see that in some (mutually opposite) directions, there is one photon, and in other directions, there's none.

Why does the measurement of the direction of photons "break" the symmetry? The reason is always the same uncertainty principle of quantum mechanics. If you measure the angle \(\theta\) of the photons away from the \(z\)-axis, the angle \(\theta\) is an observable – a Hermitian linear operator on the Hilbert space – and it just doesn't commute with \(J\) i.e. \(\theta J \neq J \theta\) for completely analogous reasons why \(xp-px\neq 0\). Because they don't commute, they can't have certain sharp values at the same moment. And because \(J=0\) has a sharp and certain value, \(\theta\) cannot have one.

But let's study the state before it decays to the photons and use our theory of nearly everything (TONE, Lisa Randall's acronym), a quantum field theory, for that. Assuming that the quantum fields are in the positronium state, are observables spherically symmetric?

Let's pick a very particular example, the energy density \(\rho\) at some distance from the center-of-mass of the positronium. And let's pick two such points in different directions. We want to look at the difference\[

\rho(r_B, 0,0) - \rho (0,r_B,0)

\] where \(r_B\) is Bohr's radius, a length constant comparable to the radius of the positronium (average distance between the electron and the positron). OK, in a quantum field theory with the electromagnetic and electron/positron Dirac field, e.g. in QED, there is an operator such as the operator above, right?

The "realists" tend to imagine that the wave function is spherically symmetric (invariant under rotations), and because everything we can measure is a function of the wave function, everything we can measure is spherically symmetric, too. Except that this opinion is completely wrong. Nothing that we can measure (except for mathematical constants whose values/outcomes are determined regardless of the state) is a function of the wave function. The assumption that the realists are making is not just slightly wrong, it is totally wrong.

In particular, \(\rho(r_B,0,0)\) is a field operator, some function of the operators \(\vec E,\vec B\), and others, that exist in QED. And you shouldn't doubt that because the field operators \(\vec E, \vec B\) etc. are completely independent at two different points of space, the difference\[

\rho(r_B, 0,0) - \rho (0,r_B,0)

\] isn't vanishing as an operator equation. OK, a patiently obnoxious realist could argue, maybe this difference doesn't vanish as an operator equation but it vanishes assuming our spherically symmetric state \(\ket\psi\). So he will say that we should have\[

[\rho(r_B, 0,0) - \rho (0,r_B,0)] \ket \psi = 0.\quad (???)

\] When the difference between the energy densities at two points – two points related to each other by a rotation – is acting on the positronium state, it has to vanish due to the spherical symmetry of the positronium. But does it vanish?

If it vanished, it would mean that \(\ket\psi\) is an eigenstate of the \(\rho_P-\rho_{P'}\) difference above corresponding to the eigenvalue \(0\), so if we measure the difference, we are 100% certain to get the result \(0\). But will we get zero?

Not at all. The energy density is fluctuating in the vacuum of QED. The operators \(\rho(r_B,0,0)\) and \(\rho(0,r_B,0)\) are commuting with each other (spacelike separation, local theory) but they express two uncertain, oscillating energy densities at two different points. (The simplest master example for the oscillations is the claim that you can't say that \(x=0\) for the ground state of the harmonic oscillator.) So the difference \(\rho-\rho'\) may be positive or negative – it's random.

If you wanted a true but similar statement, it would be\[

\bra \psi [\rho(r_B, 0,0) - \rho (0,r_B,0)] \ket \psi = 0.

\] You have to add the bra-vector \(\bra \psi\) on the left side of the product as well. And this whole matrix element vanishes. In other words, the expectation value of the difference between the energy densities at two different points related by a rotation vanishes. This identity is easily proven. You may write the second density as \(\rho' = R \rho R^{-1}\) and notice that the action of \(R^{-1}\) and \(R\) on the bra- and ket-vectors is just like the action of \(1\), so you may erase those \(R\)'s. And without \(R\)'s, the two \(\rho\)'s cancel. But that only works when it's sandwiched in between the state \(\ket\psi\) on both sides, bra and ket.

The vanishing expectation value is a much weaker statement than the previous displayed formula. The individual differences of energy densities are almost certainly nonzero, only their statistical average – when you repeat the pair-measurement on the "pure positronium" many times and average the result – converges to zero.


[\rho(r_B, 0,0) - \rho (0,r_B,0)] \ket \psi \neq 0

\] and we could describe this nonzero value of the difference acting on \(\ket\psi\) in some detail and quantitatively. For example, the expectation value of the squared difference, \([\rho(r_B, 0,0) - \rho (0,r_B,0)]^2\), is also nonzero and calculable. When you average the squared differences between the energy densities at two points related by a rotation, you will get a positive value that is greater than a certain bound – that is calculable in a similar way as in the usual proofs of the Heisenberg uncertainty inequalities.


It's important to realize that \(J=0\) only means that the wave function, i.e. a collection of probabilities or probability amplitudes, is spherically symmetric. The wave function is not observable and it is not an observable. The observables must be expressed by linear operators acting on the Hilbert space and their measurements show that they (e.g. the energy densities) are not spherically symmetric. The densities at two points related by a rotation are not equal; \(\ket\psi\) is not an eigenstate of this difference operator.

The measurements that are sensitive to directions in any way are guaranteed to break the symmetry of the wave function because they're measurements of operators that don't commute with \(\vec J\). This breaking of the symmetry by the direction-sensitive subsequent measurement isn't the evidence for some trouble in quantum mechanics. On the contrary, it's a trivial consequence or confirmation of the uncertainty principle, the main and basically only principle that distinguishes quantum mechanics from its classical limit (or counterpart).

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