How the notion of weak measurement resolves Hardy's paradox?


Dear Eelvex, first of all, the term "paradox" is misleading, much like for all "paradoxes" in physics. Quantum mechanics, when applied properly, doesn't lead to any paradox.

The right, quantum mechanical solution to the Hardy's 1992 thought experiment concludes that there is a nonzero probability that the electron and positron will be detected in two particular detectors $d^+, d^-$. As I will explain in the second part of the answer, where the whole "paradox" is clarified, the real paradox has nothing to do with any specific dynamics of antiparticles or annihilation. It's all about quantum information and the usual differences between classical physics and quantum mechanics - the same conceptual issues that arise in Bell's inequalities or the GHZ state.

Where is the paradox?

The only paradox arises if you try to imagine that the particle and the antiparticle objectively possessed some "well-defined properties" - especially the location - in the past. If you do so, the situations in which the particles end up detected in $d^+,d^-$ correspond to "histories" in which the particles had to meet and annihilate - but they didn't.

Hardy's method is just a particularly striking way to show why the realistic reasoning is inapplicable in the world of quantum phenomena. The paradox only arises because of "retrodictions" using a classical intuition. If you only apply the proper probabilistic quantum rules and never imagine that the system was represented by any "hidden variables" at each moment, and you just evolve the wave function according to the right equations and calculate the probability at the end, you will make the right prediction.

The situation has been measured experimentally. And indeed, the prediction of quantum mechanics was confirmed: the electron and the positron may be detected at "classically contradictory" pair of detectors in a certain fraction of cases. Weak measurements have been used to detect the result.

Some people have continued to assume that there could be a way how to "retrodict" what was happening according to the final results, and they found the weak measurement - producing weak values - to be useful. See e.g.

Yakir Aharonov et al.: Revisiting Hardy's Paradox: Counterfactual Statements, Real Measurements, Entanglement and Weak Values

As far as I can see, they just deduce one more "classically paradoxical" outcome for each original Hardy-like outcome.

Many wrong things have been said about Hardy's paradox during the two decades. For example, some people such as Henry Stapp have argued that even in ordinary quantum mechanics, one needs acausality to explain the observation. This assertion was showed incorrect by David Mermin:

N. David Mermin: Nonlocal character of quantum theory?

The nonlocality and acausality doesn't exist in these experiments - or any other phenomena in the Universe - and it doesn't help to explain them. Instead, the lack of "realism" (i.e. the non-existence of any sharp properties of physical systems prior to the measurement) is the right explanation why these things work as they work. By focusing on the "weak measurements", you seem to be on a wrong track.

A comprehensible explanation of Hardy's setup also appears on page 35 of this review:

Frank Laloe: Do we really understand quantum mechanics?

In particular, I like the fact that this review makes it very transparent that the thought experiment has nothing to do with any particular dynamics of particles, antiparticles, and their annihilation. It's all about the basic logic and correlations that work differently in quantum physics than in any classical or "realistic" model of physics. The electron-positron setup is just a particular realization of the "paradox".

My summary and solution of the "paradox"

Let me reformat Laloe's presentation. Imagine we have a system of two particles which are associated with letters "a" and "b", respectively. On each particle, we may perform two kinds of measurements, either A or A' for the article "a" and either B or B' for the particle "b". Imagine that the un-primed measurements are spin measurements with respect to a vertical axis and the primed measurements are with respect to a different, general axis.

Each of these measurements may produce the result $-1$ or $+1$. There are three qualitatively different types of measurements: either both "a" and "b" are tested on their "unprimed" polarization, or one of them is measured by the "primed" gadget and the other by the "unprimed" gadget, or both of them have their "primed" quantities measured.

Now, as we can easily show, it is possible to design the apparatus so that the following three conditions are guaranteed. Each condition constrains the results with respect to one of the three kinds of a measurement - either the "two unprimed measurements" or "mixed" or "two primed measurements". The conditions are:

  1. After the doubly unprimed measurements, $A=+1$ and $B=+1$ sometimes occurs
  2. After the mixed measurements, $A=+1$ and $B'=+1$ never occurs, and $A'=+1$ and $B=+1$ never occurs, either
  3. After the doubly primed measurements, $A'=-1$ and $B'=1$ never occurs.

Can you prepare the particles "a" and "b" so that regardless of the two experimenters' choice of "primed" vs "unprimed", all conditions above will be universally satisfied? The answer of any realistic theory, whether it's local or not, is "no".

The first condition guarantees that it may happen that the particles are sometimes prepared to be measured as $A=B=1$. However, for these cases, the second condition guarantees that $B'=-1$ - because $B'=+1$ is impossible because we already have $A=1$. And similarly, $A'=-1$ is guaranteed - because $A'=+1$ is impossible because we already have $B=+1$.

We just derived that in the situations in which the particles are prepared to be observed with $A=B=+1$, we have $A'=B'=-1$ which violates the third condition. It sounds like impenetrable logic. Whenever you assume that the particles already objectively have some properties prepared for any experiments that the experimenters may choose, it follows that the three rules above are incompatible.

However, in quantum mechanics, they're totally compatible, and the compatibility has been demonstrated experimentally, too. With respect to the primed bases, the right two-particle state that satisfies all the three conditions may be written as $$|\Psi\rangle = -\cos\theta \left( |A'=+1,B'=-1\rangle + |A'=-1,B'=+1\rangle \right) + \sin\theta |A'=+1,B'=+1\rangle$$ where the angle between the "primed" and "unprimed" axes was chosen to be $2\delta$ and we consider the spin equal to $1/2$. Note that one of the four basis vectors, the vector with $A'=B'=-1$, is omitted, so we automatically satisfy the third condition. By a proper transformation to the rotated bases, we may also check that the second condition is satisfied. That's why the coefficients were chosen to be cosines and sines.

However, the first condition is also satisfied because the coefficient of the $A=B=+1$ can be seen to be nonzero - after the simple rotation of both bases. So quantum mechanics implies that two particles may be arranged in a state such that all three conditions - for all three types of paired measurements that the two experimenters may decide to perform - will be satisfied. This rules out all "realistic" interpretations of reality, whether they're local or not.

What is different about the quantum logic that it doesn't allow us to derive the classical conclusion? Well, the second condition says that the state $|\Psi\rangle$ is annihilated by two projection operators, $$ P(A=1) P(B'=1)|\Psi\rangle = 0\,\mbox{ and }\,P(A'=1) P(B=1)|\Psi\rangle = 0 $$ It's also annihilated by the following operator, as dictated by the third condition, $$ P(A'=-1) P(B'=-1) |\Psi\rangle = 0. $$ Classically, we could derive that the state must also be annihilated by $$ P(A=1) P(B=1) |\Psi\rangle = 0 $$ which is a projection operator relevant in the first condition. The classical logic was showed above. However, quantum mechanically, the state doesn't have to be annihilated by the most recent projection operator because the projection operators $P(A)$ and $P(A')$ don't commute with one another, and similarly for the particle "b". This characteristic fact of quantum mechanics - that observables don't commute with each other, not even the projection operators describing Yes/No properties of systems - guarantees that we can't derive the third displayed equation above from the previous two lines.

The very Yes/No properties of a particle don't commute with each other, so we can't ever imagine that a particle is ready to react to different kinds of measurements at the same moment. In some sense, it's just another form of the uncertainty principle, optimized for projection operators and binary properties in this case.

See a blog version of this answer:


  • $\begingroup$ Is the 3rd condition supposed to be B' = -1? The third paragraph that follows the conditions and the wave function seem to indicate this. $\endgroup$ – Abhinav Apr 29 '14 at 8:42
  • $\begingroup$ @LubosMotl I read your blog where you claim BM can't reproduce P = 1/16. You say it can't because no local realist theory can. What you fail to mention/understand is that BM isn't a local realist theory, it's a nonlocal realist theory, and it has no trouble whatsoever producing P = 1/16. There's really not much new in the Hardy setup from the original Bell setup which is also clearly satisfied by BM. $\endgroup$ – user7348 May 29 '16 at 22:12
  • $\begingroup$ Why don't you either show us the calculation that produces P=1/16 or just [keep quiet]? Failing to be a local theory isn't a key towards right predictions - on the contrary, it is a lethal defect. [Comment was modified by a moderator to conform with the SE 'be nice' policy.] $\endgroup$ – Luboš Motl May 30 '16 at 4:09
  • $\begingroup$ @LubošMotl The calculation is in Bohm (1952) that I linked to you. He showed in that paper that all the same predictions between QM and dBB must agree. So, if QM predicts P = 1/16 so too does BM. $\endgroup$ – user7348 May 30 '16 at 4:33
  • $\begingroup$ @LubosMotl To be clearer, in dBB the pilot wave determines the trajectories, and in Bohm (1952) he shows that the position of the pilot wave is given by the Born rule applied to the wave itself. This guarantees that QM and dBB will always have identical experimental outcomes. This makes the wave multi-local, which you may refute for that reason. Fine, but Bohm (1952) also shows that there will never be any experimental violation of relativity. Fundamentally though, Bohm (1952) expressed the need for a preferred frame, but he showed full observational agreement with relativity. Best Wishes. $\endgroup$ – user7348 May 31 '16 at 1:09

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