More Quantum

Yesterday I saw with disappointment a new paper on the arXiv by Hossenfelder and Palmer, Rethinking Superdeterminism. There they argue that physics took a wrong turn when we immediately dismissed superdeterminism; instead it is a solution to the conundrum of nonlocality and the measurement problem.

No. It’s not. It’s a completely sterile idea. I’ll show why, by fleshing out the calculations of the smoking and cancer example they quote in the paper, and then examining the case of the Bell test.

Let’s suppose you do the appropriate randomized trial, and measure the conditional probabilities1They are close to what you’d find in reality.
\[ p(\text{cancer}|\text{smoke}) = 0.15\quad\text{and}\quad p(\text{cancer}|\neg\text{smoke}) = 0.01,\]a pretty damning result. A tobacco company objects to the conclusion, saying that the genome of the subjects was correlated with whether you forced them to smoke2Maudlin mercifully forces only rats to smoke, but I’m not so careful about the ethics of a gedankenexperiment., such that you put more people predisposed to have cancer in the smoking group.

It works like this: the law of total probability says that
\[ p(a|x) = \sum_\lambda p(\lambda|x)p(a|x,\lambda),\] where in our case $a \in \{\text{cancer},\neg\text{cancer}\}$, $x \in \{\text{smoke},\neg\text{smoke}\}$, and $\lambda \in \{\text{predisposed},\neg\text{predisposed}\}$ is the hidden variable, in this case the genome determining whether the person will have cancer anyway. The tobacco company says that your results are explained by the conspiracy $p(\text{predisposed}|\text{smoke}) = 0.15$ and $p(\text{predisposed}|\neg\text{smoke}) = 0$, from which we can calculate the actual cancer rates to be
\begin{gather*}
p(\text{cancer}|\text{smoke},\neg\text{predisposed}) = 0 \\
p(\text{cancer}|\neg\text{smoke},\neg\text{predisposed}) = 0.01,
\end{gather*}so the same data indicates that smoking prevents cancer! If you assume, though, that $p(\text{predisposed}|\text{smoke}) = p(\text{predisposed}|\neg\text{smoke})$, then the absurd conclusion is impossible.

With this example I want to illustrate two points: first, that assuming $p(\lambda|x) \neq p(\lambda)$ is just a generic excuse to dismiss any experimental result that you find inconvenient, be it that smoking causes cancer or that Bell inequalities are violated. Second, that without assuming $p(\lambda|x) = p(\lambda)$ 3Or, more generally, knowing what $p(\lambda|x)$ is. you can’t conclude anything from your data.

In their paper, Hossenfelder and Palmer dismiss this example as merely classical reasoning that is not applicable to quantum mechanics. It’s not. One can always use the law of total probability to introduce a hidden variable to explain away any correlation, whether it was observed in classical or quantum contexts. Moreover, they claim that while $p(\lambda|x) = p(\lambda)$ is plausible in classical contexts, it shouldn’t be assumed in quantum contexts. This is laughable. I find it perfectly conceivable that tobacco companies would engage in conspiracies to fake results related to smoking and cancer, but to think that Nature would engage in a conspiracy to fake the results of Bell tests? Come on.

They also propose an experiment to test their superdeterministic idea. It is nonsense, as any experiment about correlations is without the assumption that $p(\lambda|x) = p(\lambda)$. Of course, they are aware of this, and they assume that $p(\lambda|x) = p(\lambda)$ would hold for their experiment, just not for Bell tests. Superdeterminism for thee, not for me. They say that when $x$ is a measurement setting, changing it will necessarily cause a large change in the state $\lambda$, but if you don’t change the setting, the state $\lambda$ will not change much. Well, but what is a measurement setting? That’s human category, not a fundamental one. I can just as well say that the time the experiment is made is the setting, and therefore repetitions of the experiment done at different times will probe different states $\lambda$, and again you can’t conclude anything about it.

Funnily, they say that “…one should make measurements on states prepared as identically as possible with devices as small and cool as possible in time-increments as small as possible.” Well, doesn’t this sound like a very common sort of experiment? Shouldn’t we have observed deviations from the Born rule a long time ago then?

Let’s turn to how superdeterministic models dismiss violations of Bell inequalities. They respect determinism and no action at a distance, but violate no conspiracy, as I define here. The probabilities can then be decomposed as
\[ p(ab|xy) = \sum_\lambda p(\lambda|xy)p(a|x,\lambda)p(b|y,\lambda),\]and the dependence of the distribution of $\lambda$ on the settings $x,y$ is used to violate the Bell bound. Unfortunately Hossenfelder and Palmer4Or any other proponent of superdeterminism, as far as I know. do not specify $p(\lambda|xy)$, so I have to make something up. It is trivial to reproduce the quantum correlations if we let $\lambda$ be a two-bit vector, $\lambda \in \{(0,0),(0,1),(1,0),(1,1)\}$, and postulate that it is distributed as
\[p((a,b)|xy) = p^Q(ab|xy),\] where $p^Q(ab|xy)$ is the correlation predicted by quantum mechanics for the specific experiment, and the functions $p(a|x,\lambda)$ and $p(b|y,\lambda)$ are given by
\[p(a|x,(a’,b’)) = \delta_{a,a’}\quad\text{and}\quad p(b|y,(a’,b’)) = \delta_{b,b’}.\] For example, if $p^Q(ab|xy)$ is the correlation maximally violating the CHSH inequality, we would need $\lambda$ to be distributed as
\[ p((a,b)|xy) = \frac14\left(1+\frac1{\sqrt2}\right)\delta_{a\oplus b,xy}+\frac14\left(1-\frac1{\sqrt2}\right)\delta_{a\oplus b,\neg(xy)}.\]The question is, why? In the quantum mechanical case, this is explained by the quantum state being used, the dynamical laws, the observable being measured, and the Born rule. In the superdeterministic theory, what? I have never seen this distribution be even mentioned, let alone justified.

More importantly, why should this distribution be such that the superdeterministic correlations reproduce the quantum ones? For example, why couldn’t $\lambda$ be distributed like
\[ p((a,b)|xy) = \frac12\delta_{a\oplus b,xy},\] violating the Tsirelson bound?5I asked both Palmer and Hossenfelder this question. Palmer couldn’t provide an explanation, apart from his general assertion that his theory should reproduce quantum mechanics, and despite my insistence I couldn’t get a straight answer from Hossenfelder, but she seemed to believe that violating the Tsirelson bound was possible. Even worse, why should the superdeterministic distributions respect even no-signalling? What stops $\lambda$ being distributed like
\[ p((a,b)|xy) = \delta_{a,y}\delta_{b,x}?\]

In their paper, Hossenfelder and Palmer define a superdeterministic theory as a local, deterministic, reductionist theory that reproduces quantum mechanics approximately. I’m certain that such a theory will never exist. Its dynamical equations would need to correlate 97,347,490 human choices with the states of atoms and photons in 12 laboratories around the planet to reproduce the results of the BIG Bell test. Its dynamical equations would need to correlate the frequency of photons emitted by other stars in the Milky Way with the states of photons emitted by a laser in Vienna to reproduce the results of the Cosmic Bell test. Its dynamical equations would need to correlate the bits of a file of the movie “Monty Python and the Holy Grail” with the state of photons emitted by a laser in Boulder to reproduce the results of the NIST loophole-free Bell test. It cannot be done.

Last week a friend of mine, Felipe, left written the following equation in the blackboard of the coffee room: \[ \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 4,\]asking for a solution with positive integers.

It was a harmless sort of nerd sniping (as opposed to the harmful one), that he saw as a meme online. It looks innocent, but it is a rather difficult problem6Difficult for those not familiar with Diophantine equations and elliptic curves, that is, pretty much everyone.. The idea was to post it to family WhatsApp groups, as revenge for the retarded math problems with fruits as variables that come up regularly. Finding the idea hilarious, I went on and posted it to my family’s WhatsApp group.

But… how does one solve it? I had inadvertently sniped myself. Trying a little bit, I managed to show that there is no integer solution for $a=b$, and therefore one couldn’t reduce the problem from a third-degree polynomial to a second degree one, so there was no easy way out. I also realised that any integer multiple of a solution is a valid solution, so one can fix $c=1$, look for a solution for rational $a,b$, and multiply this solution by the least common denominator to solve the original problem. Fixing then $c=1$ and rewriting $a,b$ as $x,y$ for clarity, we end up with the polynomial equation
\[ x^3-3x^2-3x^2y-3x + y^3-3y^2-3y^2x-3y-5xy + 1 = 0,\]which looks pretty much hopeless. To get some intuition I plotted the curve, obtaining this:

Well great. Now I was stuck. Off to Google then. Which immediately gave me this answer, a thorough explanation of where the problem came from and how to solve it. End of story? Not really. The answer depended on two magical steps2Transforming the elliptic curve into the Weierstraß form and finding its generators. that were neither easy nor explained. No no, I want to actually find the solution, not just reproduce it using stuff I don’t understand.

As the mathematician helpfully explains there, the core idea is the centuries-old chord and tangent technique: if one draws a line through two rational points on the curve, or on a tangent point, this line will intersect the curve again on a rational point. So if we have a single rational solution to start with, we can just iterate away and produce new rational points.

Well, this I can understand and do myself, off to work then! I did everything on the programming language that’s all the rage with the kids nowadays, Julia3Crucially, Julia supports both rational numbers and arbitrarily large integers, which are necessary to solve this problem. Floating point numbers are not good enough, and the relevant integers quickly become too large for the standard types.. You can download the code here. I wrote up this solution having in mind people like me, who don’t really know anything, but have a generally favourable attitude towards math.

The first step is to brute force search for a rational solution. That’s easy enough, but gives us mostly repetitions of the points $(-1,-1)$, $(1,-1)$, and $(-1,1)$. They make some denominators of the original equation evaluate to zero, and by looking at the graph it is obvious that the point $(-1,-1)$ doesn’t work4The other two also don’t work, for less obvious reasons., so I excluded them and selected a nice-looking solution as the starting point:
\[P_0 = \left(-\frac5{11},\frac9{11}\right)\]Now, I needed to calculate the slope of the curve at this point. This lies firmly on the straightforward-but-tedious territory5It would be more convenient to do the change of variable $(x,y)\mapsto (a+b,a-b)$, but let’s stay with the original variables just to make a point., so I won’t bore you with the details. The result is that
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-3x^2+6x+6xy+3y^2+5y+3}{-3x^2-5x-6xy+3y^2-6y-3},\]and with the slope we can easily find the tangent line. Now finding the intersection of this line with the curve is even more straightforward and tedious6It’s actually very easy if the elliptic curve is in the Weierstraß form, but I’m explicitly avoiding doing the transformation., so I’m not going to even type out the answer (for the details check the Wikipedia page linked above). Anyway, now we have the intersection point
\[I = \left(\frac{5165}{9499},-\frac{8784}{9499}\right),\]but we have a problem: if we draw the line between $P_0$ and $I$ we won’t get a new point, as this line intersects the curve only at $P_0$ (twice) and $I$. We could just apply the same procedure that we did to $P_0$: calculate the tangent at $I$ and with that get a new intersection point. This works, but the integers in the numerator and denominator of the numbers we get by iterating this procedure grow hideously fast7The number of digits grows exponentially, and the computer dies before returning a solution with only positive rationals.

There is an easy way out, though: note that the equation is symmetric under exchange of $x$ and $y$, so we can simply take the new point as
\[P_1 = \left(-\frac{8784}{9499},\frac{5165}{9499}\right),\]and the line through $P_0$ and $P_1$ does intersect the curve at a new point. Taking $P_2$ to be again its flipped version
\[P_2 = \left(-\frac{396650011}{137430135},-\frac{934668779}{137430135}\right),\]we can go on.

A new question does arise. Do we take $P_3$ from the line through $P_1$ and $P_2$, or from the line through $P_0$ and $P_2$? It turns out that both methods work, but the number of digits grows much faster with the former method than with the latter8exponentially versus quadratically., so it is unwise to use the former method in general. We iterate away, and find the solution with $P_{12}$, with the resulting integers having more than 150 digits.

I have just been to Perimeter Institute, by generous invitation of Thomas Galley. I gave a talk there about my recent-ish paper, Probability in two deterministic universes. Since I have already blogged about it here, I’m not writing about it again, but rather what I discussed with Thomas about his derivations of the Born rule.

I was interested in his most recent derivation, that besides structural assumptions about measurements and probabilities, needs two substantial assumptions: no-signalling and the possibility of state estimation, or state estimation for brevity. No-signalling is well-motivated and well-understood, but I was curious about state estimation. What does it mean? How does a theory that violates it looks like?

The precise definition is that state estimation is true if there is a finite set of measurement outcomes9By focussing on measurement outcomes instead of measurements we get rid of pathological cases like a measurement with an infinite number of outcomes, or equivalently a real-number-valued measurement. whose probabilities completely determine the quantum state. Or conversely, if state estimation fails, then for any finite set of measurement outcomes there are two different quantum states that give the same probabilities for all these outcomes. This is clearly not obeyed by quantum mechanics in the case of infinite-dimensional systems – you need to know the probability at each point in space to completely determine the wavefunction, which is an infinite set of outcomes2More precisely, you need to know the overlap of the wavefunction with each member of a basis for $L^2$, so you only need a countable number of measurement outcomes, not uncountable. – so the authors require it only for finite-dimensional systems.

How bad is it to violate it for finite-dimensional systems, then? What can you learn about a quantum state with a small number of measurement outcomes? Do you get a good approximation, or would you have little idea about what the quantum state is? It seems that the former is the case. To illustrate that, we came up with a rather artificial theory where the measurements allow you to deterministically read off bits from some representation of the quantum state; for the case of a qubit $\ket{\psi}=\cos\theta\ket{0}+e^{i\varphi}\sin\theta\ket{1}$ a measurement would tell you the $n$th bit of $\theta$ or $\varphi$. It is clear that this theory violates state estimation: for any finite set of measurements there will be a largest $n$ that they can reach, and therefore any pair of quantum states that differ on bits higher than $n$ will be indistinguishable for this set of measurements. It is also clear that this violation is nothing to worry about: with only $2n$ measurements we can get a $n$-bit approximation for any qubit, which is much better than what can be done in reality! In reality we need about $2^n$ measurements to estimate the probabilities, and therefore the amplitudes, with such an accuracy.

This already tells us that state estimation is too strong; it needs at least to be qualified somehow in order to exclude the deterministic theory above. What does it mean in probabilistic theories, though? An often considered toy theory is one where the structure of quantum mechanics is kept as it is, but the exponent in the Born rule is changed from $2$ to some $n$. More precisely, let the probability of obtaining outcome $i$ when measuring the state $\psi$ in the orthogonal basis $\{\ket{i}\}$ be \[ p(i|\psi) = \frac{|\langle i|\psi\rangle|^n}{\sum_{i’}|\langle {i’}|\psi\rangle|^n}. \]An interesting feature of this theory is that a finite set of measurement outcomes can distinguish all pure states (in fact the same measurements that distinguishes them in quantum theory), so state estimation can only fail here for mixed states.

A nice example is the pair of ensembles
\[\omega_A = \{(p,\ket{0}),(1-p,\ket{1})\}\] and \[\omega_B = \{(1/2,p^\frac1n\ket{0}+(1-p)^\frac1n\ket{1}),(1/2,p^\frac1n\ket{0}-(1-p)^\frac1n\ket{1})\}.\] In quantum mechanics ($n=2$) they are equivalent, both being represented by the density matrix
\[ \rho = \begin{pmatrix} p & 0 \\ 0 & 1-p \end{pmatrix}. \] If $n\neq 2$, though, they are not equivalent anymore, even though they give the same probabilities for any measurements in the X, Y, and Z basis3The probabilities of $0,+$, and $+y$ are $p, 1/2$, and $1/2$.. To distinguish them we just need to measure the ensembles in the basis \[\{p^\frac1n\ket{0}+(1-p)^\frac1n\ket{1},(1-p)^\frac1n\ket{0}-p^\frac1n\ket{1}\}.\] The probability of obtaining the first outcome for ensemble $\omega_A$ is $p^2 +(1-p)^2$, and for ensemble $\omega_B$ it is some complicated expression that depends on $n$.

Now this is by no means a proof4I’m writing a blog post, not a paper!, but it makes me suspect that it will be rather easy to distinguish any two ensembles that are not equivalent, by making a measurement that contains one of the pure states that was mixed in to make the ensemble. Then if we divide the Bloch sphere in a number of regions, assigning a measurement to cover each such region, we do that with a good enough approximation. Unlike the deterministic theory explored above, in this toy theory it is clearly more laborious to do state estimation than in quantum mechanics, but is still firmly within the realm of possibility.

What now, then? If the possibility of state estimation is not a good assumption from which to derive the Born rule, is there a derivation in this operational framework that follows from better assumptions? It turns out that Galley himself has such a derivation, based only on similar structural assumptions together with no-signalling and purification, with no need for state estimation. But rather ironically, here the roles flip: while I find purification an excellent axiom to use, Galley is not a fan.

Let me elaborate. Purification is the assumption that every mixed state (like the ensembles above) is obtained by ignoring part of a pure state. It implies then that there are no “external” probabilities in the theory; if you want to flip a coin in order to mix two pure states, you better model that coin inside the theory, and as a pure state. Now Galley doesn’t find purification so nice: for once, because classical theories fail purification, and also because it feels like postulating that your theory is universal, which is a big step to take, in particular when the theory in question is quantum mechanics.

Well, I find that classical theories failing purification is just one more example in a huge pile of examples of how classical theories are wrong. In this particular case they are wrong by being essentially deterministic, and only allowing for probabilities when they are put in by hand. About postulating the universality of the theory, indeed that is a big assumption, but so what? I don’t think good assumptions need to be self-evidently true, I just think they should be well-motivated and physically meaningful.

Addendum: A natural question to ask is whether both no-signalling and purification are necessary in such a derivation. It turns out the answer is yes: the toy theory where the exponent in the Born rule is $n$ respects purification, when extended in the obvious way for composite systems, but violates no-signalling, and Galley’s rule respects no-signalling but violates purification.

This post hast little to do with physics, let alone quantum mechanics; I’m just writing it because I saw reports in the media about a study done by three German professors that had the incredible conclusion that electric vehicles emit more CO$_2$ than diesel vehicles. I’ll not focus in debunking this “study”, as it has already been thoroughly debunked in the newspaper articles that I’ve linked, but rather I’ll explain how the calculation is done, and how one would go about manipulating it to get the result you want. They are simple mistakes, that would have been caught by even cursory peer-review, so maybe the lesson here is that non-peer-reviewed “studies” like this one are better ignored altogether.

So, we are interested in the total amount of CO$_2$ that a vehicle emits over its lifetime. It is the emissions caused by producing it in the first place, $P$, plus the amount of CO$_2$ it emits per km $\eta$ times the distance $L$ it travels over its lifetime: $P + \eta L$. To get a number that is easier to relate with, we divide everything by $L$ and get the effective emissions per km $\frac{P}{L}+\eta$. We want to compare a diesel vehicle with an electric vehicle, so we want to know whether
\[\frac{P_E}{L}+\eta_E\quad\text{or}\quad\frac{P_D}{L}+\eta_D\] is bigger.

Assume that $P_E > P_D$, because of the extra emissions needed to produced the battery of the electric vehicle, and that $\eta_E < \eta_D$, as it is much more efficient to extract energy from oil in a big power plant than in an internal combustion engine 5Both assumptions can be false depending on the specific factories and power grid used, but they are usually true, and more germane to the argument, they are true in Germany..

Now, what could you do to make the electric vehicles look bad? Well, since their production causes more emissions, you want to emphasise that in the equation, and since they emit less CO$_2$ when running, you want to downplay that. How? We have three variables, so we have three ways of manipulating the numbers: we can multiply $P_E$ and $P_D$ by some large number $n_P$ (e.g. by assuming that the factories producing the cars are powered purely by oil shale), we can divide $\eta_E$ and $\eta_D$ by some large number $n_\eta$ (e.g. by assuming the cars are ran always at maximal efficiency), and we can divide $L$ by some large number $n_L$ (assuming that car is scrapped after a few kilometres).

What is the effect of doing that? If the real numbers say that electric vehicles are better, that is, that
\[\frac{P_E}{L}+\eta_E < \frac{P_D}{L}+\eta_D,\]which is equivalent to
\[ \frac{P_E-P_D}{L(\eta_D-\eta_E)} < 1,\]then the manipulations of the previous paragraph imply in multiplying the left hand side of this inequality by $n_Pn_Ln_\eta$; if we want to flip it we just need to make $n_P,n_L,$ and $n_\eta$ large enough so that \[ n_Pn_Ln_\eta\frac{P_E-P_D}{L(\eta_D-\eta_E)} > 1.\]

And what the authors of the study did? All of the above. Most interestingly, they used the NEDC driving cycle to calculate $\eta_D$ and $\eta_E$, a ridiculously efficient driving cycle that has been discarded in favour of the less unrealistic WLTC. They did this because WLTC numbers hadn’t yet been released for the Tesla Model 3, the electric car they used for the comparison. They claim that this is not a problem, because NEDC favours city driving, where electric cars excel, so if anything this incorrect assumption would be tilting the scales in favour of electric cars. As we have seen, though, this is not the case: pretending that the cars are more efficient than they are tilts the scales in favour of the diesels.

Another mistake the authors made is to assume that cars only last 10 years or 150.000 km before going to the junkyard, which is about half of the actual number. Again this tilts the scales if favour of the diesels, as the production of electric cars causes more emissions. The reason they made this mistake is because they assumed that the battery of an electric car would only last this much, which is false for two reasons: first because a Tesla battery retains more than 90% of its capacity after 250,000 km, hardly junkyard material, and second because batteries that have in fact degraded too much to be useful in a car, say retaining only 70% of their capacity, do not go to the junkyard, but instead are reused for applications where the energy/weight ratio doesn’t matter, like grid storage.

The third mistake the authors made is exaggerating the emissions caused by production, using a discredited study that claimed that producing the lithium-ion battery causes 145 kg CO$_2$/kWh. The peer-reviewed number I could find is 97 kg CO$_2$/kWh for production in China. Even that seems too high, though, as Tesla’s batteries are produced in the Gigafactory 1 in Nevada, which has a cleaner energy mix, and should eventually be powered fully by rooftop solar. One thing that might look like a mistake but isn’t is that the authors don’t consider the emissions caused by producing the components that are common to both electrics and diesels2The authors also don’t consider the emissions caused by producing components used only by diesels: the engine, gearbox, exhaust system, etc.: wheels, body, seats, etc. Of course, ignoring that means that the number you get is not effective emissions per km, but it doesn’t change which car is the best, as that depends only on the difference $P_E-P_D$.

With the theory done, let’s get to the numbers. The authors use $P_E = 10,875,000$ gCO$_2$, $\eta_E = 83$ gCO$_2$/km, $P_D = 0$ gCO$_2$, $\eta_D = 144$ gCO$_2$/km, and $L=150,000$ km, which results in the effective emissions
\[ E = 155\text{ gCO}_2/\text{km}\quad\text{and}\quad D = 144\text{ gCO}_2/\text{km},\]their absurd conclusion that electric vehicles emit more CO$_2$. Now what I find amazing is that this conclusion requires all three mistakes working together; correct any of the three and it flips.

First we correct $\eta_E$ and $\eta_D$ using the WLTC numbers (which are still too optimistic, but are the best I’ve got), which are already available for both the Model 3 (16 kWh/100 km) and the Mercedes C 220 d (5.1 l/100 km)3This should count as the fourth mistake, actually. The Model 3 is a mighty sports car, with 340 kW of power, whereas the C220d is not, with only 143 kW. If you compare the Model 3 to a car of similar power the conclusion also flips, even if you let all their other mistakes stay., resulting in $\eta_E = 88$ gCO$_2$/km and $\eta_D = 163$ gCO$_2$/km, and the effective emissions
\[ E = 160\text{ gCO}_2/\text{km}\quad\text{and}\quad D = 163\text{ gCO}_2/\text{km}.\] Next we keep the wrong $\eta_E$ and $\eta_D$ and just correct $L$, setting it to $250,000$ km, resulting in the effective emissions
\[ E = 126\text{ gCO}_2/\text{km}\quad\text{and}\quad D = 144\text{ gCO}_2/\text{km}.\] Next we keep the wrong $\eta_E, \eta_D$, and $L$, correcting only the emissions caused by the production of the battery. Putting 97 kg CO$_2$/kWh results in $P_E = 7,275,000$ gCO$_2$ and effective emissions
\[ E = 131\text{ gCO}_2/\text{km}\quad\text{and}\quad D = 144\text{ gCO}_2/\text{km}.\] To finalize, let’s calculate the true numbers, correcting all three mistakes at once and also taking into account the emissions caused by producing the parts common in both vehicles. I couldn’t find a good number for that, just some estimates that put it around 20 tons of CO$_2$. Using this results in $P_E = 27,275,000$ gCO$_2$, $\eta_E = 88$ gCO$_2$/km, $P_D = 20,000,000$ gCO$_2$, $\eta_D = 163$ gCO$_2$/km, and $L=250,000$ km, and effective emissions \[ E = 197\text{ gCO}_2/\text{km}\quad\text{and}\quad D = 243\text{ gCO}_2/\text{km}.\]

It doesn’t look very impressive, though. Only 19% less emissions? Is all the trouble worth it? The point is that none of the emissions of electric vehicles are necessary: as the grid cleans up both their production and operation will be CO$_2$-free. Diesels, though, will always burn diesel, so at best they will cause only the tailpipe emissions4Assuming that drilling, transporting, and refining oil is CO$_2$-free, which is obviously unrealistic., and the ultimate numbers will be \[ E = 0\text{ gCO}_2/\text{km}\quad\text{and}\quad D = 135\text{ gCO}_2/\text{km}.\] There is no need to wait, though: electric vehicles are better for the environment than diesels. Not in the future, not depending on magical technologies, not in Norway, but right here, and right now. And this is only about CO$_2$ emissions; electric vehicles also have the undeniable benefit of not poisoning the atmosphere in densely populated cities.

Philosophers really like problems. The more disturbing and confusing the better. If there’s one criticism you cannot levy at them is that they are not willing to tackle the difficult issues. I have argued with philosophers endlessly about Bell’s theorem, the Trolley problem, Newcomb’s paradox, Searle’s Chinese room, the sleeping beauty problem, etc. Which made me very surprised when I asked a couple of philosophers about objective probability, and found them strangely coy about it. The argument went along the lines of “objective probability is frequentism, frequentism is nonsense, subjective probability makes perfect sense, there’s only subjective probability”.

Which is really bizarre argument. Yes, frequentism is nonsense, and yes, subjective probability makes perfect sense. But that’s all that is true about it. No, objective probability is not the same thing as frequentism, and no, subjective probability is not the only probability that exists. Come on, that’s denying the premise! The question is interesting precisely because we strongly believe that objective probability exists; either because of quantum mechanics, or more directly from the observation of radioactive decay. Does anybody seriously believe that whether some atom decays or not depends on the opinion of an agent? There even existed natural nuclear reactors, where chain reactions occurred much before any agent existed to wonder about them.

In any case, it seems that philosophers won’t do anything about it. What can we say about objective probability, though? It is easy to come up with some desiderata: it should to be objective, to start with. The probability of some radioactive atom decaying should just be a property of the atom, not a property of some agent betting about it. Agents and bets are still important, though, as it should make sense to bet according to the objective probabilities. In other words, Lewis’ Principal Principle should hold: rational agents should set their subjective probabilities to be equal to the objective probabilities, if the latter are known5This is arguably a property that subjective, not objective probabilities should obey. But I think it is fair to include it in the list of desiderata, as a notion of objective probability that didn’t allow for such a connection would be rather undesirable.. Last but not least, objective probabilities should be connected to relative frequencies via the law of large numbers, that is, we need that
\[ \text{Pr}(|f_N-p|\ge\varepsilon) \le 2e^{-2N\varepsilon^2}, \] or, in words, the (multi-trial) probability that the frequency deviates more than $\varepsilon$ from the (single-trial) probability after $N$ trials goes down exponentially with $\varepsilon$ and $N$ 2I’m using on purpose the more complicated version of the law of large numbers to prevent people from confusing it with the nonsensical statement $p=\lim_{N\to\infty} f_N$..

I think it is also easy to come up with a definition of objective probability that fulfills these desiderata, if we model objectively random processes as deterministic branching processes. Let’s say we are interested the decay of an atom. Instead of saying that it either decays or not, we say that the world branches in several new worlds, in some of which the atom decays, and some of which it does not. Moreover, we say that we can somehow count the worlds, that is, that we can attribute a measure $\mu(E)$ to the set of worlds where event $E$ happens and a measure $\mu(\neg E)$ to the set of worlds where event $\neg E$ happens. Then we say that the objective probability of $E$ is
\[p(E) = \frac{\mu(E)}{\mu(E)+\mu(\neg E)}.\] Now, before you shut off saying that this is nonsense, because the Many-Worlds interpretation is false, so we shouldn’t consider branching, let me introduce a toy theory where this deterministic branching is literally true by fiat. In this way we can separate the question of whether the Many-Worlds interpretation is true from the question of whether deterministic branching explains objective probability.

This toy theory was introduced by Adrian Kent to argue that probability makes no sense in the Many-Worlds interpretation. Well, I think it is a great illustration of how probability actually makes perfect sense. It goes like this: the universe is a deterministic computer simulation3Ran by some advanced beings that not only have cracked the strong AI problem, but also have a truly astounding amount of computational power. Whatever, it’s not mean to be realistic. where some agents live. In this universe there is a wall with two lamps, and below each a display that shows a non-negative integer. This wall also has a “play” button, that when pressed makes either of the lamps light up.

The agents there can’t really predict which lamp will light up, but they have learned two things about how the wall works. The first is that if the number below a lamp is zero, that lamp never lights up. The second is that if the numbers are set to $n_L$ and $n_R$, respectively, and they press “play” multiple times, the fraction of times where the left lamp lights up is often close to $n_L/(n_L+n_R)$.

What is going on, of course, is that when “play” is pressed the whole computer simulation is deleted and $n_L+n_R$ new ones are initiated, $n_L$ with the left lamp lit, and $n_R$ with the right lamp lit. My proposal is to define the objective probability of some event as the proportion of simulations where this event happens, as this quantity fulfills all our desiderata for objective probability.

This clearly fulfills the “objectivity” desideratum, as a proportion of simulations is a property of the world, not some agent’s opinion. It also respects the “law of large numbers” desideratum. To see that, fist notice that for a single trial the proportion of simulations where the left lamp lights up is
\[p(L) = \frac{n_L}{n_L+n_R}.\] Now the number of simulations where the left lamp lights up $k$ times out of $N$ trials is
\[ {N \choose k}n_L^kn_R^{N-k},\] so if we divide by total number of simulations $(n_L+n_R)^N$, we see that the proportion of simulations where the left lamp lit $k$ times out of $N$ is given by \[\text{Pr}(N,k) = {N \choose k}p(L)^k(1-p(L))^{N-k}.\]Since this is formally identical to the binomial distribution, it allows us to prove a theorem formally identical to the law of large numbers:
\[ \text{Pr}(|k/N-p(L)|\ge\varepsilon) \le 2e^{-2N\varepsilon^2}, \]which says that the (multi-trial) proportion of simulations where the frequency deviates more than $\varepsilon$ from the (single-trial) proportion of simulations after $N$ trials goes down exponentially with $\varepsilon$ and $N$.

Last but not least, to see that if fulfills the “Principal Principle” desideratum, we need to use the decision-theoretic definition of subjective probability: the subjective probability $s(L)$ of an event $L$ is the highest price a rational agent should pay to play a game where they receive $1$€ if event $L$ happens and nothing otherwise. In the $n_L$ simulations where the left lamp lit the agent ends up with $(1-s(L))$ euros, and in the $n_R$ simulations where the right lamp lit the agent ends up with $-s(L)$ euros. If the agent cares equally about all its future selves, they should accept to pay $s(L)$ as long as \[(1-s(L))n_L-s(L)n_R \ge 0,\]which translates to \[s(L) \le \frac{n_L}{n_L+n_R},\] so indeed the agent should bet according to the objective probability if they know $n_L$ and $n_R$4I know, decision theory is much more subtle than this; I’m just doing the quick and dirty version of the argument here, it is done properly in the paper..

And this is it. Since it fulfills all our desiderata, I claim that deterministic branching does explain objective probability. Furthermore, it is the only coherent explanation I know of. It is hard to argue that nobody will ever come up with a single-world notion of objective probability that makes sense, but at least in one point such a notion will always be unsatisfactory: why would something be in principle impossible to predict? Current answers are limited to saying that quantum mechanics say so, or that if we could predict the result of a measurement we would run into trouble with Bell’s theorem. But that’s not really an explanation, it’s just saying that there is no alternative. Deterministic branching theories do offer an explanation, though: you cannot predict which outcome will happen because all will.

Now the interesting question is whether this argument applies to the actual Many-Worlds interpretation, and we can get a coherent definition of objective probability there. The short answer is that it’s complicated. The long answer is the paper I wrote about it =)

Philip Ball has just published an excellent article in the Quanta magazine about two recent attempts at understanding the Born rule: one by Masanes, Galley, and Müller, where they derive the Born rule from operational assumptions5Despite their provocative title, they don’t derive it from the other postulates. This is obviously impossible., and another by Cabello, that derives the set of quantum correlations from assumptions about ideal measurements.

I’m happy with how the article turned out (no bullshit, conveys complex concepts in understandable language, quotes me ;), but there is a point about it that I’d like to nitpick: Ball writes that it was not “immediately obvious” whether the probabilities should be given by $\psi$ or $\psi^2$. Well, it might not have been immediately obvious to Born, but this is just because he was not familiar with Schrödinger’s theory2He could hardly be; he was mainly working on his and Heisenberg’s matrix mechanics, in Göttingen, whereas Schrödinger was in Zürich, and had just published his famous equation.. Schrödinger, on the other hand, was very familiar with his own theory, and in the very paper where he introduced the Schrödinger equation he discussed at length the meaning of the quantity $|\psi|^2$. He got it wrong, but my point here is that he knew that $|\psi|^2$ was the right quantity to look at. It was obvious to him because the Schrödinger evolution is unitary, and absolute values squared behave well under unitary evolution.

Born’s contribution was, therefore, not mathematical, but conceptual. What he introduced was not the $|\psi|^2$ formula, but the idea that this is a probability. And the difficulty we have with the Born rule until today is conceptual, not mathematical. Nobody doubts that the probability must be given by $|\psi|^2$, but people are still puzzled by these high-level, ill-defined concepts of probability and measurement in an otherwise reductionist theory. And I think one cannot hope to understand the Born rule without understanding what probability is.

Which is why I don’t think the papers of Masanes et al. and Cabello can explain the Born rule. They refuse to tackle the conceptual difficulties, and focus on the mathematical ones. What they can explain is why quantum theory immediately goes down in flames if we replace the Born rule with anything else. I don’t want to minimize this result: it is nontrivial, and solves something that was bothering me for a long time. I’ve always wanted to find a minimally reasonable alternative to the Born rule for my research, and now I know that there isn’t one.

This is what I like, by the way, in the works of Saunders, Deutsch, Wallace, Vaidman, Carroll, and Sebens. They tackle the conceptual difficulties with probability and measurement head on3I don’t like them only because they use the Many-Worlds framework. Żurek’s work is also kind of Many-Worldsy, but he does not engage with the probability problem.. I’m not satisfied with their answers, for several reasons, but at least they are asking the right questions.

When the Frauchiger-Renner argument first came out I posted a favourable review, where I corrected the mistake in the presentation without even remarking on it. But since the authors decided to insist on the mistake, I feel the need to point it out.

The well-known argument by Frauchiger and Renner about the consistency of quantum mechanics has finally been published (in Nature Communications). With publication came a substantial change to the conclusion of the paper: while the old version claimed that “no single-world interpretation can be logically consistent”, the new version claims that “quantum theory cannot be extrapolated to complex systems” or, to use the title, that “quantum theory cannot consistently describe the use of itself”.

This is clearly bollocks. We need to find out, though, where exactly has the argument gone wrong. Several discussions popped up on the internet to do so, for example in Scott Aaronson’s blog, but to my surprise nobody pointed out the obvious mistake: the predictions that Frauchiger and Renner claim to follow from quantum mechanics do not actually follow from quantum mechanics. In fact, they are outright wrong.

For example, take the first of the predictions that appear on Table 3 of the paper. $\bar{\text{F}}$ measures $r=\text{tails}$ and claims: “I am certain that W will observe $w = \text{fail}$ at time $n$:$31$”. By assumption, though, $\bar{\text{F}}$ is in an isolated laboratory and their measurement is described by a unitary transformation. This implies that the state of lab L at time $n$:$30$ will be given either by
\[ \frac{3}{\sqrt{10}}\ket{\text{fail}}_\text{L} + \frac{1}{\sqrt{10}}\ket{\text{ok}}_\text{L}\quad\text{or}\quad\frac{1}{\sqrt{2}}\ket{\text{fail}}_\text{L} – \frac{1}{\sqrt{2}}\ket{\text{ok}}_\text{L},\]depending on the result of $\bar{\text{W}}$’s measurement. Therefore, it is not certain that W will observe $w = \text{fail}$; this will happen with probability $9/10$ or $1/2$, respectively.

To obtain the prediction the authors write in Table 3, one would need to assume that $\bar{\text{F}}$’s measurement caused a collapse of the state of their laboratory – contrary to the assumption of unitarity. In this case, the state at time $n$:$30$ would in fact be given by
\[ \ket{\text{fail}}_\text{L},\]independently of the result of $\bar{\text{W}}$’s measurement, and W would indeed observe $w = \text{fail}$ with certainty. But then W would never observe $w = \text{ok}$, and the paradox desired by the authors would never emerge.

To make this point more clear, I will describe how precisely the same problem arises in the original Wigner’s friend gedankenexperiment, so that people who are not familiar with Frauchiger and Renner’s argument can follow it. It goes like this:

Wigner is outside a perfectly isolated laboratory, and inside it there is a friend who is going to make a measurement on a qubit. Their initial state is
\[\ket{\text{Wigner}}\ket{\text{friend}}\frac{\ket{0}+\ket{1}}{\sqrt2}.\]If we assume that the measurement of the friend is a unitary transformation, after the measurement their state becomes
\[\ket{\text{Wigner}}\frac{\ket{\text{friend}_0}\ket{0} + \ket{\text{friend}_1}\ket{1}}{\sqrt2}.\]Now the friend is asked to predict what Wigner will observe if he makes a measurement on the qubit. Frauchiger and Renner claim that, using quantum mechanics, the friend can predict that “If I observed 0, then Wigner will observe 0 will certainty”4It is hard to believe that somebody that knows quantum mechanics and is aware of the experimental setup would make this incorrect prediction, but this is beside the point..

Wait, what? The quantum prediction is clearly that Wigner will observe 0 with probability 1/2. The claimed prediction only follows if we assume that the friend’s measurement caused a collapse.

And both assumptions are fine, actually. If there is no collapse, the prediction of 0 with probability 1/2 is correct and leads to no inconsistency, and if there is a collapse the prediction of 0 with probability 1 is correct and leads to no inconsistency. We only get an inconsistency if we insist that from the point of view of the friend there is a collapse, from the point of view of Wigner there is no collapse, and somehow both points of view are correct.

Update: After a long discussion with Renato, I think I understand his point of view. He thinks that this assumption of “collapse and no collapse” is just part of quantum mechanics, so it doesn’t need to be stated separately. Well, I think this is one hell of an unstated assumption, and in any case hardly part of the consensus about quantum mechanics. More technically, I think Frauchiger and Renner’s formalization of quantum mechanics — called [Q] — does not imply “collapse and no collapse”, it is too vague for that, so there is really a missing assumption in the argument.

I’ve just seen that Open Science, a new journal by the prestigious Royal Society, published the article Quantum correlations are weaved by the spinors of the Euclidean primitives, by Joy Christian. The article, as numerous others by the same author, claims that Bell’s theorem is wrong, and that one can violate Bell inequalities using a local hidden-variables model.

This is of course nonsense. Bell’s theorem is not only a rather simple piece of mathematics, with a few-lines proof that can be understood by high-school students, but also the foundation of an entire field of research — quantum information theory. It has been studied, verified, and improved upon by thousands of scientists around the world.

The form of Bell’s theorem that is relevant for the article at hand is that for all probability distributions $\rho(\lambda)$ and response functions $A(a,\lambda)$ and $B(b,\lambda)$ with range $\{-1,+1\}$ we have that
\begin{multline*}
-2 \le \sum_\lambda \rho(\lambda) \Big[A(a_1,\lambda)B(b_1,\lambda)+A(a_1,\lambda)B(b_2,\lambda) \\ +A(a_2,\lambda)B(b_1,\lambda)-A(a_2,\lambda)B(b_2,\lambda)\Big] \le 2
\end{multline*}
The author’s proposed counterexample? It’s described in equations (3.48) and (3.49): A binary random variable $\lambda$ that can take values $-1$ or $+1$, with $\rho(-1)=\rho(+1)=1/2$, and response functions $A(a,\pm1)=\pm1$ and $B(b,\pm1)=\mp1$. That’s it. Just perfectly anti-correlated results, that do not even depend on the local settings $a$ and $b$. The value of the Bell expression above is simply $-2$.

Now how could Open Science let such trivial nonsense pass? They do provide the “Review History” of the article, so we can see what happened: there were two referees that pointed out that the manuscript was wrong, one that was unsure, and two that issued a blanket approval without engaging with the contents. And the editor decided to accept it anyway.

What now? Open Science can recover a bit of its reputation by withdrawing this article, as Annals of Physics did with a previous version, but I’m never submitting an article to them.

EDIT: Now, more than two years later, the two referees that pointed out that the manuscript was nonsense came out to tell their story. You should read it there, but the tl;dr is that the editor was unsure about the paper, asked for a fifth referee report, got a very negative one, sent it to the author, who replied, and then just published the paper, without even going back to the referee! After the paper was published, they were repeatedly asked by the referees and other experts to withdraw the paper, and they refused. Frankly, that’s disgusting. Indeed, “Royal Society Open Science is not a serious journal”. But perhaps that’s being too soft on them.

I was recently leafing through the great book Quantum Computation and Quantum Information, and noticed that the version of Bell’s theorem it presents is not any of those I wrote about in my three posts about Bell’s theorem, but rather the one I like the least, the counterfactual definiteness version. Annoyed, I checked another great book, Quantum Theory: Concepts and Methods, and saw that it also uses this god-forsaken version2Unsurprisingly, since its author, Asher Peres, is the one to blame for this version of Bell’s theorem. Nevertheless, I decided that the world has bigger problems to deal with and set it aside. Until I dropped by the Quantum Information lecture given by my boss, David Gross, and saw that he also prefers this undignified version. That’s it! The rise of fascism can wait. I need to set the record straight on counterfactual definiteness.

Before I start ranting about what I find so objectionable about it, I’ll present the proof of this version of Bell’s theorem the best I can. So, what is counterfactual definiteness? It is the assumption that not only the measurement you did in fact do has a definite answer, but also the measurement you did not do has a definite answer. If feels a lot like determinism, but it is not really the same thing, as the assumption is silent about how the result of the counterfactual measurement is determined, it just says that it is. To be more clear, let’s take a look at the data that comes from a real Bell test, the Delft experiment:2The raw data looks like
2015-06-26-18:15:16.784161,1,3,5422918,1,5708337,1,1,0,10383,10372,10702,0,7587,0,0,132
2015-06-26-18:16:10.508174,1,3,5422262,1,5683582,1,0,0,10383,10372,11151,0,0,0,0,1808442
2015-06-26-18:21:20.490202,1,3,5427542,1,5675881,1,0,1,10381,10372,11169,0,0,0,0,68
2015-06-26-18:21:39.045852,1,3,5424585,1,5687937,1,1,0,10379,10367,0,0,0,0,209,628463
but who wants to see that?

The first column indicates the rounds of the experiment, the $x$ and $y$ columns indicate the settings of Alice and Bob, and the $a$ and $b$ columns the results of their measurements. If one assumes counterfactual definiteness, then definite results must also exist for the measurements that were not made, for example in the first round there must exist results corresponding to the setting $x=1$ for Alice and $y=1$ for Bob. This data would then be just part of some more complete data table, for example this:

In this table the column $a_0$ has the results of Alice’s measurements when her setting is $x=0$, and so on. The real data points, corresponding to the Delft experiment, are in black, and I filled in red the hypothetical results for the measurements that were not made.

What is the problem with assuming counterfactual definiteness, then? A complete table certainly exists. But it makes it possible to do something that wasn’t before: we can evaluate the entire CHSH game in every single round, instead of having to choose a single pair of settings. As a quick reminder, to win the CHSH game Alice and Bob must give the same answers when their settings are $(0,0)$, $(0,1)$, or $(1,0)$, and give different answers when their setting is $(1,1)$. In other words, they must have $a_0=b_0$, $a_0=b_1$, $a_1=b_0$, and $a_1 \neq b_1$. But if you try to satisfy all these equations simultaneously, you get that $a_0=b_0=a_1 \neq b_1 = a_0$, a contradiction. At most, you can satisfy 3 out of the 4 equations3If you think a little, it is easy to see that it is only possible to satisfy 3 out of 4 or 1 out of 4; it is impossible to satisfy 2 or none of the equations. Then since in every row the score in the CHSH game is at most $3/4$, if we sample randomly from each row a pair of $a_x,b_y$ we have that
\[ \frac14(p(a_0=b_0) + p(a_0=b_1) + p(a_1=b_0) + p(a_1\neq b_1)) \le \frac34,\]
which is the CHSH inequality.

But if you select the actual Delft data from each row, the score will be $0.9$. Contradiction? Well, no, because you didn’t sample randomly, but just chose $1$ out of $4^{10}$ possibilities, which would happen with probability $1/4^{10} \approx 10^{-6}$ if you actually did it randomly. One can indeed violate the CHSH inequality by luck, it is just astronomically unlikely.

Proof presented, so now ranting: what is wrong with this version of the theorem? It is just so lame! It doesn’t even explicitly deal with the issue of locality, which is fundamental in all other versions of the theorem4Some sort of locality is implicitly assumed by building such a table and letting the outcome of Alice depend only on $x$, and not on $y$.! The conclusion that one takes from it, according to Asher Peres himself, is that “Unperformed experiments have no results”. To which the man in the street could reply “Well, duh, of course unperformed experiments have no results, why are you wasting my time with this triviality?”. It leaves the reader with the impression that they only need to give up the notion that unperformed experiments have results, and they are from then on safe from Bell’s theorem. But this is not true at all! The other proofs of Bell’s theorem still hold, so you still need to give up either determinism or no action at a distance, if you consider the simple version, or unconditionally give up local causality, if you consider the nonlocal version, or choose between generalised local causality and living in a single world, if you consider the Many-Worlds version.

In today’s arXiv appeared a nice paper by Časlav Brukner, my former PhD supervisor. Its central claim is that one cannot have observer-independent measurement results in quantum mechanics, which I bet you disagree with. But if you think a bit more about it, you shouldn’t: which interpretation, after all, allows one to have observer-independent measurement results? Old-style Copenhagen definitely allows that5If any definite claim at all can be made about old-style Copenhagen, this is it., but it’s hard to find a living defender of old-style Copenhagen. Bohmian mechanics also allows that, and actually has living defenders, but it is a minority interpretation, and always appealed more to philosophers than to physicists. Collapse models also allow that, but they are on the verge on being experimentally falsified, and were never popular to start with.

What about the mainstream interpretations, then? In Časlav’s neo-Copenhagen interpretation the measurement results are observer-dependent (otherwise this would be a rather schizophrenic paper). In QBism they are explicitly subjective2One should be careful, though, to not confuse observer-dependent with subjective: the speed of a star depends on the reference frame of the observer, but is perfectly objective. What is subjective is whether you like that star. But if something is subjective, then a fortiori it is observer-dependent., as almost everything else. In Many-Worlds there isn’t a single observer after a measurement, but several of them, each with their own measurement result.

How can this be? Časlav’s argument is as simple as it gets in quantum foundations: Bell’s theorem. In its simple version, Bell’s theorem dashes the old hope that quantum mechanics could be made deterministic: if the result of a spin measurement were pre-determined, then you wouldn’t be able to win the CHSH game with probability higher than $3/4$, unless some hidden action-at-a-distance was going on. But let’s suppose you did the measurement. Surely now the weirdness is over, right? You left the quantum realm, where everything is fuzzy and complicated, and entered the classical realm, where everything is solid and clear. So solid and clear that if somebody else does a measurement on you, their measurement result will be pre-determined, right?

Well, if it were pre-determined, than people doing measurements on people doing measurements wouldn’t be able to win the CHSH game with probability higher than $3/4$, unless some hidden action-at-a-distance was going on. But if quantum mechanics holds at every scale, then again one can win it with probability $\frac{2+\sqrt{2}}{4}$.

This highlights the fundamental confusion in Frauchiger and Renner’s argument, where they consider which outcome some observer thinks that another observer will experience, but are not careful to distinguish the different copies of an observer that will experience different outcomes. I’ve reformulated their argument to make this point explicit here, and it works fine, but undermines their conclusion that in single-world but not many-world theories observers will make contradictory assertions about which outcomes other observers will experience. Well, yes, but the point is that this contradiction is resolved in many-world theories by allowing different copies of an observer to experience different outcomes, and this recourse is not available in single-world theories.