Quantum Mechanics Testability

Question

Statistics by their very nature are not definitive in their predictions for single events. We can say the probability of getting heads on a coin flip is 1/2, yet we can flip a 1000 coins and they could all come up tails. How many coins would it take to disprove the 1/2 prediction? So isn’t QM in the same situation? How can it be accurate if it can’t predict with certainty a given event?

Answer 1

For many independent measurements we are rescued by the law of large numbers. That is, it becomes exceedingly unlikely to measure such a skewed result as all tails from your example. Note that many phenomena that are described quantum mechanically involve a large number of objects (for example the number of nuclei in a sample in radio-active decay or the number of hydrogen atoms in a gas discharge lamp to measure spectra). So the answer to the question is: We can measure the probabilities of random events with sufficient precision if we do enough measurements.

Further, there are predictions that are not affected by the probabilistic nature of quantum mechanics: We know only with certain probabilities to which state an excited hydrogen atom will decay, but there are only certain discrete energy levels it can decay to, so independently of the relative brightness of the spectral lines, the presence and position of the spectral lines of some atom are prediction precisely by quantum mechanics.

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To demonstrate the law of the large numbers specifically for your classical example of a repeated coin flip we can explicitly determine the probabilities of the outcomes. If $n$ is the number of heads that turns up if we throw the coin $N$ times, we get a binomial distribution (with $p = 1/2$): $$P_N(n) = \binom{N}{n} p^n (1-p)^{N-n} = \binom{N}{n} \left(\frac 1 2\right)^N.$$ As you might know the binomial coefficients have a sharp peak around $n = N/2$, so the most likely results are the ones where the number of heads and tails is almost the same.

So how likely is the result, that $n = 0$ for $N=1000$? It is $$P_N(0) = \underbrace{\binom{1000}{0}}_{=\,1} \left(\frac 1 2\right)^{1000} = 2^{-1000}.$$ This number is so incredibly small, we cannot even begin to fathom it. If you had made one flip of coins every second since the beginning of the universe about $13 \cdot 10^9$ years ago, you would only have flipped the coins $$m = 1\,\mathrm{s^{-1}} \cdot 13 \cdot 10^9\,\mathrm{y} \cdot 365.25\,\mathrm{\frac d y} \cdot 24\,\mathrm{\frac h d} \cdot 60 \,\mathrm{\frac{min}{h}} \cdot 60\,\mathrm{\frac{s}{min}} = 4.1 \cdot 10^{17} \approx 2^{58.5}$$ times. That is the probability that in one of those flips the all tails event occurred is still negligible at $mP_N(0) \approx 2^{-941}$.

We can even quantify how probable it is to achieve a result more than a certain amount off the even result. We cite the standard deviation of the binomial probability distribution, that, intuitively, is a measure for the width of the peak of the distribution (again for $p=1/2$): $$ \sigma = \sqrt{Np(1-p)} = \sqrt{N}/2$$ So when $N$ gets large, the deviation will be of the scale of $\sqrt{N}$, in other words the relative offset from an even result will for the "average measurement" not be larger than $1/\sqrt{N}$, for $N=1000$ this is about $3\%$.

Answer 2

There is a category error in your reasoning.

That a theory is probabilistic in the sense that it explicitely uses and produce probabilities, as quantum mechanics does, does not imply that the theory itself is in any way random.

The probabilist predictions of QM are not uncertain: they are very precise probability values. What is uncertain is what you get when using these probabilities to predict a measurement outcome, which of course is what is expected from a probability value. It's kind of the whole point of what a probability is.

So, in short, probabilities are a fine tool; that a theory uses this tool does not make it inaccurate. Statistical mechanics for example works very well.