Are probabilities of a measurement in quantum mechanics absolute?

Consider an experiment in which an electromagnetic wave whose polarization is along an angle $$a$$ with the $$x$$ axis is sent through a polarizer whose polarizing direction is along $$y$$ axis. The probability that light will be transmitted is $$\cos^2 a$$, which is the ratio of transmitted and bombarded intensities.

Is this probability absolute? Because when we always measure classically the transmitted light the intensity ratio is $$\cos^2 a$$. Suppose we send in 1000 photons. The probability to get absorbed or transmitted is $$50:50$$. Does accurately 500 electrons will get transmitted or this number can vary to an extent. But if this varies why we always get intensity ratio equal to $$\cos^2 a$$.

• Note that on this site we use MathJax to format mathematics. – Emilio Pisanty Oct 26 '18 at 13:57

The probability to get absorbed or transmitted is $$50:50$$. Does accurately 500 electrons will get transmitted or this number can vary to an extent.

This reflects a misunderstanding of how probabilities work. Take a fair coin, and toss it 100 times: you will find that you will typically get anywhere between a 40:60 split through to 60:40, and only rarely will you get a completely straight 50:50 split. (The distribution for these splits is known as the binomial distribution, and for 100 throws of a fair coin, the clean 50:50 split happens some 8% of the time.) In the limit when the number of throws is larger and larger, the relative variations away from perfect fairness will diminish (so e.g. if you throw 1000 coins, you will rarely see splits far away from 460:540, and if you throw 10,000 times, you won't stray much beyond 4,900:5,100), but it is only in the limit that you will get that clean probability back.

The same thing holds in quantum mechanics, whether you're dealing with photons or electrons.

When you do this experiment classically, then at large intensities you will indeed just see the $$\cos^2(a)$$ intensity split, but as you lower the intensity, you will start to see shot noise take over as your primary source of uncertainty, and you will see increasingly large variations in the intensity split - even if you use a classical light source.

• But then why we always get intensity ratio equal to cos^2 a when we see the experiment classically . That's bothering me. Or because on classical scale there are billion of photons in a light beam that's why we get this result. But what if we repeat the experiment with 10 photons. Then will we get the same result? – Habeeb Ijaz Oct 26 '18 at 12:47
• @HabeebIjaz the deviation from the predicted outcome when the number of photons is large will become vanishingly small as you increase the number of photons. – ZeroTheHero Oct 26 '18 at 12:51
• @HabeebIjaz As the intensity goes down, you will get shot noise, and you will see increasingly large variations around the expected $\cos^2(a)$ intensity ratio, even if you're using classical light sources. At large intensities, of course, shot noise becomes smaller and smaller until it becomes negligible compared to any other sources of uncertainty in your experiment. – Emilio Pisanty Oct 26 '18 at 12:52
• slight correction to the verbiage: shot noise intensity relative to the total intensity gets smaller as the total intensity gets larger. – hyportnex Oct 26 '18 at 16:49
• @hyportnex Yes, that is an important correction. Shot noise does indeed increase with intensity (as $\sqrt{I}$) but the ratio of shot noise to intensity (which goes as $\sqrt{I}/I=1/\sqrt{I}$) decreases with intensity. – Emilio Pisanty Oct 26 '18 at 17:26

To see that the probability is not "absolute" in the way you defined it, just think about one photon, it either passes or it does not, or you can think of sending and odd number of photons, so it could never be exactly $$50:50$$. The issue you are raising is not exclusive of quantum mechanics, but about probability in general.

If you are talking about a finite set, then the probability is never exact. Just like with coin tosses. The odds of heads and tails are $$50:50$$, assuming a fair coin, but for a finite number of tosses it is not true that you will get exactly half of it.

For the other part, where you ask about the intensity of light detected, that is only "exact" because the number of photons in the electromagnetic passing through the polarizer is enourmous, so in average you get the desired answer. Just like if you toss a coin $$10^{big~~number}$$ of times you will get a distribution that is extremelly close to $$50:50$$, and the bigger the "big number" the closer it is the probability distribution mentioned.

No, 50% probability doesn't mean that exactly 500 of 1000 electrons would pass through.

Imgine it was true. We make an experiment with 1000 electrons. 999 already emitted. 500 of them absorbed, 499 - transmitted. And now we have the last electron, and it must go through so that we get 500:500. Somehow it must know that 1000 electrons participate in our experiment and results for all the previous electrons. No, quantum world is weird, but not that weird.

All the particles in our experiment are independent. Results are also independent as in case you toss a coin. If you toss a coin 1000 times it doesn't mean exactly 500 results would be heads and 500 tails. Most probably there will be a difference. But the ratio (difference)/(total number of experiments) decrease as the number of experiments grows.

If we deal with light the total number of experiments (photons) is very large and the relative difference is very small.

• You didn't resolve my issue. I also know that. The thing that is confusing me is why the math we perform gives us intensity ratio equal to cos ^2 a . On classical scale there are billion of photons so I can understand this outcome of cos^2 a. But what if we repeat the experiment with 10 photons. Then will we get this number . Or this math we performed gave us this number because we are seeing things classically. – Habeeb Ijaz Oct 26 '18 at 12:55

Exactly correct with 10 photons you will see the difference. In fact what you are taking about with photons is known as photon shot noise which is equal to the square root of the signal (for 1 std deviation), so if you try and measure 10 you will typically get 7 or 13 photons. Most of the variation comes from the light source itself, if you excite an atom you can not accurately predict when exactly it will emit, it's QM. A light source is very statistical.