I could prove: Copenhagen QM is NOT causal. Is my argument acceptable?

-1

0

A quantum system consists of some states. It is in all of the states but beCAUSE of the act of the measurement, the wavefunction collapses into one state. I am really confused because I can repeat the same measurement and get different results at each time! The probability to get a specific outcome is considered to be a function of the wavefunction, but this is not what I mean. This is OK, but when it comes to predicting the outcome of each measurement, it becomes acausal in my opinion, but I am not sure. I have tried to bring my confusion from my mind to stack exchange and this is the result: If performing exactly similar observations yields to different outcomes, so it means that the cause of each outcome (=observation) is not enough to determine the outcome. So, what else contributes in determining the outcome to be what it is after each measurement? There is no other cause (that we don't know) because if there was any, it would be that hidden variable that realists like Einstein believed to exist and we are assuming the orthodox interpretation here.Therefore, the only other thing that can contribute in determining the outcome is the system itself. For instance, the electron decides to which state to go! This means that the rest of this causality must be filled by the system itself, so the system must change itself. This is impossible because of the same reason that I cannot pull myself up the ground by pulling my hair upward! (or in mechanics, the third law of Newton explains why something can't change itself by exerting a net force on itself!).

MY QUESTION: So, here is my question: I assumed causality and orthodox viewpoint and arrived at either 1- rejecting orthodox by accepting realism or 2- accepting that systems can cause themselves to be changed. None is logical as the first is in contradiction to my assumption and second is in contrast with the third law of Newton. So, the two assumptions I made are incompatible, that is, they cannot coexist. So, the Copenhagen interpretation and causality cannot be both true, and so the orthodox viewpoint is acausal. Now, is this argument true? If no, why and if yes, which one do you accept to be true: the causality or the Copenhagen interpretation of QM?

seyed sepehr mousavi

Posted 2019-12-19T13:27:27.817

Reputation: 135

I see you've asked an apparently related question at https://physics.stackexchange.com/questions/520423/ But it seems to me that question might be better suited for this forum, whereas this question might be better suited for that one.

– None – 2019-12-19T16:21:24.560

Yeah that's right. There are many physics questions so most are not answered properly, including mine. So, I asked my question also here; maybe I would find my answer! – seyed sepehr mousavi – 2019-12-19T17:05:51.043

1@JohnForkosh So the OP's questions are entangled? – user4894 – 2019-12-19T21:54:34.740

"There is no other cause... Therefore, the only other thing that can contribute in determining the outcome is the system itself" does not follow. There needs to be no other cause contributing, electron does not "decide" anything, it just happens, uncaused. The principle of sufficient cause can simply be false, this is called indeterminism. And indeterminism is perfectly compatible with the momentum conservation law, which is what "the third law of Newton" amounts to, see Quantum Explanation of Newton's Third Law of Motion.

– Conifold – 2019-12-20T01:20:01.143

@conifold, thanks for the comment. There is a problem about your statement I think. This "happens uncaused ", does it mean that no cause was behind the choice of the electron to spin up but not down? If yes, then it is exactly what I tried to show. If no, please tell me why and I think it would be the very last question in that case! – seyed sepehr mousavi – 2019-12-20T07:39:46.130

My impression from your phrasing (like "electron decides", "filled by the system itself") was that you take it as self-caused, i.e. still caused, rather than uncaused. Also, you seem to call "causality" what is usually called "determinism", and "non-causal" what is usually called indeterministic. Copenhagen is universally acknowledged to be that, but it is still causal. It offers causal explanations, even though causes are not sufficient, and adds that no further (hidden) causes exist. – Conifold – 2019-12-20T09:41:40.383

Thanks Conifold, my question is solved! Look at John Forkosh answer below to see how. But I might be defining causality in a wrong way, and I still don't know. However, my question is solved fortunately! – seyed sepehr mousavi – 2019-12-20T13:54:15.403

I doubt it. "I can repeat the same measurement and get different results at each time" is usually not interpreted as repeating the measurement on the system after the first measurement is made. Rather it means imagining that one goes "back in time", when the system was still in the mixed state, and performs the same masurement the "second time". According to Copenhagen, the outcome may well be different "each time". And what happens upon repeating afterwards has little to do with either determinism or causality. – Conifold – 2019-12-21T05:57:00.377

Answers

1

Excuse me for being a little nasty, but your profile says A-level student, whereas I think you just made an F-level remark in your premise

      "I can repeat the same measurement and get different results at each time"

Wanna try again, i.e., repeat your statement and try getting a different result:)?

When you >>first<< make a measurement, then the outcome is indeed stochastic. But as a result of that measurement process, not only do you get a value for the measured observable, but the system also collapses (typically from a mixed state) into a corresponding eigenstate of that observable.

So that means if you immediately >>repeat<< the same measurement, with no intervening disturbance to the system, then the same outcome is guaranteed with probability 1.

user19423

Posted 2019-12-19T13:27:27.817

Reputation:

I agree that if you measure again simultaneously, the next outcome is predictable with certainty. However, that is another assumption you are making which I didn't. In other words, I did not assume in my argument that the measurements are simultaneously after each other. – seyed sepehr mousavi – 2019-12-19T17:10:26.447

By the way, A level is the last two years of highschool education in UK, and there is no F level here, so I didn't quite understand what you said. – seyed sepehr mousavi – 2019-12-19T17:12:49.540

1@seyedsepehrmousavi What I said is, "no intervening disturbance", and >>that's<< the fundamental requirement; the interval between measurements could conceivably be a million years, as long as you can guarantee there's >>no intervening disturbance<<. And if there is a disturbance, then that's the >>cause<< of any difference between measurement outcomes. But anyway, everyone here and in physics.se told you you're wrong, which you seem disinclined to accept. So go write a paper and submit it to Physical Review, or elsewhere. If you're indeed right, then it's a great and important discovery. – None – 2019-12-20T01:33:19.040

1So if the act of measurement has finished, and my system is isolated and so there is no disturbance, is the next outcome the same as the previous one? If yes, my question is solved! – seyed sepehr mousavi – 2019-12-20T07:48:25.627

1@seyedsepehrmousavi Guess what?... Yes, so your question is indeed solved. The observable corresponding to a measurement apparatus is represented by a Hermitian operator. So once the measured system's state is an eigenfunction of that operator, then repeated applications of that same operator to that state (i.e., repeated measurements) leave the state unchanged, whereby you repeatedly measure exactly the same eigenvalue outcome. Ask for further clarification back on one of those related physics.se posts you started. That's where this question belongs. – None – 2019-12-20T08:36:11.337

1Thank you so much, John. I really appreciate your help. I will try to understand what is a hermitian operator and eigenfunction so that I can understand QM more deeply. Meanwhile, I would ask my question from you and other kind people at physics.se for clarifications. Thanks! – seyed sepehr mousavi – 2019-12-20T13:51:01.833

@seyedsepehrmousavi I'm not quite sure what book to recommend for "the last two years of highschool". But think of a projection operator instead. That would be like the Sun in the sky casting your shadow on the ground -- a two-dimensional projection of your three-dimensional body. Now, consider the Sun's position in the sky as an observable, i.e., different position=different observable. If you repeat the shadow-measurement with the Sun in the same position, you'll necessarily measure the same shadow again and again. But in a different position, then you'll get a different shadow-outcome... – None – 2019-12-20T15:04:13.603

@seyedsepehrmousavi ...But that's a deterministic shadow-outcome. I can tell you the outcome beforehand if I know your body-shape=state-of-system and I know the Sun's position=observable. The non-determinism stuff involves more complicated math than the toy-model shadow can illustrate (and I'm not offhand thinking of any simple toy model that does illustrate it). Anyway, keep having your ideas. As many as you can. But just don't get carried away with them. Think them through, work them out, but realize it takes a long, long time before you'll stumble across a really, really good one. – None – 2019-12-20T15:13:27.327

Thank you! I will continue as you recommended. – seyed sepehr mousavi – 2019-12-21T16:46:57.957