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It seems to be a wide impression that quantum mechanics is not deterministic, e.g. the world is quantum-mechanical and not deterministic.

I have a basic question about quantum mechanics itself. A quantum-mechanical object is completely characterized by the state vector. The time-evolution of state vector is perfectly deterministic. The system, equipment, environment, and observer are part of the state vector of universe. The measurements with different results are part of state vector at different spacetime. The measurement is a complicated process between system and equipment. The equipment has $10^{23}$ degrees of freedom, the states of equipment we neither know nor able to compute. In this sense, the situation of QM is quite similar with statistical physics. Why can't the situation just like statistical physics, we introduce an assumption to simply calculation, that every accessible microscopic state has equal probability? In QM, we also introduce an assumption about the probabilistic measurement to produce the measurement outcome.

PS1: If we regarded non-deterministic is intrinsic feature of quantum mechanics, then the measurement has to disobey the Schrödinger picture.

PS2: The bold phase argument above does not obey the Bell's inequality. In the local hidden variable theory from Sakurai's modern quantum mechanics, a particle with $z+$, $x-$ spin measurement result corresponds to $(\hat{z}+,\hat{x}-)$ "state". If I just say the time-evolution of universe is $$\hat{U}(t,t_0) \lvert \mathrm{universe} (t_0) \rangle = \lvert \mathrm{universe} (t) \rangle.$$ When the $z+$ was obtained, the state of universe is $\lvert\mathrm{rest} \rangle \lvert z+ \rangle $. Later the $x-$ was obtained, the state of universe is $\lvert\mathrm{rest}' \rangle \lvert x- \rangle $. It is deterministic, and does not require hidden-variable setup as in Sakurai's book.

PS3: My question is just about quantum mechanics itself. It is entirely possible that the final theory of nature will require drastic modification of QM. Nevertheless it is outside the current question.

PS4: One might say the state vector is probabilistic. However, the result of measurement happens in equipment, which is a part of total state vector. Given a probabilistic interpretation in a deterministic theory is logical inconsistent.

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user26143
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    Quantum mechanics is deterministic, but it is also probabilistic -- i.e. you can deterministically calculate the probability of a random event happening. This is to distinguish it from non-deterministic (i.e. stochastic) systems where you do not generally have "one" solution but an entire family of solutions depending on random variables. – webb May 02 '14 at 22:33
  • If I know the wavefunction, or state vector, more generally, of the universe, then I don't need the probability anymore – user26143 May 03 '14 at 07:34
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    If you know the state vector of the universe, then this still doesn't give you information about exact outcome of any quantum experiment — only probabilities. – Ruslan May 03 '14 at 07:56
  • If the equipment and system are governed by the Schrodinger picture, there is no (strict, by means of not in the sense happened in statistical mechanics) probability. If there is (strict) probability, then the Schrodinger picture is incomplete. – user26143 May 03 '14 at 08:22
  • Concerning completeness. – Ruslan May 03 '14 at 08:46
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    It is not clear what you are asking. Quantum theory is non-deterministic in the sense that it works with objects ($\psi$ functions, kets) that can be used to calculate probabilities, not the actual results. It is the same as in statistical physics, only probabilistic statements can be derived. – Ján Lalinský May 03 '14 at 09:53
  • My question is why people regard quantum mechanics is non-deterministic. The postulate of QM is: a system is described by state vector, not about the usage of state vector/wave function. The experimental object, equipment, environment, and observer are all QM systems. The state vector for them is perfectly deterministic. In statistical physics, the underlying equation of motion may be deterministic, e.g. Newtonian particles, the probability enters for practical, not fundamental, reason: the system has vast degree of freedom. – user26143 May 03 '14 at 10:02
  • @user26143 in the future could you condense your editing so as to keep the total number of edits down? – David Z May 04 '14 at 00:26
  • @DavidZ, Yes. I will try. Sorry for that. – user26143 May 04 '14 at 05:51
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    If you only hold the algebraic structure of QM you do not have a physical theory. How this mathematical structure are connected with the experiments? You need to link that. This is science. And this is because QM is non-deterministic. QM is not only the mathematical structure but also the Born principle, that connects mathematical objects with experimental results. You can think that exist an underlying mathematical structure that explain Born principle in a "deterministic" way, but this is purely metaphysical reasoning because is not sensible. – Nogueira Dec 27 '15 at 09:11
  • In statistical mechanics we have, at least in principle, the power to measure the microstates. So your comparison is wrong. – Nogueira Dec 27 '15 at 09:14
  • You said, "The time-evolution of state vector is perfectly deterministic". This is not true. In the copenhagen interpretation, where wave-function collapse is a non-deterministic process. – Brian Moths Feb 13 '16 at 14:36

7 Answers7

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I agree with much of what you write in your question. Whether quantum mechanics is considered to be deterministic is a matter of interpretation, summarised in this wiki comparison of interpretations. The wiki definition of determinism is this context, which I think is entirely satisfactory, is

Determinism is a property characterizing state changes due to the passage of time, namely that the state at a future instant is a function of the state in the present (see time evolution). It may not always be clear whether a particular interpretation is deterministic or not, as there may not be a clear choice of a time parameter. Moreover, a given theory may have two interpretations, one of which is deterministic and the other not.

In, for example, many-worlds interpretation, time evolution is unitary and is governed entirely by Schrödinger’s equation. There is nothing like the "collapse of the wave-function" or a Born rule for probabilities.

In other interpretations, for example, Copenhagen, there is a Born rule, which introduces a non-deterministic collapse along with the deterministic evolution of the wave-function by Schrödinger’s equation.

In your linked text, the author writes that quantum mechanics is non-deterministic. I assume the author rejects the many-worlds and other deterministic interpretations of quantum mechanics. Aspects of such interpretations remain somewhat unsatisfactory; for example, it is difficult to calculate probabilities correctly without the Born rule.

innisfree
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    The problem is that in many-worlds interpretation there is no deterministic connection between the state and the observed behavior since theoretically all branches co-exist, but in practice only one is observed. "Determinism" is a linguistic sleight of hand. Bohmian mechanics is indeed deterministic, but it involves faster than light signals and ephemeral entities (Bohmian particles) unobservable in principle, like ether. For that matter Everett's branches are much like ether as well, and play the same role as Bohmian particles. – Conifold Apr 28 '16 at 01:46
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The difference between statistical physics and quantum mechanics is that, in statistical physics, it is always reasonable to either measure a quantity, or demonstrate that the effect of that quantity can be bundled into an easy to work with random variable, often through the use of the Central Limit Theorem. In such situations, it can be shown that the answer will be a deterministic answer plus a small perturbation from the random variables with a 0 expectation and a very small variance.

In quantum mechanics, the interesting properties show up in situations where its not possible to measure a quantity and not plausible to bundle it up into a random variable using the central limit theorem. Sometimes you can, of course: in particular, this approach works well in modeling an quantum mechanic system which is already well modeled in classical physics. For the most part, we don't observe many quantum effects in day to day life! However, quantum mechanics is focused on the more interesting regions where those unmeasurable quantities have an important impact on the outcome of the system.

As an example, in many entanglement scenarios, you can get away with ignoring the correlation between the states of the particles. This is good, because in theory, there's some small level of entanglement between all particles that have interacted, and its good to know that we can often get away with ignoring this, and treating the values as simple independent and identically distributed variables. However, in the entanglement cases quantum mechanics are interested in, we intentionally explore situations where the entanglement is strong enough that that correlation can't just be handwaved away and still yield experimentally validated results. We are obliged to carry it through our equations if we want to provide a good model of reality.

There are many ways to do this, and one of the dividing lines regarding the topic is the line drawn between the different interpretations of QM. Some of them hold to a deterministic model, others hold to non-deterministic arguments (the Copenhagen interpretation being an example). In general, the models which are deterministic have to give up something else which is valued by physicists. The many-worlds theory gets away with being deterministic by arguing that every possible outcome of every classical observation occurs, in its own universe. This is consistent with the equations that we believe are a good model of quantum mechanics, but comes with strange side effects when applied to the larger world (quantum suicide, for instance). The Copenhagen interpretation is, in my opinion, the most natural interpretation in that it dovetails with the way we do classical physics smoothly, without any pesky alternate realities. I have found that mere mortals are most comfortable with the intuitive leaps of the Copenhagen interpretation, as compared to the intuitive leaps of other interpretations. However, the Copenhagen interpretation is decidedly non-deterministic. Because this one seems easier to explain to many people, it has achieved a great deal of notoriety, so its non-determinism gets applied to all of quantum mechanics via social mechanisms (which are far more complicated than any quantum mechanisms!)

So you can pick any interpretation you please. If you like determinism, there are plenty of options. However, one cannot use many of the basic tools of statistical mechanics to handle quantum scenarios because the basic physics of quantum mechanics leads to situations where the basic assumptions of statistical mechanics become untenable. Your example of the result of the measurement happening in the equipment is an excellent example. Like in statistical physics, the state of the measurement equipment can be modeled as a state vector, and it turns out that it's a very reasonable assumption to assume that it is randomly distributed. However, equipment designed to measure quantum effects is expressly designed to strongly correlate with the state of the particle under observation before measurement began. When the measurement is complete, the distribution of the state of the measurement equipment is decidedly poorly modeled as a state plus a perturbation with a small variance. The distribution is, instead, a very multimodal distribution, because it was correlated to the state of the particle, and most of the interesting measurements we want to take are those of a particle whose [unmeasured] state is well described by a multimodal distribution.

Cort Ammon
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Quantum mechanics is non deterministic of actual measurements even in a gedanken experiment because of the Heisenberg Uncertainty Principle, which in the operator representation appears as non commuting operators. It is a fundamental relation of quantum mechanics:

If you measure the position accurately, the momentum is completely undefined.

The interpretation of the solutions of Schrodinger's equation as predicting the behavior of matter depends on the postulates: the state function determined by the equation is a probability distribution for finding the system under observation with given energy and coordinates. This does not change if large ensembles are considered except computationally. The probabilistic nature will always be there as long as the theory is the same.

anna v
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  • The problem is one only looked at the wavefunction of the system, but not equipment. Hypothetically, if (1) we are able to know and compute the wavefunction of system+equipment+enviroment+observer, if (2) they obey Schrodinger picture, then there is no room for any uncertainty. Take the $[x,p]=i$ for instance, once the state of universe is $|rest\rangle | p \rangle $. The experimentalist decided to measure position, the state becomes $|rest'\rangle | x \rangle $ with a particular position. It is again deterministic. – user26143 May 03 '14 at 11:20
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    You are wrong. The HUP is not optional. The total universe obeys the HUP postulate so as far as the theory of quantum mechanics goes, which is what you are asking, it will always be indeterminate by construction of the theory. It was constructed to fit observations and if you extrapolate to the total universe it makes no difference. (You said you are not considering other theories ) – anna v May 03 '14 at 12:02
  • (1) Heisenberg uncertainty principle ($\Delta x \Delta p \geq \hbar/2$) is not a postulate, it is derived from commutator plus measurement postulates. (2) How a theory first constructed does not imply we should understand it in this line, like Lorentz thought Lorentz transformation. (3) What HUP predicts, is we have a collection of particle at $|p \rangle$. We measure the position of each particle, we got $x_1$, $x_2$, $x_3$,... looks randomly. However, the whole universe obeys Schrodinger picture. The result could be interpreted as the state of universe evolves into – user26143 May 03 '14 at 12:33
  • $| rest \rangle | x(1)_1 \rangle$ at the time one measured the position of the first particle. Then $| rest' \rangle | x(2)_2 \rangle$ at the time one measured the second particle. etc. The experimental results are exactly the same. The only reason to abandon deterministic, as far as I concern, is to admit there is some boundary between system and equipment. Outside this boundary (like what happened in measurement), the Schrodinger picture does not apply. This approach is even more weird, since QM applies to macroscopic system as well. – user26143 May 03 '14 at 12:38
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    When measuring one particle's x then going to the next, their momentum will be indeterminate and "next" will have a whole phase space to be chosen from because momentum determines the next probability of x, not a point but a probability of being found at that point, whether 1 2 3 or infinite number of particles. – anna v May 03 '14 at 12:48
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    The HUP is a postulate incorporated into the mathematics of commutators. – anna v May 03 '14 at 12:50
  • (1) HUP, with the general expression, $\Delta A \Delta B \geq 1/2 |<[A,B]>|$, is derived from commutator and Cauchy-Schwartz inequality, e.g. Sakurai's modern quantum mechanics rev edi (1.4.53), and postulates that measurement obtain eigenvalue with probability of coefficient square. (2) Whatever the momentum of the particles are, they obey the Schrodinger picture. If one admits the Schrodinger picture govern the particles and equipment, there is no place to let the intrinsic probability come into.(3) I thought the measurement is just like statistical physics in throw dice, simply calculation. – user26143 May 03 '14 at 13:04
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    No, the schrodinger picture gives a probability of finding any measurement value, not a fixed value of the momentum. One has to operate on the schrodinger state function, with the momentum operator to get the momentum, and the operation.measurement will give a value within the probability envelope. – anna v May 03 '14 at 13:05
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    Let us suppose, the whole universe evolves under Schrodinger picture. Everything is perfectly deterministic. There is no room to give probability. The Schrodinger picture of the universe cannot give probability. – user26143 May 03 '14 at 13:07
  • Lets stop this. You are completely wrong, the schrodinger picture is a probabilistic picture because it is a quantum mechanical theory. It determines the state function, not the measurable quantities, which are given as probability distributions. – anna v May 03 '14 at 13:09
  • I think, it is helpless that I repeat my opinion. And you are free to stop. I am trying a different statement, from Weinberg's Einstein's mistake: "Physicists and their apparatus must be governed by the same quantum mechanical rules that govern everything else in the universe. But these rules are expressed in terms of a wavefunction that evolves in a perfectly deterministic way. So where do the probabilistic rules of the Copenhagen interpretation come from? Considerable progress has been made in recent years toward the resolution of the problem, which I cannot go into here" – user26143 May 03 '14 at 13:18
  • According to your opinion, the standard probability interpretation already solve the problem. Then Weinberg's later work arXiv:1109.6462 is not motivated at all (He modified Schrodinger picture, if I understood correctly). Anyway, I would like to know your criticize about Weinberg. – user26143 May 03 '14 at 13:20
  • If it is correct, it is another isomorphic construct of quantum mechanics (like bohmian qm). I have not heard of any predictions being verified or even tested.Anyway, "collapse", "entanglement" are, imo, fancy words to describe what are simple mathematics and measurement interpretation rules and leave me indifferent. – anna v May 03 '14 at 14:09
  • HUP is not a postulate of QM; it is a derived result. – innisfree May 03 '14 at 18:00
  • @innisfree I guess I was counting on "principle". A principle is stronger than a postulate. The mathematical formulation chosen to describe the quantum mechanical state, have to obey it, otherwise different mathematical forms would have been sought. This history is interesting http://www.aip.org/history/heisenberg/p08.htm . I will edit – anna v May 03 '14 at 19:04
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    HUP isn't critical to determinism, the key point is born rule/wave function collapse. i think this answer is off target. – innisfree May 03 '14 at 19:11
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Forget interpretations. The predictions of quantum mechanics - which agree with all interpretations (by definition of 'interpretation')- does not allow prediction of experimental/observational outcomes no matter how much information is gathered about initial conditions. (You can't even get the classical information needed in classical physics because of the uncertainty principle). None of the interpretations challenge this, not even in principle. According to the math, which is wildly successful in it's predictions, a given present does not determine the future. That's why quantum mechanics is said to be indeterministic, not because of any interpretation. It doesn't matter if you believe in wave function collapse or not or other worlds or not or whatever. Saying the theory is deterministic because of some math involved in the calculation isn't related to the fact that experimental outcomes cannot be predicted, The present does not determine the future.

Vector Shift
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  • What makes you say that we can never gather enough information about a state? We are perfectly capable of engineering finite-dimensional quantum states, for example specific entangled qubit states. The evolution of these states is then entirely deterministic (in fact, quantum computing wouldn't be possible otherwise). Similarly, we may not be capable of measuring, say, the position and the momentum of a particle, but why would your definition of a quantum system require the existence of such observables anyway? – Alex Bogatskiy May 17 '21 at 19:01
  • To continue, in principle there is nothing stopping us from engineering states that are arbitrarily close to, say, a plane wave with $\Delta p\to 0$ and $\Delta x\to \infty$. The fact that $\Delta x$ is large in no way means that we "don't know" what the state is — we do! It's a plane wave $e^{i p x}$! – Alex Bogatskiy May 17 '21 at 19:05
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If you learn Quantum Mechanics you will see that the observables of any quantum system depend on the state of the system(final, initial, ground state or excited state). In theory, there are a number of interpretations of Quantum Mechanics wiki, link.

The mathematical formulation of quantum mechanics is built onto the notions of an operators. When you do a measurement you perturb the system state by applying an operator on it. The eigenvalue of the operator corresponds to the measured value of the system observable. However, each eigenvalue have a certain probability, and therefore by measuring(applying) an operator on the state system there will be a finite(or infinite) number of final states, each of them with a given probability. This is the essence of non-deterministic in quantum mechanics.

The next question arises:how the non-deterministic applies on large scale universe and the "length" of the not-deterministic" phenomena in the universe?

Because in classical theory(like general relativity, electromagnetism), you have for example the Einstein equations which govern the dynamics and they are full deterministic.

Nikey Mike
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The quantum state of a system is completely characterized by a state vector only when the system is a pure state. The state vector evolves in two different ways described by two postulates: the Schrödinger postulate (valid when there is no measurements) and the measurement postulate. The Schrödinger postulate describes a deterministic and reversible evolution $U$. The measurement postulate describes a non-deterministic and irreversible evolution $R$.

$R$ is not derivable from $U$. In fact $R$ is incompatible with $U$, and that is the reason why the founder fathers introduced two evolution postulates in QM. Indeed, assuming an initial superposition of two states for the composite supersystem (system + apparatus + environment)

$$|\Psi\rangle = a |A\rangle + b |B\rangle $$

the result of a measurement is either $|A\rangle$ or $|B\rangle$, but because these states are orthogonal, they cannot both have evolved from a single initial state by a deterministic, unitary evolution, since that $|A\rangle = U |\Psi\rangle$ and $|B\rangle = U |\Psi\rangle$ implies $\langle A|B\rangle = \langle\Psi |U^{*} U | \Psi\rangle = 1$, which is incompatible with the requirement of ortohogonality.

So, if the result of the measurement was $|B\rangle$, the evolution was $|B\rangle = R |\Psi\rangle$.

juanrga
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The fact that QM is probabilistic and not deterministic is forced by the 4 rules stated below. This rules can not coexist logically to provide determinism. They lead without effort to the probablistic interpretation.

Yes, unfortunately (for me) I am not a physicist. So take this with a grain of salt.

Some thinking about this puzzling issue will make you have these conclusions based on well-known facts:

@Quantum world:

1) Entities have a 'spread' existence. (A kind of 'field of energy' which tries to 'fill' all space).

2) Entities have some 'oscilatory' existence. (Which gives rise to 'interference' phenomena).

3) Interactions between entities are 'discrete'. (They exchange 'quanta' of somestuff).

4) Interactions use the 'minimum amount' of some 'energy stuff'.

The interplaying of these facts is what gives rise to the non-determinism (probability) in QM.

Let's think of a simple example:

Suppose you have 3 entities A, B and C (a 1 sender & 2 receivers scenario), where A is the source of some perturbation to be sent to B and C at the 'same time'. Let's think of the perturbation in practical terms (i.e.: money) and assign it a unit of measure (dollars).

Now how would A send 2 dollars total to both of them (B & C)?

Well, A should give them 1 dollar each and problem solved!!!. However, there is a constraint here (remember #4) and that is: Interactions are only done with minimun currency!!!'.

With that in mind, how can A give B and C one cent (minimun currency) at the same time? Well, .. It can't!!!

At each time (interaction) A must choose between B or C to give away every cent until completes the 2 dollars to both of them. And if you think a little bit about it, you realize that the only objective solution for A must be to throw an imaginary coin each time to decide whom will receive the 1 cent!. [Of course, for this 1 sender & 2 receivers situation, a coin with 2 faces fits rigth!. But for others scenarios, the coin or dice will have to change.]

In the analog world of classical mechanics, A would send an infinite small amount of money to both of them (no minimum currency constraint and at the same time!) and what we will see is a beautiful continuous growing of B and C money pockets. No need to deal with probabilities!!!!.

If you think carefully, in plain simple terms, probability arise from the discrete nature of interactions between entities. This is the real deal which turns everything so strange and interesting.

[Hope this general and somewhat vague answer gives you a clue about why probability arise in the description offered by QM]

The question now is: Why it has to be like that?

fante
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