Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
1  
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 2 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 3 at 7:34
    
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 3 at 7:56
    
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 3 at 8:22
    
Concerning completeness. –  Ruslan May 3 at 8:46
show 4 more comments

2 Answers 2

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.

share|improve this answer
add comment

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.

share|improve this answer
    
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 3 at 11:20
    
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 3 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 3 at 12:33
1  
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 3 at 13:07
1  
HUP isn't critical to determinism, the key point is born rule/wave function collapse. i think this answer is off target. –  innisfree May 3 at 19:11
show 11 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.