I don't get this part of quantum mechanics.

I get the part that you can't observe particles and not affect their behavior because you are shooting photons to them while you are observing them, but how can this show that while you are not observing them, they behave in indeterministically and that's a feature of nature.


4 Answers 4


The reason for this seeming leap is that the principles of logical positivism, which is the founding philosophy of Heisenberg, Bohr, and all physicists really. This states that if a question cannot be answered even in principle by some sort of experiment, then it is not a valid question, the question is just gibberish.

Consider the following question:

  • In a Hydrogen atom in its ground state, where is the electron in its orbit?

Superficially, it seems sensible, doesn't it?

But how would you formulate an experiment to determine the answer? Now it isn't so clear. Suppose you shine light on the electron to try to find out where it is, then you excite the atom, it is no longer in its ground state. Suppose you shine very low wavelength light, so as not to excite the atom. Then the light scatters off the atom as a whole, and is useless for answering the question.

If you try to use a hard X-ray to localize the electron precisely, you ionize the atom. So this question is impossible to answer by experiment, and now it doesn't look so sensible. It is a valid act of positivism to assert that this question is, in fact, meaningless. The electron does not have a position when the atom is in its ground state.

But lets say you ignore the positivism, and you suppose that the electron has a secret position which is varying in time, as Bohr often did. You might believe that the orbit is periodic, so that the Fourier transform of the orbit has integer multiples of a given frequency. The observed frequency of light emitted by a moving classical object is in integer multiples of the fundamental frequency, the inverse orbital period. So you expect that the light emitted by the atom to come in multiples of the orbital period.

But the atomic transitions have frequencies which are not integer multiples of anything. So they cannot be the description of a classical periodic trajectory. But they correspond to these periodic trajectories when the quantum number is large, when the electron is orbiting far away from the proton. This much was understood by Bohr.

But for Pauli and Heisenberg, who were more radically positivist (at first, Bohr was the most positivist of all later in life), the difficulty of finding an effective procedure led them to renounce the question entirely. They rejected the idea that the electron had a position to be determined. Heisenberg then developed a detailed mathematical theory of the transitions between different atomic levels, and this theory was able to answer questions of the form "if I shine light on the atom in the ground state, what spectral intensities come out?" But this theory could not answer the question "where is the electron", because it did not have an electron position variable which made a sharp classical orbit.

The modern perspective is just an elaboration of this position. the wavefunction describes the probability of a position measurement experiment to find a given answer, or the probability of getting the answer to an energy measurement. It does not represent the position of an electron, or any other classical quantity.

The reason people believe that there are no fundamental classical quantities evolving deterministically is because of the logical positivist position that they couldn't define them operationally. Logical positivism fell out of favor in the 1970s, for stupid reasons, so it is no longer a prominent position defended in humanistic intellectual circles. This is very sad for most physicists, who are just as positivist as ever, especially considering string theory, holography, and the positivist resolution of the information loss puzzle.

  • $\begingroup$ Why is it no longer a prominent position? By whom and what came instead? $\endgroup$
    – Nikolaj-K
    Jan 15, 2012 at 11:59
  • 3
    $\begingroup$ @Nick: It was a product of the 1970s. Marijuana made the world more spiritual, and people had more respect for unseeable stuff and interpersonal differences in perception. These things are generally (wrongly) thought incompatible with positivism. As for what came instead: Searle's room, religious revivalism, midieval thinking. The philosophers got stoned; out went the positivism. The physicists got stoned; out went the S-matrix theory. This is basically everybody in the 1970s. I think this is a tragedy, and it is ever so slowly being reversed by people who do not get stoned. Don't get stoned. $\endgroup$
    – Ron Maimon
    Jan 15, 2012 at 14:50
  • $\begingroup$ @Nick: I was informed by a person involved with philosophy that the stoner rejection of positivism is pretty much restricted to France and continental philosophy. In analytic philosophy, the idea that there exists an objective world was the motivating thing. So that Bohr style positivism, which rejects the idea that there is anything real, was the target. I suppose this is plausible, but the whole thing smells like marijuana to me. $\endgroup$
    – Ron Maimon
    Feb 29, 2012 at 16:50
  • $\begingroup$ I advocate some sort of positivistic position. I think I tend to wanting to ultimately formulate everything in term of what is relevant for the thinking individuals and their perception and I see no necessarity for objective things. I wish there was some reliable procedure to find out which terms, concepts or ways to think about stuff are more misleading or useless. I'm not so much talking about physics here anymore btw. $\endgroup$
    – Nikolaj-K
    Mar 1, 2012 at 8:23
  • $\begingroup$ @Nick: I understand. I think the point is that we can come to agree using only positivistic notions about external objective stuff. In mathematics, we can all agree on ordinals less than CK. In physics, I think we can all agree somehow that Bohr, Everett, Pauli, and Wigner are all talking about the same thing, and something different than Bohm. I think of Positivism as a definition of equality: two theories are the same if all their sensory testable propositions are indistinguishable. It is just a law of thought that you always need to use, otherwise you argue over nonsense. $\endgroup$
    – Ron Maimon
    Mar 1, 2012 at 15:46

There is no indeterminacy in QM. The principles are:

  • The states $|\psi\rangle$ live in Hilbert space.
  • Measurements are represented by Hermitian operators $O$.
  • QM only gives answers to questions that can be posed using an eigenvalue equation for a Hermitian operator. If one has $O|\psi\rangle=\lambda|\psi\rangle$ then the state $|\psi\rangle$ has the property represented by $O$ with sharp value $\lambda$. This is all QM tells us and we have to frame all our questions about the world in this terse language.

These principles say nothing about indeterminacy or probability or collapse. The Born rule for probabilities appears as follows.

Consider a spin 1/2 system in a state $|\psi\rangle=\alpha|+\rangle+\beta|-\rangle$. Set up an ensemble of $N$ copies of the system. The ensemble is described by a tensored state with $N$ factors, \begin{eqnarray*} |\Psi\rangle&=&(\alpha|+\rangle+\beta|-\rangle)\otimes\ldots \otimes(\alpha|+\rangle+\beta|-\rangle)\\ &=&\sum_{k=0}^{N}\alpha^{N-k}\beta^{k}(|+\ldots-\ldots\rangle+\ldots) \end{eqnarray*} The bracket has $^{N}C_{k}$ states each with $N-k$ "+" labels and $k$ "-" labels in various arrangements. This symmetric combination of $^{N}C_{k}$ states is a spin $s=N/2$ eigenstate with $J_{z}$ eigenvalue $m=N/2-k$. With the correct normalization, the tensored state for the ensemble is, \begin{eqnarray*} |\Psi\rangle=\sum_{k=0}^{N}\alpha^{N-k}\beta^{k} \sqrt{^{N}C_{k}}|s=N/2,m=N/2-k\rangle \ . \end{eqnarray*} Now set up a Hermitian operator $\frac{N_{+}}{N}$ that counts the fraction of $+$ states. The action of the operator is, \begin{eqnarray*} \frac{N_{+}}{N}|s=N/2,m=N/2-k\rangle=\frac{N-k}{N}|s=N/2,m=N/2-k\rangle \end{eqnarray*} The state $|\Psi\rangle$ is not an eigenstate of $\frac{N_{+}}{N}$ for finite $N$ so QM cannot say anything sharp about the fraction of "+" states in the ensemble. However, as $N\rightarrow\infty$ the amplitudes peak sharpy around a single spin state. To find the value of $k$ at which the amplitude peaks, one solves, \begin{eqnarray*} 0=\frac{\partial}{\partial k}\frac{|\alpha|^{2(N-k)}|\beta|^{2k}N!}{(N-k)!k!} \end{eqnarray*} and by using Stirling's approximation, the solution for $k$ at the peak amplitude is, \begin{equation} \frac{N-k}{k}=\frac{|\alpha|^{2}}{|\beta|^{2}} \end{equation} So, for an infinite ensemble, $$\lim_{N\rightarrow\infty}|\Psi\rangle=\alpha^{N-k}\beta^{k} \sqrt{^{N}C_{k}}|s=N/2,m=N/2-k\rangle$$ with $k$ given by the peak solution. The infinite ensemble is an eigenstate of operator $\frac{N_{+}}{N}$, $$\frac{N_{+}}{N}|\Psi\rangle=\frac{N-k}{N}|\Psi\rangle$$ and substituting for the peak $k$, the eigenvalue is $(N-k)/N=|\alpha|^{2}$. So, the tensored state for the infinite ensemble is an eigenstate of operator $\frac{N_{+}}{N}$ that counts the fraction of "+" states and the eigenvalue is $|\alpha|^{2}$. This means that QM can give a sharp answer to the question about the fraction of "+" states in the ensemble; it's $|\alpha|^{2}$ which is exactly the same as the Born rule for probabilities. This calculation shows,

  • There is nothing probabilistic or indeterministic about QM.
  • QM has nothing to say about the single system $\alpha|+\rangle+\beta|-\rangle$.
  • QM only says something sharp about an infinite ensemble of systems.

The derivation of the Born rule in this answer is outlined in lecture http://pirsa.org/10080035 .

Edit to answer Bruce Greetham.

Bruce Greetham suggests that by deleting the Born rule as an axiom, I'm forced to take Everett's many worlds position.

Consider a spin 1/2 system $|\psi\rangle=\alpha |+\rangle +\beta |-\rangle$. Let's ask QM the sharp question "Is the system in the + state?". The Hermitian operator that asks this question is $\hat{O}(\text{'is it +?'})=|+\rangle\langle+|$ because $\hat{O}(\text{'is it +?'})|+\rangle=|+\rangle$ and $\hat{O}(\text{'is it +?'})|-\rangle=0$. The system $|\psi\rangle=\alpha |+\rangle +\beta |-\rangle$ is not an eigenstate of $\hat{O}(\text{'is it +?'})$ so $|\psi\rangle$ is not in the $|+\rangle$ state.

Now ask the sharp question "Is the system in the + state or the - state?". The operator that asks this question is $\hat{O}(\text{'is it +?'})+\hat{O}(\text{'is it -?'})=|+\rangle\langle+|+|-\rangle\langle -|$. This operator is the identity operator $1$ and the state $|\psi\rangle$ is an eigenstate of this operator with eigenvalue 1. In other words QM is saying that $|\psi\rangle$ is + or - which, I'm forced to admit, does suggest that the world has branched. This is an intellectually uncomfortable position to be in, and I mostly try not to think about it and concentrate on understanding more about group representations and QFT.

  • $\begingroup$ Where is the rest of the ensemble? are you talkin about MWI? $\endgroup$ Jan 17, 2012 at 6:18
  • $\begingroup$ @SchroedingersGhost : The ensemble of $N$ systems is in the experimentalist's lab. I'm not talking about MWI. $\endgroup$
    – user7154
    Jan 17, 2012 at 19:55
  • $\begingroup$ I really like this answer and voted it up, but note that the explanation you give is exactly the frequentist interpretation of probability: if the fraction of N identical experiments gives a definite answer as N goes to infinity, then that is the frequentist definition of the probability of the outcome. $\endgroup$
    – arsmath
    Nov 16, 2013 at 10:26
  • $\begingroup$ @arsmath : Thanks for your kind words and for pointing out it implies QM sits in the frequentist camp in the frequentist/calculus of consistent belief dichotomy. $\endgroup$
    – user7154
    Nov 16, 2013 at 16:31
  • $\begingroup$ But surely, for a single experiment, quantum mechanics is still indeterministic. You cant just say we cant ask about the result of a single experiment. We can do a single experiment and get a result. The notion of indeterminism in this answer doesn't seem to match common use of this term. In Arkani-Hamed lecture you reference he goes on to marvel at the indeterminacy of QM. What he is arguing though is that the Born rule is not a separate postulate of QM. $\endgroup$
    – isometry
    Sep 24, 2016 at 17:38

How can this show that while you are not observing the particles, they behave indeterministically and that's a feature of nature?

What makes you word this question like that?

If you know the full starting configuration as well as the Hamiltonian (Energy Operator) of the system, then according to quantum mechanics you can determine the time evolution of the system. In this sense it is possible to know it. It's in principle possible to compute and know the propagating wave function for later times and therefore you can predict an expectation value for every time, not more or less. At later times we might decide to make a measurement. And the fact that the measured probabilities turn out to be in accordance with what that theory (the one which is build around that wave function) suggests that we are up to something.

However, the purpose of the wave function is "just" to be the tool to predict what happens if there actually is interaction with the system. So the background to your question is of ontological/epistemological nature. What does it mean to know something which is beyond measurement? Maybe every three seconds during its unperturbed propagation, the wave function turns into a pink tea drinking elephant, logs into StackExchange Physics under someones user account and answers questions about Quantum Mechanics, and then turns back into a wave function as if nothing happened. That's not a very useful theory though.

  • $\begingroup$ Contrary to Nick's implication, I think the question is perfectly worded and it is in fact THE question of quantum mechanics. The uncertainty is frequently described in terms of a mere practical issue; even Feynmann does this in explaining the double slit. EPR insists the uncertainty goes much deeper than this, and Bell makes it a practical issue subject to experiment. If the OP doesn't understand this, he's in good company. $\endgroup$ Jan 15, 2012 at 2:37
  • $\begingroup$ @Marty Green: I wasn't implying that the wording of the question is bad, I was asking where OP got his information from. $\endgroup$
    – Nikolaj-K
    Jan 15, 2012 at 10:37

There are two different aspects to the question:

  • Can we predict, at least in principle, the outcome of any experiment?
  • Is nature actually indeterministic?

The first question is a physical one: We can answer it with experiments. This may come to a surprise, because experiments always come with experimental errors, but the point is that experiments never really prove something anyway, but they support or reject a theory. And all experiments support quantum mechanics, where we cannot tell, even in principle, what result we will get. Note that this is true even for those interpretations of quantum mechanics which are fundamentally deterministic (like Bohm's mechanics), because we cannot, not even in principle, access the full state (this is why such theories are called "hidden variable theories", because there are some aspects of the state we cannot access, that is, they are "hidden".

The second one is largely a philosophical one. For a positivist, the fact that experimentally you cannot predict the result already tells you that nature is indeterministic. However, positivism is a philosophical position which cannot be proved experimentally (that is, it is not part of physics; this of course doesn't mean a physicist cannot be a positivist, but it means that a physicist advocating positivism is acting outside his field ― note however that there's a fine line between what might be called "practical positivism", refusing to say anything about what we cannot measure, because it would be outside the context of physics, and real positivism, claiming that there is nothing besides what we can observe).

For a realist, the standard quantum mechanics is not satisfying because either one assumes the wave function is real, but then the collapse of the wave function would also be real, but this would give a special status of observations, and thus consciousness, which most people don't believe in nowadays (and those people who do are usually more in the esoteric fields than in the scientific). Therefore realists seek explanations which go in one way or another beyond standard quantum mechanics. However, there's a problem with this: Locality. Quantum mechanics can be (experimentally!) shown to be non-local (at least under the "one world" assumption, one reason why the many-worlds interpretation is popular for some people), but non-locality is at odds with Relativity, because the latter denies an absolute simultaneity. In short, for a realist interpretation of quantum mechanics, one has to drop at least one other assumption typically given as granted.

Now on the subject of determinism, in the realist camp there are both deterministic and indeterministic interpretations of quantum mechanics. The deterministic can be grouped into two categories in their explanation of why we observe indeterminism:

  • Hidden variables: Here the indeterminism is classic lack of knowledge (combined with the inability to obtain the knowledge, even in principle). However this comes at the cost of physical nonlocality (i.e. true physical interactions going infinitely fast over arbitrary distances).
  • Everettian interpretations (variations of Many Worlds): Here, indeterminism is a result of "splitting worlds" where we only observe "our" world. The result is not predictable because all possible results occur, but each result comes with another copy of "you" which observes just that result.

If you accept neither physical nonlocality nor many worlds, you have to accept fundamental indeterminism.


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