Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Why do quantum physical properties come in pairs, governed by the uncertainty principle (that is, position and momentum?)

Why not in groups of three, four, etc.?

share|cite|improve this question
up vote 10 down vote accepted

The duality is a duality because of the notion of canonical conjugation in classical mechanics.

The reason people say that they come in pairs has nothing to do with quantum mechanics, but with the structure of classical mechanics. In classical mechanics, to give you the initial conditions for a system, you need to give the initial position of everything, and also the initial momentum. The classical variables come in pairs. These pairs are called canonically conjugate, because they have the property that their time rate of change of one is given by the derivative of the energy with respect to the other one.

The quantum mechanical description is only for wavefunctions which vary over values of one of the two canonically conjugate pairs. The other one is not freely specified, the wavefunction which gives its quantum description can be derived from the first.

People express the failure of classical mechanics by saying that half of all the initial data is related by uncertainty to the other half. This is the quantum duality. This idea was important historically, becuase it could explain how the classical equations could be taken up in quantum mechanics unchanged, while the predictions became probabilistic. People made analogies with the case where you have a classical particle whose position and momentum are unknown, and obey the Heisenberg relation. Quantum mechanics is completely different in that the presence at different position states is parametrized by probability amplitudes, not probabilities. But the description is always on half the phase space variables.

The uncertainty in position/momentum is directly analogous to the uncertainty in angular position/angular momentum, in the uncertainty of phase of a field mode and particle number of that mode, and in every other canonically cojugate pair. States of definite position are also uncertain energy, because energy and position do not commute, but nobody calls this a duality, because energy and position are not canonically conjugate.

I should also point out the energy/time uncertainty principle, which is hard to think of in terms of canonical pairs in the usual formulations of QM, because particles don't have a time associated to them, but all particles have a global time. In Schwinger/Feynman particle formalisms this is taken care of, but the uncertainty relation can be worked out in any formalism of course. This uncertainty relation might not be called a duality by some, I don't know.

share|cite|improve this answer

Good question! For the properties related by the uncertainty principle, there are two reasons why they come in pairs:

  1. Intuitively, the uncertainty principle relates the variance of a function to the variance of its Fourier transform. And, up to a couple of numerical factors, the Fourier transform of a Fourier transform is the original function. (Mathematicians will balk at this statement because it's not technically true, but conceptually it's accurate enough for my purposes here.) So the process of Fourier conjugation leads you through a cycle of two functions.

  2. Mathematically, the uncertainty principle is based on commutators, namely that

    $$\sigma_A \sigma_B \ge \biggl|\frac{1}{2i}\langle[A,B]\rangle\biggr|$$

    and it doesn't really make as much sense to compute the commutator of three or more operators since there are multiple ways in which you can rearrange them.

share|cite|improve this answer
indeed. I agree mostly with you, but still one can avoid to think that for instance, we also have vector spaces and their duals, but have not more complex tuples of vector spaces. What is called again an algebraic object with ternary operators beside the usual binary operators that define groups? we don't have it because we don't think in those terms, not necessarily because it might not make sense – lurscher Dec 14 '11 at 19:32
I agree with the first of the arguments. As for the second, one could easily generalize commutators to n tuples of operators in a manner similar to antisymmetric tensors (i.e. sum over all permutations employing permutation sign). See Levi-Civita symbol. – Adam Zalcman Dec 14 '11 at 20:10
exactly, maybe there is some generalization out there for 'Lie thing of ternary operators' using such symbols as infinitesimal generators, but maybe such math is too weird to fit in our thinking models – lurscher Dec 14 '11 at 20:53
Yeah, I've seen things like this on occasion (like the Nambu bracket Qmechanic mentioned), but they're not as generally useful as the commutator. Albeit perhaps only because the math is more complicated and less intuitive. – David Z Dec 16 '11 at 20:35

1) There are already many good answers explaining the conventional theory and observations. Nevertheless, related to comments by lurscher and Adam Zalcman, it seems appropriate to mention the Nambu bracket, which is a Poisson-like bracket

$$ \{ f,g,h \} $$

with 3 function entries, originally invented by Nambu in 1973, purportedly in a failed attempt to explain the $SU(3)$ symmetry of quarks. Already Nambu discusses an operator 3-bracket

$$ [\hat{f},\hat{g},\hat{h}], $$

and one can imagine some kind of uncertainty relation associated to this, where canonical variables come in triples. Unfortunately, the subject so far has remained just theoretical speculations. [Authors even don't agree what should replace the Jacobi identity for the Poisson bracket $\{ f,g\}$, although most think it should be the so-called Filippov fundamental Identity (FI).]

2) Recently in 2008, the Nambu-bracket has been used in the Bagger–Lambert–Gustavsson M2 brane proposal.

3) The exist various generalizations to higher Nambu brackets

$$ \{ f_1,\ldots ,f_n \} $$

with $n$ entries.

4) For more information, see this recent review.

share|cite|improve this answer

I think this has to do with the fact that one is the infinitesimal generator of the other (i.e p is the generator of position). If I remember correctly, commutators arise naturally when we generate a Lie Algebra (not entirely sure about this).

share|cite|improve this answer

That's not necessarily the case. For instance, besides position and momentum, there are also other observables like energy (and no, time is a parameter and not an operator). More strikingly, for spin, there is no natural pair. There is an angular momentum operator corresponding to the component in each spatial direction, but directions don't come naturally in pairs.

share|cite|improve this answer

[Peter: You asked this some time ago, but it's just too delightful of a question to pass over without adding my two conjugated cents... :) You already have several good answers (I liked David Zaslavsky's in particular), so this one will be concept-heavy and math-light.]

One ways to visualize why quantum uncertainty occurs in pairs is to picture the universe as having two spaces that share a symmetric, two-way relationship.

The first of these spaces is the one we call spacetime, consisting of xyz (space) plus t (time). For brevity I'll call this xyzt space "length space", and will refer to it even more briefly as just x. Length space is of course the space we know best, and is the space in which most classical physics phenomena take place.

The second of spaces is technically an abstraction, but in terms of measurable effects it's about as real of an abstraction as you are likely to encounter. It's called momentum space, and it too has four dimensions: $p_{x}p_{y}p_{z}$, the momentum axes, plus $E$, the energy axis. I'll call $p_{x}p_{y}p_{z}E$ "momentum space," and will also refer to it as just p.

Momentum and energy are of course very much part of classical mechanics. However, in classical mechanics, x dominates over p in much the same way that air dominates over water in a cloud by breaking up the water into many tiny and isolated droplets. This structural domination of x over p means that p shows up mostly in the dynamic properties of how large object move and interact in x, rather than as a space.

However, for objects that are very small, very low in energy, or very low in momentum, that situation changes. For example, there are cases where ordinary matter can "condense" or form fluids that are most easily described as residing in momentum space. You might thing that such bizarre fluids would be rare and exotic, but that’s not the case at all. For example, if you look in a mirror or at a shiny piece of metal you are looking directly at the surface of a type of p space condensate called a Fermi sea. The Fermi sea is composed of conduction electrons, and only electrons at the very top of that sea are capable of reflecting light.

It's often useful to think of p as approximately the place where the odd behaviors of quantum mechanics take place. Why that is I'll explain below, since it's directly related to your question about quantum uncertainty.

Now to the meat of your question: For reasons I think is best described as the way our universe happens to works, there exists an extraordinarily deep and mostly symmetric relationship between x and p. I can describe that relationship this way: Each point in each space x and p behaves a tuned radio receiver for some frequency along the equivalent axis in the other space.

For example, picture the $p_x$ axis of p as an old-style analog radio tuner, the kind with a little red dot indicating the frequency you have selected. Sliding the red dot up to the position labeled “103.5 megahertz” would create in a sinusoidal wave with just that frequency over on the $x$ axis of x. That infinitely long sinusoidal wave then becomes another way to "view" the red dot, since each precisely defines the other.

That’s all well and good, but it’s also important to realize that this works both ways. That is, if you instead decide to select a “dial point” along the length space axis $x$, that too will select a pure frequency along the corresponding axis of the other space, that is, along the $p_x$ of p. That's a nice symmetry!

It's a symmetry with some profound consequences. Remember, if you select a compact point in p you get an very long sinusoidal frequency over in x, and vice-versa. So let's say you don't want to have your particle scattered about so much in x, and you decide to pull it all in to one compact point in x. Oops! The moment you do that, the symmetric relationship causes the particle's representation in p to explode into an indefinitely long sinusoidal wave.

Now, since x corresponds to length (and time) and p to momentum (and energy), what does this curious see-saw relationship remind you of? If you answered quantum uncertainty, you are exactly correct. In fact, that is what quantum uncertainty is, at its deepest levels: the impossibility of representing the same particle compactly in both x and p space at the same time, due to their symmetric "tuning" relationship to each other.

The radio-dial like frequency tuning of these spaces has a more precise name: It's called a Fourier transform. So, if you have an old-style radio receiver and use it to move a dot back and forth along a linear scale, you can correctly say that you are doing a Fourier transform from frequency of the radio wave to the linear position on your radio dial.

So, here at last is the answer to your original question quantum uncertainty always seems to express itself in pairs, rather than in triples or some other combination.

The answer is that uncertainty is defined by pairs of spaces connected in both directions by Fourier transforms, which in turn just seems to be how our universe is constructed. To create a quantum uncertainty triplet, you would need to postulate a new universe in which there was a three-way Fourier-like relationship, presumably extending across three parallel spaces. Ouch! That’s not intuitive at all.

But also, it’s not necessarily impossible. Some of the most interesting and important insights in both mathematics and physics have come about from people trying to do nominally “impossible” things that turned out to have interesting and non-trivial solutions. Dirac’s prediction of antimatter certainly comes to mind, since in that case he dug in and got stubborn about wanting only “linear” solutions for some equations that didn’t have linear solutions. By using complex matrix mathematics, he finally found a way to do it -- and simultaneously uncovered one of the deepest equations in physics. His instincts, and his stubborn insistence in following through on them, ended in a unique and spectacular success.

(And warning to the wise: Very few people indeed have physics instincts like Dirac did, and even Dirac got stubborn in some quite wrong-headed ways in his later years. So if anyone wants to try exploring ideas like these for fun, please use a bit of caution on adopting the "stubborn” part of the Dirac insight recipe!)

And finally: I said earlier that x and p are not exactly symmetric. What did I mean by that? The back-and-forth Fourier relationship is exactly symmetric, as best we can tell. So where does the breakdown in symmetry come in?

The main break is this: While adding size or distance costs nothing x (length) space, adding size in p (momentum) space turns out to be very costly. This means that there is no problem with placing a particle very precisely in p, even if it results in an absolutely enormous (e.g. light years across) wave function in ordinary x space. Since adding more size to the x wave function costs nothing, it can just keep spreading until something disrupts it.

If instead you try to stuff the particle into a very precise location in x, the wave function in p again become enormous. However, since momentum cost energy, a large wave function in p quickly becomes hugely expensive in terms of the mass-energy needed to sustain it. That’s one reason why particle accelerators have to pile so much energy into their particles. The extra energy expands their energy-intensive p space wave until they are large enough to make their sizes in ordinary x space very small indeed. Such additions of energy can for example be used to locate electrons precisely enough in space for them to "see" the motions of individual quarks inside protons and neutrons.

So, massive overkill, but again: That was a good question. And who knows? Perhaps some clever mathematically inclined person will someday come up with an unexpected n-space generalization of the Fourier 2-space relationship that seems to guide our universe. As the old saying goes, the only sure way to fail is never to try.

share|cite|improve this answer

The generalized uncertainty principles relates any two observables (see eg the Schrödinger uncertainty relation).

It probably can be further generalized to more than two observables, possibly by starting from $$ \langle x,x\rangle\langle y,y\rangle\langle z,z\rangle \geq \frac13 \left(\langle x,x\rangle|\langle y,z\rangle|^2 + \langle y,y\rangle|\langle x,z\rangle|^2 + \langle z,z\rangle|\langle x,y\rangle|^2\right) $$ instead of straight Cauchy-Schwarz in case of three observables, but this is just a guess on my part - I did not check any literature.

Related to this are canonically conjugate observables like $x,p_x$, for which the classical uncertainty principle holds because they are governed by the commutation relation $$ [x,p_x]=i\hbar $$ corresponding to the fundamental Poisson bracket $$ \{x,p_x\}=1 $$ of classical mechanics.

However, these observables are not ordinary: In particular, the observable $x$ corresponds to a coordinate of the underlying configuration space.

In classical mechanics, the existence of a conjugate momentum follows from the symplectic structure of phase space, in quantum mechanics, it follows from the properties of the Fourier transform on our state space $L^2$.

share|cite|improve this answer

It's because physical observables are operators in quantum mechanics. For two operators $F,G$, one may always compute the commutator $$ [F,G] = FG-GF $$ which is the simplest quantity that should be zero in classical physics but it is nonzero in quantum mechanics. In general, the commutator is another observable (it may be a $c$-number, too). In the leading approximation, the commutator is equal to $i\hbar$ times the classical "Poisson bracket" of the classical observables with the same name.

But if you define your classical initial state by the values of observables $Q_i$ (includes both positions and velocities or momenta in the familiar case), you may study configurations close to $Q_i$. In this way, you linearize the phase space and the commutator $$ [Q_i,Q_j] = \omega_{ij} $$ may be approximated by a constant antisymmetric matrix that I called $\omega$. (It becomes a nonconstant function of $Q_i$ if you allow large deviations from the predetermined point.) It defines the so-called symplectic structure on the phase space. In this linearized approximation where the commutator is constant, the symplectic structure may always be diagonalized by choosing better combinations of $Q_i$, so that $\omega$ becomes a block diagonal matrix composed of $$ \left ( \begin{array}{rr}0&+1\\-1&0\end{array} \right ) $$ blocks, and these blocks guarantee that the observables defining the initial conditions - coordinates on the phase space - may be divided to pairs. In a simpler way, your question is really "because $\omega_{ij}$ has two indices".

The rearrangement of a larger number of operators, e.g. $XYZT-YTZX$, may always be reduced to transpositions of the neighbors i.e. to commutators of two operators. That's why the ordinary commutators are always more important. In a group-theory context of mathematics, the commutators define the so-called "Lie algebra". Generalizations of commutators with more than 2 objects in the brackets also exist but they seem to be much less relevant physically. In particular, they're not naturally defined for operators - and physical observables become operators in QM.

share|cite|improve this answer

Your Answer


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.