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Usually in all the standard examples in quantum mechanics textbooks the spectrum of the position operator is continuous.

Are there (nontrivial) examples where position is quantized? or position quantization is forbidden for some fundamental reason in quantum mechanics (what is that reason?)?

Update: By position quantization I mean, if position (of a particle say) is measured we get only a discrete spectrum (say 2.5 cm and 2.7 cm, but nothing in between, just in the same way that energy levels can be discrete). In that sense interference patter of photons on a photographic plate cannot be considered as position quantization because the probability density varies "continuously" from maximum to zero (or am I wrong?)

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position quantization is equivalent to periodic momentum, and this happens in a crystal. Position is quantized in a crystal, on a grid. –  Ron Maimon Jul 23 '12 at 19:29
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5 Answers

Position quantization in vacuum is forbidden by rotational, translational, and boost invariance. There is no rotationally invariant grid. On the other hand, if you have electrons in a periodic potential, the result in any one band is mathematically the theory of an electron on a discrete lattice. In this case, the position is quantized, so that the momentum is periodic with period p.

Fourier duality

The quasimomentum p in a crystal is defined as i times the log of the eigenvalue of the crystal translation operator acting eigenvector. Here I give the definition of crystal position operators and momentum operators, which are relevant in the tight-binding bands, and to describe the analog of the canonical commutation relations which these operators obey. These are the discrete space canonical commutation relations.

In 1d, consider a periodic potential of period 1, the translation by one unit on an energy eigenstate commutes with H, so it gives a phase, which you write as:

$$ e^{ip}$$

and for p in a Brillouin zone $-\pi<0<\pi$ this gives a unique phase. The p direction has become periodic with period $2\pi$. This means that any superposition of p waves is a periodic function in p-space.

The Fourier transform is a duality, and a periodic spatial coordinate leads to discrete p. In this case, the duality takes a periodic p to a discrete x. Define the dual position operator using eigenstates of position. The position eigenstate is defined as follows on an infinite lattice:

$$ |x=0\rangle = \int_0^{2\pi} |p\rangle $$

Where the sum is over the Brillouin zone, and the sum is over one band only. This state comes with a whole family of others, which are translated by the lattice symmetry:

$$ |x=n\rangle = e^{iPn} |x=0\rangle = \int_0^{2\pi} e^{inp} |p\rangle $$

These are the only superpositions which are periodic on P space. This allows one to define the X operator as;

$$ X = \sum_n n |x=n\rangle\langle x=n| $$

The X operator has discrete eigenvalues, it tells you which atom you are bound to. It only takes you inside one band, it doesn't have matrix elements from band to band.

The commutation relations for the quasiposition X and quasimomentum P is derived from the fact that integer translation of X is accomplished by P:

$$ X+ n = e^{-inP} X e^{inP}$$

This is the lattice analog of the canonical commutation relation. It isn't infinitesimal. If you make the translation increment infinitesimal, the lattice goes away and it becomes Heisenberg's relation.

If you start with a free particle, any free $|p\rangle$ state is also a quasimomentum p state, but for any given quasimomentum p, all the states

$$ |p + 2\pi k\rangle $$

have the same quasimomentum for any integer k. If you add a small periodic potential and do perturbation theory, these different k-states at a fixed quasimomentum mix with each other to produce the bands, and the energy eigenstates $|p,n\rangle$ are labelled by the quasimomentum and the band number n:

you define the discrete position states as above for each band

$$ |x,n\rangle = \int_0^{2\pi} e^{inp} |p,n\rangle $$

These give you the discrete position operator and the discrete band number operator.

$$ N |x,n\rangle = n |x,n\rangle $$

if you further make the crystal of finite size, by imposing periodic boundaries in x, the discrete X become periodic and p becomes the Fourier dual lattice, so that the number of lattice points in x and in p are equal, but the increments are reciprocal.

This is what finite volume discrete space QM looks like, and it does not allow canonical commutators, since these only emerge at small lattice spacing.

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Do you mean the "momentum" with period p or the position? –  anna v Jul 23 '12 at 15:26
    
@annav: I mean the momentum periodic with period p which is the same as the position quantized is a lattice of size 2pi/p. –  Ron Maimon Jul 23 '12 at 18:39
    
"Position quantization in vacuum is forbidden by rotational, translational, and boost invariance" Wow... This is an awesome insight that I never thought of. I always wondered myself if space was quantized, like at plank distance, for example. But if there was a "universal grid of spacetime" then you would loose translational and rotational symmetry of natural laws. And that would cost you conservation of linear and angular momentum respectively. en.wikipedia.org/wiki/… –  John Jul 24 '12 at 14:15
    
Actually, if posistion in space was quantized, in order for translational and rotational symmetry to hold, and therefore, for the conservation laws to hold, wouldn't momentum need to be quantized? If mass, time, and energy were all quantized, wouldn't that be possible? Or no? –  John Jul 24 '12 at 14:23
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@John: I don't know if you're serious, but yes, you lose conservation laws. It's not completely trivial in field theory, because you can have emergent rotational invariance at long distances, like an atomic lattice can have rotationally invariant sound waves. This doesn't work because the corrections are going to be too big from natural lattice scale at the Planck scale. It also doesn't work in string theory, where the space is holographicaly reconstructed from the symmetries of the boundary, SUSY and spacetime, and you don't have a lattice, and the symmetry is the fundamental ingredient –  Ron Maimon Jul 24 '12 at 14:34
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So, suppose you have an eigenstate of $\hat{x}$: $$\hat{x}|\psi\rangle = x|\psi\rangle$$ Now let us act with $\hat{x}$ on $e^{i\hat{p}\delta}|\psi\rangle$, and use this formula (I have $\hbar=1$): $$\hat{x}e^{i\hat{p}\delta}|\psi\rangle=e^{i\hat{p}\delta}\left(\hat{x}+i\delta[\hat{p},\hat{x}]+\frac{i\delta}{2!}[\hat{p},[\hat{p},\hat{x}]]+\frac{i\delta}{3!}[\hat{p},[\hat{p},[\hat{p},\hat{x}]]]+...\right)|\psi\rangle=$$ $$=e^{i\hat{p}\delta}(\hat{x}+\delta)|\psi\rangle=(x+\delta)e^{i\hat{p}\delta}|\psi\rangle$$ So you have another eigenstate, shifted at arbitrary distance $\delta$.

It seems that this result holds for any pair of observables with constant commutator.

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From this derivation it seems it works for $[P,X]=f(X)$, where $f(X)$ might be more complicated than $1$. Then you don't get the $+\delta$, but a shift by a function of $\delta$ and the eigenvalue $x$. (I don't know about convergence and reality of the series though.) –  NiftyKitty95 Jul 23 '12 at 20:46
    
This also assumes constant commutator. The commutation relation has to fail for discrete space time--- it fails for position and momentum operators for quasiparticles in the lattice, hence you get Brillouin zones. Reducing the dynamics of each band, you get the right lattice QM formulation. It's only physical problem is lack of symmetry, no rotations and no boosts. –  Ron Maimon Jul 24 '12 at 1:50
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@RonMaimon I'm really puzzled about this crystal thing everyone is talking about. As far as I remember the commutation relations for crystal modes are still $[x_k,q_l]=i\delta_{kl}$, aren't they? –  Kostya Jul 24 '12 at 9:05
    
@RonMaimon: I have the same problem as Kostya and have asked for references on details of the whole definition of the so called x/p operators in the case of crystals. I want to see please their deduction and interpretation. However discrete lattice anyone talks about, that cannot be fundamental, as atoms vibrate around equilibrium positions coming from main expectation values of quantized phase space position, thus subject to ordinary canonical commutation relations and the uncertainty principle. Misunderstanding a discrete model for collective latticed equilibrium averages with real position. –  Javier Álvarez Jul 24 '12 at 9:20
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@JavierÁlvarez: I dislike it not because of "conceptualization of scientific models of theories", but because the assumption of canonical commutation is exactly equivalent to the assumption of continuous space, and it is obvious that one implies the other. If someone is asking why X is discrete, one cannot use canonical commutators to argue it, because it is an equivalent assumption. You need to understand what a discrete X QM looks like, and argue that it doesn't correspond to nature, not that it is forbidden by a mathematical identity which is derived from continuity assumption. –  Ron Maimon Jul 24 '12 at 11:38
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The answer is essentially what Kostya has pointed out:

Position is quantized but has a continuous spectrum of (generalized) eigenvalues because the canonical commutation relations on position and momentum forbid that both of them be bounded operators (and act on finite-dimensional state spaces) by Stone-von Neumann theorem. This means that, given general constraints, at least one of them must be unbounded and thus must have a nonemtpy continuum spectrum, implying that the resolution of the identity must be in terms of an integral over non physical states. This is why in general a specific point-like position is not a physical observable eigenstate of a system, and must be spread over a dense interval (which is related to the instrumental resolution of the measurement): e.g. particles can have as much localized wave packets as our resolution allows, but they have no defined positions even in principle (at least in standard quantum mechanics). Thus the canonical commutation relations forbid you have discrete position if momentum is discrete by boundary conditions (e.g. particle in a box). This may seem trivial because position and momentum are the generators of translations of each other, but the point of the noncommutativity and the theorem is that one of them can indeed have discrete spectrum (usually boundary conditions discretize momentum, so position generalized eigenvalues are continuous).

Concering semantics of "quantized", some observable property is quantized if it is an operator in the quantum physical state of the system, regardless of the discreteness or continuity of its spectrum of eigenvalues because, even in the continuum case, it is subject to the formalism of quantum mechanics: noncommutativity, uncertainty, probabilistic expected values ... Quantum jumps in possible values of discrete spectra are just the most remarkable property of the quantum world, without analogue in the classical world, hence the name quantum mechanics, but that is not a necessary condition. Since there are purely quantum degrees of freedom (e.g. spin), quantization is not fundamental, but they are nevertheless quantum observables and not classical in both senses: spin observables belong to an operator algebra formalism, they are not a function on classical phase space, and have discrete spectrum, thus being quantized (or more properly, "quantum-mechanical") in any meaning of the word.

UPDATE on crystal lattices: what Ron is calling realization of discrete space might be very misleading. Any nonrelativistic space lattice model (i.e., in non quantum gravity theories) is an effective discrete model for constrained main expectation values of real position observables. The quasi-position/momentum of crystals, and thus crystal lattices of condensed matter, are an emergent property out of the symmetries of the approximate arrangement of equilibrium positions of many atoms. Any position measurement on any atom is however not point-like defined, but just highly localized around the lattice nodes. From a very rigorous point of view, one should distinguish the fundamental degrees of freedom of the system from effective quantities emergent from the system's symmetries. In the case of solid state matter, the system is composed of a huge collection of atoms which whose collective interactions constrain their localization around well-defined points of equilibrium. Therefore at the structural level we can talk about quasi-classical atoms at fixed positions (the maxima of their position probability distributions), thus creating an effective lattice of discrete quasi-position where the rest of the constraints and properties of the system (periodic potentials, momentum...) give an emergent model for quasi-particles which may seem like discretized space and momentum simultaneously. I defend that thinking of this as purely quantum mechanical position is misleading, because the lattice as a whole is classical, although discrete, as its structure is not subject to superpositions in a Hilbert space, to the noncommutativity and uncertainty principle, and its (constant) structure is not a static solution of a collective system's Schrödinger equation. The fundamental degrees of freedom are the quantum position and momentum operators of each atom, always subject to canonical commutation relations and thus continuum spectrum of one of the two, along with uncertainty in position-momentum. A quasi-position and quasi-momentum of a quasi-particle on a crystal lattice is NOT realized by the eigenstates of any actual quantum particle. In this sense, crystal lattices do not serve as example of "quantized position" (meaning discrete spectrum), as long as one retains the noun "position" to mean ordinary position. Real atoms in any crystal have nonzero bounded temperature, forbidding them to have definite constant positions: upon measuring position of any atoms in a crystal, the atom will appear localized with a distribution probability around the lattice node, but the real observable position of the atom need not be the node itself (in fact is always a region, since point-like positions are not eigenstates, only wave-packets).

The quantum model Ron talks about in his answer is useful and nice for the effective lattice I talked about above. I just do not call position to something which is not what we observe when measuring position on quantum particles, that is why I talk about quasi-position in that case. In absence of evidence for a discrete quantum spacetime, quantized phase space is what is traditionally called in theoretical physics as far as I know. Therefore, any real quantum position and momentum, in the sense of the quantum operators whose expectation values "obey" (in the sense of Ehrenfest Theorem) classical Hamilton equations, and whose Schrödinger evolution "obey" (in the sense of eikonal approximations) classical Hamilton-Jacobi equations, are not simultaneously discrete.

The only theories where quantum position gets discretized spectrum is in quantum gravity theories like loop quantum gravity, where the fundamental degrees of freedom of space (and time) itself get quantized (do not confuse quantum position as before, which is newtonian position related to reference bodies, with position as understood in general relativity, where spacetime is the gravitational field). There you get a granular graph of nodes forming space itself, and our traditional continuum positions are just approximations (even at the atomic level) of the relational position given by a flat gravitational field.

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I stated clearly what I meant by the word quantization in the question. I did not mean the quantization "procedure" by which one promote the classical momentum for example into an operator which equals $-i \hbar \nabla$. In that sense Spin is not quantized because it has no classical analog. Anyway, again, I was talking about the quantization of spectrum. –  Revo Jul 23 '12 at 17:31
    
I have added more to my answer at the beginning, hoping to answer more concretely the discrete spectrum issue for position due to the canonical commutation relations. –  Javier Álvarez Jul 23 '12 at 18:32
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-1: This answer is basically saying [x,p]=i means no quantized x. This is mathematically true, but both obvious and ridiculous--- the very idea of an infinitesimal generator of x translations means x is continuous from the beginning. The deduction is false for the example of a quasiparticles on a crystal lattice of finite volume where the quasiparticle momentum and position are both discrete. It's still QM, but canonical commutation fails, you get the discrete version with exponentials of P instead, and this is the right mathematical statement of discrete x. The rest is wordy and irrelevant. –  Ron Maimon Jul 23 '12 at 19:33
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On the other hand, disqualifying other people's answer as wordy and irrelevant is subjective and unnecessary. Indeed it was a wordy answer (now trimmed to the obvious and ridiculous) but not irrelevant at all if the person who wants to know about quantization, spectra and position likes a deeper interrelation of ideas. For example a careful distinction between quantized observable and discrete spectrum and its relation to the different mathematical setting of classical and quantum mechanics. However, to satisfy users like Ron Maimon, some of us prefer to remain brief, obvious and ridiculous. –  Javier Álvarez Jul 23 '12 at 20:10
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@JavierÁlvarez: Ok, I read the extended answer--- I agree with the additions for the most part, but there is one point which I disagree very strongly--- even if you quantize the atomic lattice, so that you consider quanutm nuclei interacting with quantum electrons, you still get the discrete lattice emerging in a band. This holds even at finite temperature, up to the melting temperature, it's an example of spontaneous symmetry breaking of translation symmetry, and it doesn't depend on having an external potential. The only thing it requires is a very large system, so that you approximate SSB. –  Ron Maimon Jul 24 '12 at 12:27
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It depends on what one defines as "position".

In crystals, for example, there exists a three dimensional grid on which the atoms are allowed , stacked in unit cells, so there is quantization in space to be observed, and quantum mechanical solutions are involved . More numerous are the interference solutions of qm waves which also display a quantization of space, where some positions are more probable than othes. So it is not true that position is not quantizable.

All the solutions where the energy is quantized involve matter and potentials also. A free particle does not display a quantization of energy as it does not display quantization of position.

If the question is addressing whether intrinsically space is quantized, as it would be also if one asked if energy is intrinsically quantized, i.e. comes in a minimum packet, that is another question.

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One must take into consideration the fact that the notion of 'particle position' in quantum mechanics is meaningless. One cannot talk about the position of a particle just like one cannot talk about a specific path taken by the particle. The position of a particle in quantum mechanics is not a dynamic variable like it is in Newtonian mechanics, it does not exist as such. The commutation relation [p,x]=ih(bar) used by some tells us, that it is not possible to measure both x and p with arbitrarily high accuracy, hence Heisenberg's uncertainty principle results from it. A good clue whether position is quantised or not can be found by observing the TDSE (time dependent Schrodinger equation.) The mathematical structure of the equation, being a differential equation in x,y,z and t, requires that both position and time must be continuous variables. This requirement is a necessity for the definition of the wave function, but nobody knows where the particle described by the wave function actually is, let alone whether its position is quantised or not! One should not confuse the quantisation of the electronic orbitals in atoms with position quantisation.

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In non-relativistic quantum mechanics position is a dynamical variable, according to the usual definition: an operator that evolves according to the equation of motion $\dot{x} = i [H,x]$ in the Heisenberg picture. It is only in QFT that position is demoted to a continuous index, like time in non-relativistic QM. Also, it is not "meaningless" to talk about the position of a particle, it is simply that you cannot know the position to arbitrary precision, due to the commutation relations as you mentioned. –  Mark Mitchison Jan 9 '13 at 22:26
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Thanks for you response. The question is both mathematical and physical. Ehrenfest's theorem can give the evolution of the mean position, <x>, of the particle accurately. If as you say "..., it is simply that you cannot know the position to arbitrary precision, ..." then, is it still meaningful asking where the particle is? The impossibility of knowing exactly where the particle is, is not subject to improvents by higher precision measurements, but it requires a compromise in the precision of the momentum. Only the wave function can tell us where the particle can be, but probabilistically. –  JKL Jan 13 '13 at 12:02
    
I would argue that position is a meaningful concept, it just has to be revised from the naive classical concept. It is surely meaningful to ask where an electron is. For example, Young's double slit experiment relies on distinguishing where electrons land on a screen (obviously with some intrinsic uncertainty); this is how one builds up the famous interference pattern. Etcetera etcetera. But anyhow, this is more a matter of terminology/philosophy than physics, I just wanted to bring to your attention the possibility that your phrasing could be a bit misleading if taken literally. –  Mark Mitchison Jan 13 '13 at 13:42
    
Thanks for that. Bohr and Einstein argued, about the meaning of QM and whether objective reality of things makes any sense, untill the end of their lives! The thing that is important is what is demonstrated by the experiment. To say that the particle has gone through one slit or the other in Young's DSE is one thing. When you do the experiment to test this out, you soon find that the interference pattern is gone and for a good reason. The spot of the electron position on the screen is only a result of its interaction, w-f collapse as they call it. So far Bohr seems to have won the argument! –  JKL Jan 13 '13 at 20:32
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