Why position is not quantized in quantum mechanics?

Question

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?)

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.