I Why is position-space favored in QM?

[LarryS]

In non-relativistic QM, when one speaks of a "wave function" it is understood that one is referring to the position-space version of the wave function. Even if the observable being measured is other than position, like momentum or energy, the associated eigenfunctions are always from the position-space point-of-view. Why is that? It seems to me that it would be simpler to express the wave function in the same "space" as the observable - that way the eigenfunctions would always be just Dirac delta functions.

 

[dRic2]

I think you choose the space to get the "nicest" equation possible. Usually the differential equation resulting from choosing the X basis (position) is the simplest to solve. That is just my opinion though.

 

[ajeet mishra]

Because most of the times it is easier to solve Schrodinger equation in position basis. As we know Hamiltonian H is given by
H= ^p2/2m + V(x)
Now when V(x) is simple function of x ( like harmonic oscillator ) then both position and momentum basis can be used to solve the problem with almost the same effort but if the potential is complicated function of x( like exp(x)) then use of position basis will simply change it into eigenvalue equation saving a lot of effort and time

 

[bhobba]

Because Schrodingers equation is usually written in wave-function form - the wave-function technically is the expansion of the Ket in position eigen-kets.

Thanks
Bill

 

[atyy]

There is not much reason, except habit.

One excuse is that ultimately, even when we measure momentum, for example, the final measurement or reading is a position measurement of another observable (eg. the position of the dial on the measurement apparatus).

However, you are absolutely right that it is a good computational trick to use the same basis as the observable - this is called "diagonalization".
https://ocw.mit.edu/courses/physics...all-2013/lecture-notes/MIT8_05F13_Chap_05.pdf (see section 5: 5 Diagonalization of Operators)https://ocw.mit.edu/courses/physics...all-2013/lecture-notes/MIT8_05F13_Chap_05.pdf
https://en.wikiversity.org/wiki/Quantum_Simulation/Exact_diagonalization

 

[PeterDonis]

that way the eigenfunctions would always be just Dirac delta functions.

This is true for continuous observables, but not for discrete ones like spin.

 

[vanhees71]

The deeper reason for why the position representation in non-relativistic QM is so useful is the fundamental structure of this particular realization of a QT, which is governed to a large extent by the underlying symmetry principles, most importantly of space-time. In this case it's the Galilei group, which is a bit tricky and from a mathematical point of view even more complicated than the Poincare group underlying special-relativistic spacetime. The analysis of the ray representations of the Galilei group or, (nearly) equivalently, the unitary representation of the central extensions of its covering group (which adds to the classical Galilei Lie algebra the mass as a non-trivial central charge and substitutes the rotation group SO(3) by its covering group SU(2)), leads to the typical structure of the Hamiltonian in terms of kinetic energy plus a potential that is only a function of the position variables, where one should be aware that the position variables are constructed from the representations of the quantum Galilei group for $m \neq 0$; the case $m=0$, i.e., the unitary representations of the classical Galilei group do not lead to useful quantum dynamics (Wigner and Inönü).

In the relativistic case the Poincare group has no non-trivial central extensions, and the mass squared is a Casimir operator of its Lie algebra. So far only the cases $m^2 \geq 0$ lead to successful models for physics. For $m^2>0$ one can always construct a position observable from the Lie algebra of the (positive orthochronous) Poincare group, but in the relativistic regime particles are not as localizable (i.e., it's position cannot be determined to be in a small volume) that easily. Losely speaking in an attempt to squeeze the particles into a small volume you need pretty high energies, which rather leads to the creation of new particles than a more accurate localization of the present ones.

For massless quanta of spin $\geq 1$ a position observable cannot even be fully defined. As with classical light (em. waves) photons can only be localized in some sense in transverse direction but not in beam direction. In lack of a proper definition of a position observable, what's meant by "localization" is rather a detection probability distribution that is narrowly peaked in some volume. The detection probability distributions are essentially what defines the observable facts about photons. In the most adequate formalism, which is QED, it's well defined by adequate correlation functions of field operators.

 

[LarryS]

Interesting arguments. The one common thread seems to be that the potential energy is usually a function of position only.

 

[Urs Schreiber]

Curiously, in quantum field theory it is much more popular to work in momentum space instead.

Ironically, here, too, working in position space (e.g. renormalizing in position space) is actually more robust, because position-space methods in QFT generalize to QFT on curved spacetimes (e.g. CMB fluctuations, Hawkind radiation), where the Fourier isomorphism is in general not available, globally.

(The position-space formulation of renormalized perturbative QFT goes back to Epstein-Glaser 73.)

 

[vanhees71]

But in QFT time and space coordinates are just parameters. Time is a parameter in QT anyway (in order to have stable groundstates, as was made very clear already by Pauli in his very readable encyclopedia article on wave mechanics) and space coordinates are labels of the continuous number of degrees of freedom of a field. For free fields and thus the interaction picture it's often convenient to build up the Fock raum in terms of the momentum-energy-spin/helicity eigenbasis. For massless particles with spin $\geq 1$ there's not even a position observable and thus also no position representation. To use the momentum-energy-spin/helicity eigenbasis to build up the Fock space is convenient in perturbation theory because of energy-momentum conservation and the simplicity of the free propagators. You can, of course, formulate everything in position space parametrization.

 

[LarryS]

But in QFT time and space coordinates are just parameters.

I've read this before. But how can that be if QFT is relativistic? In relativity, time and space are much more than just abstract indexes. They form a unified space-time with the Minkowski metric.

 

[PeroK]

I've read this before. But how can that be if QFT is relativistic? In relativity, time and space are much more than just abstract indexes. They form a unified space-time with the Minkowski metric.

It's often said that understanding that coordinates are arbitrary in General Relativity is hard for students to grasp. A case in point is the singularity (in Schwarzschild coordinates) at the event horizon of a black hole. If you change your coordinates, the singularity disappears (e.g. using the wonderfully named Eddington-Finkelstein coordinates).

 

[microsansfil]

Hello,

There is two notions of time : proper time (invariant relativist) and coordinate times (arbitrary). In special relativity particles are described as points, as in classical mechanics. A “particle at a given instant” (proper time) will be represented by a point in the spacetime E (an event), and the “successive positions” of the particle will draw a one-dimensional curve in the affine space E (worldline of the particle).

Relativity has banished the concept of absolute time. It however introduces along each worldline a privileged time: that given by the metric tensor according to : (2.6).

what kind of "time" is it question ?Patrick

 

[PeterDonis]

how can that be if QFT is relativistic? In relativity, time and space are much more than just abstract indexes. They form a unified space-time with the Minkowski metric.

He didn't say "time and space", he said "time and space coordinates". The coordinates are parameters; each unique set of 4 coordinates labels a distinct event in spacetime, so spacetime can be thought of as a family of events parameterized by the coordinates. And each event has its own set of quantum field operators in QFT; so the coordinates parameterize the operators just as they parameterize the events.

 

[LarryS]

He didn't say "time and space", he said "time and space coordinates". The coordinates are parameters; each unique set of 4 coordinates labels a distinct event in spacetime, so spacetime can be thought of as a family of events parameterized by the coordinates. And each event has its own set of quantum field operators in QFT; so the coordinates parameterize the operators just as they parameterize the events.

Okay. Yet the only way to compute the (Minkowski) spacetime separation between any two of these spacetime events is by using the values of their corresponding coordinate/parameters. Is that correct?

 

[PeterDonis]

the only way to compute the (Minkowski) spacetime separation between any two of these spacetime events is by using the values of their corresponding coordinate/parameters.

Yes. Spacetime intervals are invariants. But that does not mean coordinates aren't parameters.

 

[Demystifier]

But in QFT time and space coordinates are just parameters. Time is a parameter in QT anyway (in order to have stable groundstates, as was made very clear already by Pauli in his very readable encyclopedia article on wave mechanics) and space coordinates are labels of the continuous number of degrees of freedom of a field. For free fields and thus the interaction picture it's often convenient to build up the Fock raum in terms of the momentum-energy-spin/helicity eigenbasis. For massless particles with spin $\geq 1$ there's not even a position observable and thus also no position representation. To use the momentum-energy-spin/helicity eigenbasis to build up the Fock space is convenient in perturbation theory because of energy-momentum conservation and the simplicity of the free propagators. You can, of course, formulate everything in position space parametrization.

For some reason, particle physicists cannot accept that
$$QFT\neq RQFT$$
i.e. that quantum field theory is not the same as relativistic quantum field theory. Your claims above are true in relativistic QFT, but not necessarily in non-relativistic QFT.

 

[Demystifier]

Because most of the times it is easier to solve Schrodinger equation in position basis. As we know Hamiltonian H is given by
H= ^p2/2m + V(x)
Now when V(x) is simple function of x ( like harmonic oscillator ) then both position and momentum basis can be used to solve the problem with almost the same effort but if the potential is complicated function of x( like exp(x)) then use of position basis will simply change it into eigenvalue equation saving a lot of effort and time

Exactly, the x-representation is preferred because V is a simple function of x.

 

[Physics Footnotes]

In non-relativistic QM, when one speaks of a "wave function" it is understood that one is referring to the position-space version of the wave function. Even if the observable being measured is other than position, like momentum or energy, the associated eigenfunctions are always from the position-space point-of-view. Why is that? It seems to me that it would be simpler to express the wave function in the same "space" as the observable - that way the eigenfunctions would always be just Dirac delta functions.

Because our direct empirical experience of the world is ultimately spatio-temporal in character (dots, flashes, tracks, etc.). Physical quantities like energy, momentum, and spin, are conceptual tools to organize that experience.

Take a Stern-Gerlach experiment, for example. You don't actually measure the spin of a silver atom. What you measure is the location of a dot on a plate. You infer the spin from a conceptual/mathematical analysis, but the empirical test is always a matter of gathering spatio-temporal information and comparing it with a mathematical prediction. And that means that, at the end of the day, if you want to test a theory encoded in the language of energy/momentum/spin-space, you're going to have to translate your mathematics into a form which predicts a spatio-temporal probability distribution.

 

[vanhees71]

I've read this before. But how can that be if QFT is relativistic? In relativity, time and space are much more than just abstract indexes. They form a unified space-time with the Minkowski metric.

Of course, in quantum theory, space and time (or Minkowski spacetime of special relativity) are just taken as the classical concepts you learn about from the first mechanics lecture on.

Quantum theory deals with the description of matter in spacetime, not of spacetime itself. I think, it's the most puzzling unsolved question of physics to find a quantum theory of spacetime which might then include the gravitational interaction in quantum theory, as the gravitational interaction within classical physics is described as spacetime as a dynamical entity within the General Theory of Relativity (Einstein 1915).

At present thus we have a classical spacetime within which the matter is dynamically described by relativistic quantum field theory. Now within this theory you can indeed define a position observable for any massive particle, which implies that you can determine a massive particle's position by some measurement. This is not possible for massless particles with a spin $\geq 1$. Indeed, e.g., photons cannot be localized, and thus something like a photon wave function does not exist. You can only determine the probabilities for detecting a photon with the detector put at some place, and these probabilities are given by certain expectation values of field-operator products, the socalled Wightman functions.