I Why QFT still goes well while it lacks the notion of wave function?

[A. Neumaier]

Interesting, thanks. I would guess things get hairier in cases where one doesn't have any sense of a weak-coupling, like certain CFTs. I would also guess that 2d CFTs are at least better understood since mathematicians have spent a lot of time with them, but as far as I'm aware their Hilbert space is largely studied on the cylinder where the spectrum is gapped anyways.

The 2D case is quite well understood, as there are many exactly solvable models in 2 spacetime dimensions, and many others. Many exactly solvable models do not have a Fock space like structure, and cannot be deformed into Fock spaces.

But 2D is also very exceptional, featuring things like nonstandard statistics, bosonization and fermionization, and quantum group symmetries, all of which are absent in 4D. Thus extrapolating from 2D to 4D is a dangerous business.

 

[pavsic]

The absence of Lorentz invariance:
Let be a position eigenstate. Then

Position can be defined with respect to a reference frame, i.e., on a "simultaneity hypersurface" $\Sigma_\mu$. When passing from one reference frame to another, the hypersurface $\Sigma_\mu$ transforms as a Lorentz 4-vector. Position on the hypersurface can be denoted by a 4-vector ${\bar x}\equiv {\bar x}^\mu$, $\mu=0,1,2,3$.
The quantum position states are $|{\bar x} \rangle$, and they satisfy
$$\langle {\bar x}_1|{\bar x}_2 \rangle = \delta^4 ({\bar x}_1 - {\bar x}_2) ,$$
which is a covariant relation. Namely, it has the same form in all Lorentz reference frames, i.e., in another frame it reads
$$\langle {\bar x}'_1|{\bar x}'_2 \rangle = \delta^4 ({\bar x}'_1 - {\bar x}'_2),$$
where the primed position 4-vectors are Lorentz transforms of the unprimed ones.
This was just a quick, superficial explanation aiming at pointing out the difference between invariance and covariance, giving an idea that all this subject of relativistic position states can be formulated covariantly. More explanation can be found in the article
Manifestly Covariant Canonical Quantization of the Scalar Field and Particle Localization
and in the chapter "Particle Position in Quantum Field Theories" of the book
https://www.worldscientific.com/worldscibooks/10.1142/11738

 

[PeterDonis]

The quantum position states are $|{\bar x} \rangle$, and they satisfy
$$\langle {\bar x}_1|{\bar x}_2 \rangle = \delta^4 ({\bar x}_1 - {\bar x}_2) ,$$
which is a covariant relation.

But this doesn't work as soon as you have more than one particle. There is no way to covariantly write down a state with two particles at different positions.

 

[Demystifier]

But this doesn't work as soon as you have more than one particle. There is no way to covariantly write down a state with two particles at different positions.

Why not
$$\langle x_1,x_2|x'_1,x'_2\rangle =\delta^4(x_1-x'_1) \delta^4(x_2-x'_2)?$$
Do you see anything wrong with that?

 

[vanhees71]

Well, as usual the solution is the quantum-field theoretical approach. There you also clearly can answer the question whether there exist position operators in the usual sense or not. The result is that for free (and thus also asymptotic free) fields, which have a clear "particle interpretation" all fields corresponding to a massive representation of the proper orthochronous Poincare group allow for a position operator all massless fields with spin $\geq 1$ don't.

In QFT you naturally get a Poincare invariant description. The physics is contained in the appropriate autocorrelation functions of field-operator products. In the usual case of "vacuum QFT", used to describe scattering reactions in terms of the S-matrix, you need the time-ordered products; for thermal-equilibrium many-body theory you can work with the Matsubara formalism and use analytic continuation to retarded propagators to at least cover linear-response theory (via the famous Green-Kubo relation); for the general non-equilibrium case you need all kinds of $N$-point functions of the Schwinger-Keldysh real-time contour (or equivalently other time contours, representing e.g., the thermofield formalism). For reviews, see

N. P. Landsmann and C. G. van Weert, Real- and
Imaginary-time Field Theory at Finite Temperature and
Density, Physics Reports 145, 141 (1987),
https://doi.org/10.1016/0370-1573(87)90121-9.

K. Chou, Z. Su, B. Hao and L. Yu, Equilibrium and
Nonequilibrium Formalisms made unified, Phys. Rept. 118, 1
(1985), https://doi.org/10.1016/0370-1573(85)90136-X.

P. Henning, Thermo field dynamics for quantum fields with
continuous mass spectrum, Phys. Rept. 253, 235 (1995),
https://dx.doi.org/10.1016/0370-1573(94)00083-F.

 

[pavsic]

But this doesn't work as soon as you have more than one particle. There is no way to covariantly write down a state with two particles at different positions.

Of course it works. As stated by Demystifier, one can write a two particle state as
$$|{\bar x}_1,{\bar x}_2 \rangle ,$$
where ${\bar x}_1$ and ${\bar x}_2$ are the positions on a given simultaneity hypersurface $\Sigma_\mu .$

A generic state of $N$ particles is a superposition of basis states
$$
|\Psi \rangle = \sum_{r=1}^N d \Sigma_1 d\Sigma_2 ...d \Sigma_r \psi(s,\bar x_1,\bar x_2,...,\bar x_r)
a^\dagger (\bar x_1) a^\dagger (\bar x_2) ... a^\dagger (\bar x_r) |0\rangle.$$
Here $\psi(s,\bar x_1,\bar x_2,...,\bar x_r)$, $r=1,2,3,...,N$, are multiparticle wave functions, and $a^\dagger (\bar x_r)$ the operators that create particles at positions ${\bar x}_r$ on the hypersurface $\Sigma$. Notice the "bar" over $x$, which denotes that the $\bar x$ is on the hypersurface.

A wave function in QFT is thus a multicomponent object, namely, a set of wave functions $\psi(s,\bar x_1,\bar x_2,...,\bar x_r)$ for $r=1,2,3,..,N$. They are "coefficients" of the expansion of a state $|\Psi \rangle$ in terms of the basis states
$$|{\bar x}_1,{\bar x}_2, ...,\bar x_r \rangle = a^\dagger (\bar x_1) a^\dagger (\bar x_2) ... a^\dagger (\bar x_r) |0\rangle . $$
In the presence of an interaction, there are in general transitions between those basis states (of the Fock space). The normalization of such multicomponent wave function is over the whole Fock space. A single component $\psi(s,\bar x_1,\bar x_2,...,\bar x_r)$ for a fixed $r$ is not normalized; its absolute square can change during the evolution of the system.

Position operator can be defined covariantly as
$${\hat {\bar x}}^\mu = \int d \Sigma \, a^\dagger ({\bar x}) {\bar x}^\mu a({\bar x}).$$
In a particular Lorentz reference frame in which ${\bar x}^\mu = (0,x^1,x^2,x^3) \equiv (0, {\bf x})$, which is then written as ${\bf x}$, we obtain the usual non covariant definition of the position operator.
Again, I was superficial in this explanation. For a more exhaustive discussion see https://inspirehep.net/literature/1667079 and
https://www.worldscientific.com/worldscibooks/10.1142/11738

 

[PeterDonis]

The quantum position states are $|{\bar x} \rangle$, and they satisfy
$$\langle {\bar x}_1|{\bar x}_2 \rangle = \delta^4 ({\bar x}_1 - {\bar x}_2) ,$$
which is a covariant relation.

Why not
$$\langle x_1,x_2|x'_1,x'_2\rangle =\delta^4(x_1-x'_1) \delta^4(x_2-x'_2)?$$
Do you see anything wrong with that?

Actually, I was misreading before. I see now that you are using $\delta^4$ functions. In that case, I don't understand what these states are supposed to mean. They seem to describe events in spacetime, not particles--the first state describes one event, and the second describes two. How are these "particle" states at all?

 

[vanhees71]

They cannot be particle states at all since this means that the particles are popping up for an infinitesimal small time intervall at $t_1'$ and $t_2'$ at places $\vec{x}_1'$ and $\vec{x}_2'$ respectively. That doesn't make sense. One should also be aware that "particle interpretations" in relativistic QFT have a well-defined meaning only in asymptotic free states.

 

[Tendex]

Pavsic and Demystifier are discussing particles quantum position operators, there seems to be some clutter here by adding unrelated "particle states" to the discussion.

 

[PeterDonis]

Pavsic and Demystifier are discussing particles quantum position operators

How are they "particle position operators"? That's the part I don't get. They seem like operators that pick out a particular event in spacetime, not the position of a particle in space.

 

[Tendex]

How are they "particle position operators"? That's the part I don't get. They seem like operators that pick out a particular event in spacetime, not the position of a particle in space.

I'm not sure why you bring up events here but anyway these are quantum fields being discussed, we have certain number of particles (not just a-particle which wouldn't make much sense with identical particles in second quantization to decipher what state corresponds to what particle, as mentioned by vanhees71 above "particle" states don't make much sense in this context) assigned position state/s in the wavefunction view of free quantum fields. Just that. Unless they were referring to something else.

 

[PeterDonis]

I'm not sure why you bring up events here

Because that's what it looks like the states given in post #32, #34, and #36 describe. It doesn't look to me like those states describe "particle positions", but that's what they are claimed to describe in those posts.

 

[Demystifier]

They seem to describe events in spacetime, not particles--the first state describes one event, and the second describes two. How are these "particle" states at all?

How about an event in which the particle is detected at the spacetime point $x$?

 

[Demystifier]

They cannot be particle states at all since this means that the particles are popping up for an infinitesimal small time intervall at $t_1'$ and $t_2'$ at places $\vec{x}_1'$ and $\vec{x}_2'$ respectively.

Recall that Schwinger makes sense of such states in his proper time method. Essentially, one introduces an additional time variable: the proper time $s$. Classically, it corresponds to a particle having the spacetime position $x$ at any value of proper time $s$. See the attached classic paper by Schwinger, especially Eq. (2.40).

 

[vanhees71]

Pavsic and Demystifier are discussing particles quantum position operators, there seems to be some clutter here by adding unrelated "particle states" to the discussion.

To define eigenstates of an observable operator you first need an observable. For massive particles you can define a position operator and also (generalized) position eigenstates.

Time is never an observable in quantum theory, because otherwise you'd never have a theory with a stable ground state. That's why the only relevant relativistic QT is in terms of relativistic QFT, and you have to construct the position operator using the quantum fields. For a review, see

https://arnold-neumaier.at/physfaq/topics/position.html

 

[vanhees71]

Recall that Schwinger makes sense of such states in his proper time method. Essentially, one introduces an additional time variable: the proper time $s$. Classically, it corresponds to a particle having the spacetime position $x$ at any value of proper time $s$. See the attached classic paper by Schwinger, especially Eq. (2.40).

I know the world-line formalism, but this doesn't make time an observable!

 

[pavsic]

How are they "particle position operators"? That's the part I don't get. They seem like operators that pick out a particular event in spacetime, not the position of a particle in space.

They pick out the position of a particle in space. Recall what is "space". It is just a time slice, i.e., a hypersurface $\Sigma_\mu$ oriented so the that $\Sigma_\mu = (0,\Sigma_i)$, $i=1,2,3$. But it can be oriented differently (or observed from another reference frame), so that also the component $\Sigma_0 \neq 0$. A position on such hypersurface is given by the projection of coordinates $x^\mu$ onto $\Sigma_\mu$ by means of the projection operator ${P^\mu}_\nu={\delta^\mu}_\nu- n^\mu n_\nu$, where $n^\mu$ is the normal unit vector to $\Sigma_\mu$. Position on such generally oriented hypersurface $\Sigma_\mu$ is then
$${\bar x}^\mu = {P^\mu}_\nu x^\nu.$$
That is the tangential (to the $\Sigma_\mu$) part of the spacetime position $x^\mu$.
The normal part (to the $\Sigma_\mu$) is
$$s=x^\mu n_\mu.$$
When we write an action, e.g., for a scalar field, we integrate the Lagrangian over the spacetime volume element $d t \,d^3 {\bf x}$.
This 4-volume element is in the particular reference frame in which the normal to $\Sigma_\mu$ is $n^\mu = (1,0,0,0)$ so that $\Sigma_\mu = (0,d^3 {\bf x})$. In general, the volume element is $d s \, d \Sigma$, where $d \Sigma = n^\mu \epsilon_{\mu \nu \rho \sigma} d \bar x^\nu d \bar x^\rho d \bar x^\sigma$.

One can then repeat in such a covariant manner the whole procedure of the quantiztion of the scalar field, write the Newto-Wigner position operator, position creation operator, etc. We then arrive at the covariant form of the wave function $\psi(s,\bar x^\mu)$, where instead of $t$ stands $s$. It is covariant, because if you change the coordinates (reference frame) we have the relation $\psi(s,\bar x^\mu)=\psi'(s',{\bar x}'^\mu)$.

Here I wanted to formulate the usual QFT and the role of particle position. The other thing is the Stueckelberg theory (or Fock-Schwinger formalism) that employs the invariant evolution parameter $\tau$. In that formalism $\tau$ is not observable, but the spacetime coordinates $x^\mu$ are considered as observables. In that formalism, the wave function is $\psi(\tau,x^\mu).$

 

[PeterDonis]

How about an event in which the particle is detected at the spacetime point $x$?

First, that's not what @pavsic is claiming:

They pick out the position of a particle in space.

Second, while individual particle detections happen at particular spacetime points, that's what quantum field operators are for--those operators are "attached" to particular points in spacetime. But this thread is about wave functions, which aren't. And this particular subthread is (I thought) about "particle positions", which aren't attached to particular points in spacetime either.

 

[PeterDonis]

They pick out the position of a particle in space.

No, they don't. You aren't using 3-D delta functions $\delta^3$. You are using 4-D delta functions $\delta^4$. That picks out points in spacetime, not space.

Recall what is "space". It is just a time slice, i.e., a hypersurface oriented so the that , . But it can be oriented differently (or observed from another reference frame)

What you appear to be trying to do here is pick out the hypersurface and hold that constant while making a Lorentz transformation. That's not what changing frames does. When you change frames, you change hypersurfaces: a different set of hypersurfaces become the "surfaces of constant time" that particle positions are supposed to be points in.

 

[vanhees71]

First, that's not what @pavsic is claiming:
Second, while individual particle detections happen at particular spacetime points, that's what quantum field operators are for--those operators are "attached" to particular points in spacetime. But this thread is about wave functions, which aren't. And this particular subthread is (I thought) about "particle positions", which aren't attached to particular points in spacetime either.

Well, but that's the problem. Wave functions do not make too much sense for relativistic QT, at least not the same sense as they make in non-relativistic QM, and that's the problem here. All you can say from relativistic QFT are detection probabilities of a particle (i.e., an asymptotic free state, where a particle interpretation is possible) with a detector at a given position. This information is formally encoded in the field-operator autorcorrelation functions (aka $N$-point Green's functions).

 

[andresB]

https://arnold-neumaier.at/physfaq/topics/position.html

Quoting from the link: " Therefore, if we have a quantum system defined in an arbitrary Hilbert space in which a momentum operator is defined, the necessary and sufficient condition for the existence of a spatial probability interpretation of the system is the existence of a position operator with commuting components which satisfy standard commutation relations with the components of the momentum operator and the angular momentum operator. "

I can see the sufficient part, it's what we have in nonrelativistic QM (though, In the non-relativistic case, I think demanding the right commutation relations with the Galilean boost is also required).

But I fail to see why it is a necessary condition.

 

[pavsic]

What you appear to be trying to do here is pick out the hypersurface and hold that constant while making a Lorentz transformation. That's not what changing frames does. When you change frames, you change hypersurfaces: a different set of hypersurfaces become the "surfaces of constant time" that particle positions are supposed to be points in.

I wrote: "Recall what is "space". It is just a time slice, i.e., a hypersurface oriented so the that , . But it can be oriented differently (or observed from another reference frame)."
Here I meant the active (passive) Lorentz transformation.

Active LT: In a given reference frame one can have a surface $\Sigma$ and change it into another surface $\Sigma'$ related to $\Sigma$ by a Lorentz transformation. That is, the object of observation, namely, the hysperfusrface changes under the transformation.

Passive LT: The same object is observed from different reference frames. In one reference frame the surface components are $\Sigma_\mu$, in another reference frame the components of the same hypersurface are $\Sigma'_\mu = {L_\mu}^\nu \Sigma_\nu.$

To sum up, in the active case, the object changes, while the reference frame remains the same. In the passive case, the object remains the same, and the reference frame changes.

 

[pavsic]

No, they don't. You aren't using 3-D delta functions $\delta^3$. You are using 4-D delta functions $\delta^4$. That picks out points in spacetime, not space.

I am using $\delta^4 ({\bar x}' - {\bar x})$, where ${\bar x} \equiv {\bar x}^\mu$, and, as I explained in the previous reply, $\bar x^\mu= {P^\mu}_\nu x^\mu$ is the projection of the spacetime point onto the hypersurface $\Sigma$. This projection is also a spacetime point, it is given by four coordinates, therefore the 4-D delta function, but restricted to the hypersurface. Therefore, more precisely it should be written, e.g.,
$$\delta^4 ({\bar x}' - {\bar x})|_\Sigma.$$
Explicitly, it is
$$\delta^4 ({\bar x}' - {\bar x})|_\Sigma = n_\mu {\epsilon^\mu}_{\nu \rho \sigma} \delta (x'^\nu-x^\nu)\delta (x'^\rho-x^\rho)\delta (x'^\sigma-x^\sigma)$$
This is a covariant expression that replaces the non covariant 3-D delta function. In a particular reference frame in which the normal to $\Sigma$ is $n^\mu= (1,0,0,0)$ we have $\delta^4 ({\bar x}' - {\bar x})|_\Sigma =\delta^3 ({\bf x}'-{\bf x})$.

Thank you for pointing out that my notation was incomplete and that in fact it is a 3D delta-function, but, as I explained above, written covariantly, by using all four spacetime coordinates to designate position on the hypersurface $\Sigma .$

To your next comment I will reply in a separate post, because I do not know how to do with multiquotes.
I also intend to discuss why this "minithread" on position is necessary, and how it will lead us to the concept of wave function in QFT. For the time being, see my post number 36.

 

[PeterDonis]

Here I meant the active (passive) Lorentz transformation.

If we are talking about "position in space", which is what you claim to be trying to capture, then "space" means the appropriate hypersurface for a particular observer. Neither of your definitions captures this. If two observers are in relative motion, they each choose a different hypersurface as "space" (more precisely, "space" at an instant of their time). Typically they also each choose a different frame to describe what they observe, the frame in which their chosen "space" is a surface of constant coordinate time. So if we transform from one observer's description to the other, both the frame (coordinate chart) and the hypersurface change. Neither of your definitions captures this.

position on the hypersurface $\Sigma .$

But which hypersurface depends on the observer. Different observers will disagree about which hypersurface is "space" (at an instant of their time). That was my original objection, and you are not addressing it at all.

 

[Tendex]

depends on the observer. Different observers will disagree about which hypersurface

What definition of observer are you using and what change of observer are you referring to?
These are free quantum fields at a fixed time in the Schrodinger picture, the states don't change and don't pick specific particles positions.

 

[pavsic]

Well, but that's the problem. Wave functions do not make too much sense for relativistic QT, at least not the same sense as they make in non-relativistic QM, and that's the problem here.

Wave function makes sense in relativistic quantum field theory. Starting from the scalar quantum field $\varphi(x)$, where $x\equiv x^\mu = (t,{\bf x})$, we can construct the operators (see Jackiw: Diverse topics in theoretical and mathematical physics)
$$a(t, {\bf x}) = \frac{1}{\sqrt{2}} \left ( \sqrt{\omega_{\bf x}}\varphi + \frac{i}{\sqrt{\omega_{\bf x}}} \Pi \right) $$
$$a^\dagger(t, {\bf x}) = \frac{1}{\sqrt{2}}\left ( \sqrt{\omega_{\bf x}}\varphi - \frac{i}{\sqrt{\omega_{\bf x}}} \Pi \right) $$
which stisfy
$$[a(t,{\bf x}),a^\dagger (t,{\bf x}') = \delta^3 ({\bf x},\bf x'),$$
where $\Pi$ is the canonical momentum conjugated to $\varphi$ and $\omega_{\bf x}= \sqrt{m^2 -\nabla^2}.$
The above operators are just like the annihilation and creation operators of the harmonic oscillator. But here we have uncountabley infinite number of operators.
A state with with a definite ${\bf x}$ at time $t$, say $t=0$ is given by
$$a^\dagger({\bf x})|0\rangle,$$
where $a^\dagger({\bf x})\equiv a^\dagger(0,{\bf x}).$
A generic one particle state is
$$|\Psi \rangle = \int d^3 {\bf x}\, \psi (t,{\bf x}) a^\dagger ({\bf x})|0 \rangle.$$
Here $\psi (t,{\bf x})$ is a single particle wave function, satisfying the relativistic Schroedinger equation:
$$i \frac{\partial \psi}{\partial t}= \sqrt{m^2 -\nabla^2} \psi.$$
This is the Foldy equation. It comes from the Schroedinger equation for the state $|\Psi \rangle$:
$$i \frac{\partial |\Psi \rangle}{\partial t} = H |\Psi \rangle,$$
where $H$ is the usual Hamiltonian composed of the quantum field operators.

Position operator is
$$ \hat{\bf x}= \int d^3 {\bf x} \,a^\dagger ({\bf x}) {\bf x} \,a({\bf x}).$$.
Its eigenstates are $|{\bf x} \rangle \equiv a^\dagger ({\bf x}) |0\rangle.$

All those expressions can be cast into covariant forms. Namely,
${\bf x}$ are coordinates of a point on the hypersurface $t=constant.$ In a subsequent post I will discuss this in more detail and also address the point raised by PeterDonis.

 

[vanhees71]

You are dealing with field operaters, which of course makes sense. These are no wave functions though!

 

[Tendex]

You are dealing with field operaters, which of course makes sense. These are no wave functions though!

Sometimes Fock states in free quantum fields in the Schrodinger picture are referred to as "wave functions". These are not of course the same as the NRQM wavefunctions!

 

[pavsic]

You are dealing with field operaters, which of course makes sense. These are no wave functions though!

In my post 56, $a({\bf x})$ and $a^\dagger ({\bf x})$ are field operators, whilst $\psi (t,{\bf x})$ are complex valued wave functions ("wave packet profiles"), not operators.

 

[pavsic]

Sometimes Fock states in free quantum fields in the Schrodinger picture are referred to as "wave functions". These are not of course the same as the NRQM wavefunctions!

The Foldy equation of my post 56
$$i \frac{\partial \psi}{\partial t}= \sqrt{m^2 -\nabla^2} \psi$$,
in non relativistic approximation becomes the Schroedinger equation.