Particles and Fields – a neverending story

[reilly]

Sorry, I haven't a clue to do anything but directly cite post 32 for the algebra that I can't get to show up here. Sorry 'bout that.
Regards,
Reilly Atkinson

 

[strangerep]

One small quibble, however.

Normally the unitary transformation operator, U, will act on the "a" operators and the vacuum state. U will commute with p and the volume measure in p-space.

Yes -- I was using an "abuse of notation". The $U_\Lambda$ outside the integral was meant to denote
the abstract Lorentz transformation, but that's confusing of course. Here we need 3
different notations: one to denote the abstract Lorentz transformation, a second to denote
the representation of the LT acting on 3-momenta, and a third to denote the unitary
representation of the LT acting in the Fock space.

Since the physical (relativistic) Hilbert space is a subset of a dumb 4-momentum space,
any summations or integrals over the momentum parameters must be constructed
carefully so that they remain in the Hilbert space. Hence the integral measure must
be considered appropriately when performing LTs. That's what I was trying to explain,
but I agree the notation could be improved.

 

[jostpuur]

I realized that if the Hamiltonian is known, the state $$|\psi\rangle$$ immediately fixes the time evolution $$|\psi(t)\rangle$$ for all $$t\in\mathbb{R}$$. Consequently the boost $$|\psi\rangle\mapsto \Lambda|\psi\rangle$$ should exists. Could it be, that you are implicitly using the H in those transformations?

For example with one particle states you have the (spin-0, scalar) wave function $$\psi(t,x)$$, and the boosted wave function is defined with a relation

$$\psi'(t',x')=\psi(t,x).$$

Surely we all agree, that if the wave function is given only on a fixed time in some fixed frame, $$\psi(x)$$, there is no chance of transforming it into $$\psi'(x')$$. Use of more abstract notation with kets $$|\psi\rangle$$, or use of more general multi-particle states, is not going to change this fact. If you define some transformation with fixed time, you are defining something else than a Lorentz transformation. Only way to perform Lorentz boost is to first solve the state for all times, and then perform the boost.

If we did not know anything about the H, then moving onto the momentum space would not change the fact that transformation cannot be carried out with fixed time, but it happens to be that the momentum eigenstates are also the energy eigenstates, so it seems that the proper dealing with the time evolution could be inserted/hidden into those transformations that on first sight appear to be performed on fixed time.

 

[strangerep]

I realized that if the Hamiltonian is known, the state $$|\psi\rangle$$ immediately fixes the time evolution $$|\psi(t)\rangle$$ for all $$t\in\mathbb{R}$$. Consequently the boost $$|\psi\rangle\mapsto \Lambda|\psi\rangle$$ should exists. Could it be, that you are implicitly using the H in those transformations?

I've assumed all along that we're talking about a relativistic quantum theory. That means
we must have already constructed a Hilbert space carrying a unitary representation of the
Poincare group. So indeed, all the generators $J_{\mu\nu}, P_i, H$ of the Poincare group
are represented there as operators on the Hilbert space. In particular, if $|\psi\rangle$
is a state vector in the Hilbert space, then so is $|\phi\rangle := e^{i\tau H} |\psi\rangle$ .

Surely we all agree, that if the wave function is given only on a fixed time in some fixed frame, $$\psi(x)$$, there is no chance of [...]

If that's all you've got, then you don't have a relativistic quantum theory.
(Sorry, but I don't know how else to respond to this.)

 

[jostpuur]

I've assumed all along that we're talking about a relativistic quantum theory. That means
we must have already constructed a Hilbert space carrying a unitary representation of the
Poincare group. So indeed, all the generators $J_{\mu\nu}, P_i, H$ of the Poincare group
are represented there as operators on the Hilbert space. In particular, if $|\psi\rangle$
is a state vector in the Hilbert space, then so is $|\phi\rangle := e^{i\tau H} |\psi\rangle$ .

This doesn't yet immediately imply that the Hamiltonian is used in the definition of the generators of the Lorentz subgroup (or in the definition of boosts) which is what I asked about.

 

[koolmodee]

I thought i'd add a classic thought experiment about localization often taught in early QFT courses (eg Coleman's lectures). This very much relates to the nonvanishing of the propagator outside lightcone and things like that, as well as the early tensions between field theory and relativistic QM.

If you take a one particle state in a box (defined at the moment, purely in terms of relativistic qm, not field theory), and you squeeze it past the compton wavelength of the particle. That is exactly the regime where pair creation processes start becoming kinematically important on purely dimensional analysis grounds. So the more you squeeze, the more the occupancy number becomes a bad 'state', and you are forced to no longer deal with 1 particle, but rather many. You've essentially lost localization and your would be position operator becomes mathematically illdefined.

Enter field theory, which is a theory with infinite degrees of freedom, and the (somewhat sad) requirement to only use momentum operators.

You are free to try to sidestep this somehow (see Newton Wigner) but it starts becoming more and more intractable as you add interaction terms, and the technical and conceptual difficulties start increasing (the photon for instance becomes the source of much debate).

I subscribe to this view, which I believe is the standard one.

 

[strangerep]

This doesn't yet immediately imply that the Hamiltonian is used in the definition of the generators of the Lorentz subgroup (or in the definition of boosts) which is what I asked about.

I guess I don't the question, then. The Lorentz generators (in isolation) are defined by
their abstract commutation relations
$$[J_{\mu\nu} , J_{\rho\sigma}] ~=~ \eta_{\mu\rho} J_{\nu\sigma} - \eta_{\mu\sigma} J_{\nu\rho} - \eta_{\nu\rho} J_{\mu\sigma} + \eta_{\nu\sigma} J_{\mu\rho}$$
(No Hamiltonian there.)

But you said Lorentz subgroup -- so I'm guessing you mean "Lorentz generators in the
context of the Poincare group". Then one must adjoin the extra commutation relations
$$[J_{\mu\nu}, P_{\rho}] ~=~ \eta_{\mu\rho} P_{\nu} - \eta_{\nu\rho} P_{\mu}$$
which involve the Hamiltonian $P_0$.

This is all a bit like asking whether rotations somehow depend on translations for their
definition. The short answer is that they are not independent concepts (because their
respective generators do not commute). Nevertheless, one can speak of "rotational
invariance" by restricting one's attention to the rotation group only.

Similarly, when one talks about "Lorentz invariance", one usually means an
invariance in the absence of spacetime translation transformations.

 

[strangerep]

(Haelfix):
If you take a one particle state in a box (defined at the moment, purely in terms of relativistic qm, not field theory), and you squeeze it past the compton wavelength of the particle. That is exactly the regime where pair creation processes start becoming kinematically important on purely dimensional analysis grounds. So the more you squeeze, the more the occupancy number becomes a bad 'state', and you are forced to no longer deal with 1 particle, but rather many. You've essentially lost localization and your would be position operator becomes mathematically illdefined.

I subscribe to this view, which I believe is the standard one.

It should probably be noted that this semi-heuristic argument is at least partly a
consequence of defining "particle" as corresponding to a free energy-momentum
eigenstate. In field theory, the number operator is something like
$$N ~:=~ \int_k \dots a_k^\dagger \, a_k$$
and the momentum operator is like
$$P_i ~:=~ \int_k \dots \, k_i \, a_k^\dagger \, a_k$$
But we know that position and momentum do not commute, so it should
be no surprise that position and particle-number don't commute either.
Therefore, acting on a particle-number eigenstate with a position operator
inevitably yields a state whose particle-number is now non-deterministic.
That's another way of saying:

you are forced to no longer deal with 1 particle, but rather many.

 

[meopemuk]

But we know that position and momentum do not commute, so it should
be no surprise that position and particle-number don't commute either.

This doesn't look convincing. It is known that Newton-Wigner position operators for individual particles do commute with particle number operators. So, it should be possible to measure particle positions as precisely as one wants without disturbing the number of particles in the system.

Eugene.

 

[strangerep]

Hi Eugene,

Nice to hear from you again. I didn't think you frequented these parts anymore, and I
wondered whether you had disappeared from the face of the Earth. :-)

But we know that position and momentum do not commute, so it should
be no surprise that position and particle-number don't commute either.

This doesn't look convincing. It is known that Newton-Wigner position operators for individual particles do commute with particle number operators. So, it should be possible to measure particle positions as precisely as one wants without disturbing the number of particles in the system.

By restricting to NW operators for single particles, there's an implicit projection
operator to the 1-particle sector floating around, in whose presence I'm not sure that
your larger conclusion necessarily follows.

But... (well done...) -- you've made me feel a need to go back and double-check a few
things in the literature before I say any more.

 

[meopemuk]

By restricting to NW operators for single particles, there's an implicit projection
operator to the 1-particle sector floating around, in whose presence I'm not sure that
your larger conclusion necessarily follows.

Hi strangerep,

I am doing fine, thanks. Though I don't visit physicsforums.com as frequently as before. As you know, relativistic QM and position operators is my passion. So I couldn't resist...

Operators of single particle observables (including position) are defined not just in 1-particle sectors of the Fock space. For example, one can define 1-electron position operator in all sectors, which contain at least one electron (plus anything else). Such a definition should be possible in any consistent quantum theory, because experimentally one can measure electron's position in many-particle systems. Therefore, there should exist single-electron Hermitian position operator in the Hilbert space (or Fock space sector) of such a system.

These operators of 1-particle observables commute with the particle number operators simply by construction.

Eugene.

 

[jostpuur]

strangerep, I understand that it would be strange to ask if rotations depend on translations, but my question that does the boost of a state $$\psi$$ depend on the time evolution of the state was not equally dumb. I believe I understand what you re talking about, and that I can make this all clearer with a following calculation now.

At any instant in some particular frame the spatial and momentum space representations are related by equations

$$\psi(t,\boldsymbol{x}) = \int\frac{d^3p}{(2\pi)^3}\hat{\psi}(t,\boldsymbol{p})e^{i\boldsymbol{x}\cdot\boldsymbol{p}},\quad\quad \hat{\psi}(t,\boldsymbol{p})=\int d^3x\;\psi(t,\boldsymbol{x})e^{-i\boldsymbol{x}\cdot\boldsymbol{p}},$$

and the spatial representations of the state, in the spacetime, between two different frames are related by equation

$$\psi'(t',\boldsymbol{x}') = \psi(t,\boldsymbol{x}) = \psi\big(\frac{t' + u(x')^3}{\sqrt{1-u^2}}, (x')^1, (x')^2,\frac{(x')^3 + ut'}{\sqrt{1-u^2}}\big) = \psi((\Lambda^{-1})^0{}_{\mu} (x')^{\mu},\ldots, (\Lambda^{-1})^3{}_{\mu} (x')^{\mu}).$$

In any particular frame, with initial configuration $$\psi(0,\boldsymbol{x})$$, $$\hat{\psi}(0,\boldsymbol{p})$$, the time evolution is fixed with (by definition equivalent) equations

$$\psi(t,\boldsymbol{x}) = e^{-it\sqrt{-\nabla^2 + m^2}}\psi(0,\boldsymbol{x}),\quad\quad \hat{\psi}(t,\boldsymbol{p})=e^{-it\sqrt{|\boldsymbol{p}|^2 + m^2}}\hat{\psi}(0,\boldsymbol{p}).$$

These are my assumptions. Having the assumptions fixed, I can now solve how the momentum space representation transforms under a boost.

$$\hat{\psi}'(t',\boldsymbol{p}') = \int d^3x'\;\psi'(t',\boldsymbol{x}')e^{-i\boldsymbol{x}'\cdot\boldsymbol{p}'} = \int d^3x'\;\psi\big(\frac{t' + u(x')^3}{\sqrt{1-u^2}}, (x')^1, (x')^2, \frac{(x')^3 + ut'}{\sqrt{1-u^2}}\big) e^{-i\boldsymbol{x}'\cdot\boldsymbol{p}'}$$

$$=\int d^3x'\;\Big(\int\frac{d^3p}{(2\pi)^3}\hat{\psi}\big(\frac{t' + u(x')^3}{\sqrt{1-u^2}},\boldsymbol{p}\big)e^{i((x')^1,(x')^2,((x')^3 + ut')/\sqrt{1-u^2})\cdot\boldsymbol{p}}\Big)e^{-i\boldsymbol{x}'\cdot\boldsymbol{p}'}$$

$$=\int\frac{d^3x'\;d^3p}{(2\pi)^3}\hat{\psi}\big(\frac{t' + u(x')^3}{\sqrt{1-u^2}},\boldsymbol{p}\big) e^{-i(\Lambda^{-1})^i{}_{\mu}(x')^{\mu} p_i - i\boldsymbol{x}'\cdot\boldsymbol{p}'} = \cdots$$

$$\hat{\psi}\big(\frac{t' + u(x')^3}{\sqrt{1-u^2}},\boldsymbol{p}\big) = e^{-i((t' + u(x')^3)/\sqrt{1-u^2})\sqrt{|\boldsymbol{p}|^2 + m^2}}\hat{\psi}(0,\boldsymbol{p}) = e^{-i(\Lambda^{-1})^0{}_{\mu}(x')^{\mu}p_0}\hat{\psi}(0,\boldsymbol{p})$$

$$\cdots = \int\frac{d^3x'\;d^3p}{(2\pi)^3} \hat{\psi}(0,\boldsymbol{p}) e^{-i(\Lambda^{-1})^{\nu}{}_{\mu}(x')^{\mu}p_{\nu} - i\boldsymbol{x}'\cdot\boldsymbol{p}} = \cdots$$

At this point a change of variable

$$\bar{\boldsymbol{p}} = \Lambda(\boldsymbol{p})\quad\Leftrightarrow\quad \bar{p}^{\alpha}=\Lambda^{\alpha}{}_{\nu} p^{\nu}\quad\Leftrightarrow\quad p_{\nu} = (\Lambda^{-1})_{\nu}{}^{\alpha}\bar{p}_{\alpha}$$

comes handy.

$$\cdots = \int\frac{d^3x'\;d^3\bar{p}}{(2\pi)^3}\frac{E_{\Lambda^{-1}(\bar{\boldsymbol{p}})}}{E_{\bar{\boldsymbol{p}}}} \hat{\psi}(0,\Lambda^{-1}(\bar{\boldsymbol{p}})) e^{-i(\Lambda^{-1})^{\nu}{}_{\mu}(\Lambda^{-1})_{\nu}{}^{\alpha}(x')^{\mu}\bar{p}^{\alpha} - i\boldsymbol{x}'\cdot\boldsymbol{p}'} = \int d^3\bar{p}\;\frac{E_{\Lambda^{-1}(\bar{\boldsymbol{p}})}}{E_{\bar{\boldsymbol{p}}}} \hat{\psi}((0,\Lambda^{-1}(\bar{\boldsymbol{p}})) e^{-i(x')^0\bar{p}_0} \int\frac{d^3x'}{(2\pi)^3} e^{i\boldsymbol{x}'\cdot(\bar{\boldsymbol{p}} - \boldsymbol{p}')}$$
$$=\frac{E_{\Lambda^{-1}(\boldsymbol{p}')}}{E_{\boldsymbol{p}'}} \hat{\psi}(0,\Lambda^{-1}(\boldsymbol{p}')) e^{-it'E_{\boldsymbol{p}'}}.$$

With $$t'=0$$ that means

$$E_{\boldsymbol{p}'}\hat{\psi}'(0,\boldsymbol{p}') = E_{\Lambda^{-1}(\boldsymbol{p}')}\hat{\psi}(0,\Lambda^{-1}(\boldsymbol{p})) = E_{\boldsymbol{p}}\hat{\psi}(0,\boldsymbol{p}).$$

If we now define

$$\phi(\boldsymbol{p}) = E_{\boldsymbol{p}}\hat{\psi}(\boldsymbol{p}),$$

we can write the spatial wave function like

$$\psi(\boldsymbol{x}) = \int\frac{d^3p}{(2\pi)^3}\frac{1}{E_{\boldsymbol{p}}}\phi(\boldsymbol{p})e^{i\boldsymbol{x}\cdot\boldsymbol{p}},$$

and by the previous calculation, it is clear in which sense the right side is Lorentz invariant. The boosted wave function

$$\psi'(\boldsymbol{x}')=\psi'(0,\boldsymbol{x}') = \int\frac{d^3p'}{(2\pi)^3}\hat{\psi}'(0,\boldsymbol{p}')e^{i\boldsymbol{x}'\cdot\boldsymbol{p}'} = \int\frac{d^3p'}{(2\pi)^3}\frac{1}{E_{\boldsymbol{p}'}} \phi'(\boldsymbol{p}')e^{i\boldsymbol{x'}\cdot\boldsymbol{p}'}$$

could as well be obtained by substituting $$\phi'(\boldsymbol{p}')=\phi(\boldsymbol{p})$$ and performing the change of integration variable $$\boldsymbol{p}\to\boldsymbol{p}'$$ as usual, with the Jacobian determinant. Now it makes sense to let the boosting operators act on the superpositions of the momentum basis vectors like this:

$$U(\Lambda)|\psi\rangle = \int\frac{d^3p}{(2\pi)^3}\frac{1}{E_{\boldsymbol{p}}}\phi(\boldsymbol{p})|\Lambda(\boldsymbol{p})\rangle = \int\frac{d^3p'}{(2\pi)^3}\frac{1}{E_{\boldsymbol{p}'}} \underbrace{\phi(\Lambda^{-1}(\boldsymbol{p}'))}_{=\phi'(\boldsymbol{p}')} |\boldsymbol{p}'\rangle$$

I believe that the approach where one sets the spatial wave function satisfy the relativistic Schrödinger equation $$i\partial_t\psi = \sqrt{-\nabla^2 + m^2}\psi$$ is equivalent with the approach where one postulates Lorentz transformations to to the basis vectors $$|\boldsymbol{p}\rangle$$ and then writes the state $$|\psi\rangle$$ as superposition with a Lorentz invariant integral. Assuming we know what we mean by the notation, of course. I tried to demostrate the equivalence now here. If you think there is still fundamental problems, feel free to explain.

I prefer starting with the spatial representation, because it is somehow more concrete, and then proceed to solve transformations in other representations. I suppose some people prefer starting things which are invariant, or have elegant transformation properties, like in this case the certain ket-vectors. In this case, IMO, the ket-vector approach has the disadvantage that is is more abstract, and it seems people don't know how to derive the spatial space properties out of it.

Finally... Strangerep, the clash with the role of Hamiltonian in the boost rose from this fact: If one starts with the spatial space approach, it does not make sense to perform boosts without already knowing the time evolution, and thus the Hamiltonian is necessary for boosts. If one starts with the momentum ket-vector approach, one can immidiately define boosts for the ket-vectors without time evolution just like one can define rotations withouth translations. There is no contradiction in this: The time evolution is needed anyway when one wants to prove the two approaches equivalent.

 

[meopemuk]

At any instant in some particular frame the spatial and momentum space representations are related by equations

$$\psi(t,\boldsymbol{x}) = \int\frac{d^3p}{(2\pi)^3}\hat{\psi}(t,\boldsymbol{p})e^{i\boldsymbol{x}\cdot\boldsymbol{p}},\quad\quad \hat{\psi}(t,\boldsymbol{p})=\int d^3x\;\psi(t,\boldsymbol{x})e^{-i\boldsymbol{x}\cdot\boldsymbol{p}},$$

and the spatial representations of the state, in the spacetime, between two different frames are related by equation

$$\psi'(t',\boldsymbol{x}') = \psi(t,\boldsymbol{x}) = \psi\big(\frac{t' + u(x')^3}{\sqrt{1-u^2}}, (x')^1, (x')^2,\frac{(x')^3 + ut'}{\sqrt{1-u^2}}\big) = \psi((\Lambda^{-1})^0{}_{\mu} (x')^{\mu},\ldots, (\Lambda^{-1})^3{}_{\mu} (x')^{\mu}).$$ (1)

In any particular frame, with initial configuration $$\psi(0,\boldsymbol{x})$$, $$\hat{\psi}(0,\boldsymbol{p})$$, the time evolution is fixed with (by definition equivalent) equations

$$\psi(t,\boldsymbol{x}) = e^{-it\sqrt{-\nabla^2 + m^2}}\psi(0,\boldsymbol{x}),\quad\quad \hat{\psi}(t,\boldsymbol{p})=e^{-it\sqrt{|\boldsymbol{p}|^2 + m^2}}\hat{\psi}(0,\boldsymbol{p}).$$

These are my assumptions.

jostpuur,

your "manifestly covariant" transformation in the (x,t) space (I marked it by (1)) is perfectly OK if $$\psi(t,\boldsymbol{x})$$ denotes quantum field. However, I would disagree if you applied it to particle wavefunctions in the position representation. Here is my argument:

Your formula implies that the time parameter (t) is affected by the boost transformation. It then follows that t must be an eigenvalue of some Hermitian operator (the operator of time), which does not commute with the operator of boost. However, it is well-know that it is not possible to introduce the time operator in relativistic quantum theory (at least, I haven't seen any successful attempt to do so). From basic postulates of quantum theory it follows that time is just a classical parameter, and NOT an eigenvalue of some Hermitian operator.

There is a more consistent way to derive boost transformations of particle wavefunctions in the position representation:

1. Define particle's Hilbert space as the space of irreducible unitary representation of the Poincare group.

2. Define the Newton-Wigner position operator in this Hilbert space. Eigenvectors of this operator define the orthonormal "position" basis in the Hilbert space.

3. Then the position-space wavefunction of any state is just the set of expansion coefficients with respect to the "position" basis, and boost transformation laws are calculated as usual in QM, by knowing the commutators of the Newton-Wigner operator with boost generators.

If you follow these rules you'll obtain transformation formulas, which are different from your eq. (1).

Eugene.

 

[jostpuur]

your "manifestly covariant" transformation in the (x,t) space (I marked it by (1)) is perfectly OK if $$\psi(t,\boldsymbol{x})$$ denotes quantum field. However, I would disagree if you applied it to particle wavefunctions in the position representation.

Do you also believe that there is something wrong in defining momentum eigenstates $$|p\rangle$$, which can be acted on by Lorentz transformation operators? I believe I just showed how this approach can be made equivalent with the approach where the Lorentz transformation is postulated with $$\psi'(t',x')=\psi(t,x)$$. Logically, you should either believe that the (t,x) transforming approach is right, or that the |p> transforming approach is wrong, or then that there is something wrong with my calculation.

 

[jostpuur]

About the old problem of time and space being more equal in relativity than in QM:

Might sound funny, but IMO the right solution is to not worry too much, and not get too philosophical. Of course if one, in any fixed frame, defines the time evolution with the Schrödinger equation where time is fundamentally different from spatial coordinates, it might appear that we are not getting a consistent theory, because we get infinitely ways of calculating the time evolution: one for each frame. But if it turns out in the end, that the procedure is Lorentz invariant, and that it actually doesn't matter which frame is chosen for calculation of time evolution, why worry about it more?

 

[meopemuk]

Do you also believe that there is something wrong in defining momentum eigenstates $$|p\rangle$$, which can be acted on by Lorentz transformation operators? I believe I just showed how this approach can be made equivalent with the approach where the Lorentz transformation is postulated with $$\psi'(t',x')=\psi(t,x)$$. Logically, you should either believe that the (t,x) transforming approach is right, or that the |p> transforming approach is wrong, or then that there is something wrong with my calculation.

Hi jostpuur,

I don't see anything wrong in your calculation. However, I would be more convinced if you performed some consistency checks. For example, your transformation (1) must conserve the norm of the wavefunction. In other words, boost transformation must be unitary. Can you prove that?

Eugene.

 

[strangerep]

[...]
In any particular frame, with initial configuration $$\psi(0,\boldsymbol{x})$$, $$\hat{\psi}(0,\boldsymbol{p})$$, the time evolution is fixed with (by definition
equivalent) equations

$$\psi(t,\boldsymbol{x}) = e^{-it\sqrt{-\nabla^2 + m^2}}\psi(0,\boldsymbol{x}),\quad\quad \hat{\psi}(t,\boldsymbol{p})=e^{-it\sqrt{|\boldsymbol{p}|^2 + m^2}}\hat{\psi}(0,\boldsymbol{p}).$$

These are my assumptions. [...]

Those equations above, and your later one:

$$i\partial_t\psi = \sqrt{-\nabla^2 + m^2}\psi$$

are just a representation of the Poincare Casimir operator
$M^2 := H^2 - P^2$.

By assuming $m^2$ to be a scalar constant in this way,
one is already assuming special relativity.

I guess it's a matter of personal taste which formalism
one follows -- provided one ends up at the correct destination.
(I've found that my own tastes tend to change over time.)

BTW, Eugene's suggested exercise of showing that the boosts
preserve the norm is a good one (and an important one).

 

[Hans de Vries]

This debate keeps popping up on these boards, and I'll side with Strangereps point of view on the subject.

I thought i'd add a classic thought experiment about localization often taught in early QFT courses (eg Coleman's lectures). This very much relates to the nonvanishing of the propagator outside lightcone and things like that, as well as the early tensions between field theory and relativistic QM.

If you take a one particle state in a box (defined at the moment, purely in terms of relativistic qm, not field theory), and you squeeze it past the compton wavelength of the particle. That is exactly the regime where pair creation processes start becoming kinematically important on purely dimensional analysis grounds. So the more you squeeze, the more the occupancy number becomes a bad 'state', and you are forced to no longer deal with 1 particle, but rather many. You've essentially lost localization and your would be position operator becomes mathematically illdefined.

A closely related version of this is the vacuum polarization which occurs below
the Compton radius. The divergence of the electric field is not zero anymore so
there remains an extended charge density (from virtual pairs) with a constant size
even if the particle itself becomes confined to a smaller and smaller volume.


Regards, Hans