Particles and Fields – a neverending story

[ledamage]

It is often said that the field operator \hat{\phi}(x) of some free field theory, e.g. Klein-Gordon or Dirac, acting on the vacuum state, creates "one particle localized at x" or "in a position eigenstate at x".

For example, from the Fourier expansion of the real Klein-Gordon field operator, we get

\hat{\phi}(x) |0\rangle = \int \mathrm{d}^3 \tilde{p} \: f_p^*(x) a^\dagger(p) |0\rangle \ ,

where \mathrm{d}^3 \tilde{p} \equiv \mathrm{d}^3 p/2p^0 is the Lorentz covariant measure and f_p(x) are the plane wave solutions to the Klein-Gordon equation.

Now, this state could be regarded as a localized one-particle state if f_p^*(x) is the ordinary QM position space wave-function of the particle in momentum eigenstate |\mathbf{p}\rangle, namely f_p^*(x)=\langle \mathbf{p} |\mathbf{x} \rangle. Writing a^\dagger(p) |0\rangle \equiv |\mathbf{p}\rangle, the above equation would turn into

\hat{\phi}(x) |0\rangle = \int \mathrm{d}^3 \tilde{p} \: |\mathbf{p} \rangle \langle \mathbf{p} |\mathbf{x} \rangle \ ,

which would somehow be related to |\mathbf{x}\rangle since the first part resembles an identity operator. But two problems occur:

1. What is |\mathbf{x}\rangle? While I can define the state of "one particle with momentum p" to be a^\dagger(p) |0\rangle \equiv |\mathbf{p}\rangle, I have no idea how a state like |\mathbf{x}\rangle can canonically arise in a QFT. My only idea is to promote the equation

f_p^*(x)=\langle \mathbf{p} |\mathbf{x} \rangle

to a defining axiom for |\mathbf{x}\rangle, that is, postulating that the free field solutions to the real Klein Gordon equation can be interpreted as one-particle wave functions in the ordinary QM sense. But this would mean you have to add a new axiom relating QM and QFT?!

2. I have some doubts that I can regard \int \mathrm{d}^3 \tilde{p} \: |\mathbf{p} \rangle \langle \mathbf{p} | as an identity operator because of the covariant measure.

Can anyone help me? Thanks!

 

[Demystifier]

The problem with the states
|\mathbf{x}\rangle = \hat{\phi}(\mathbf{x},0)|0\rangle
is that they are not orthogonal.
Namely,
\langle\mathbf{x}'|\mathbf{x}\rangle \neq \delta^3(\mathbf{x}-\mathbf{x}')
Note, however, that they become orthogonal in the non-relativistic limit.

 

[ledamage]

Thanks for the reply! I a book about group theory, the basic relation

\psi(x) = \langle 0 | \hat{\psi}(x)| \mathbf{p}\rangle

is used to obtain the transformation behavior of the field operator from the transformation behavior of the single-particle wave function. In a way, I use the representation theory of the Poincaré group to construct QM single-particle basis states and relate their transformation behavior to field operators of QFT by the above equation. Is this right?

But I don't get the connection; I think this is a subtle point: from the viewpoint of second quantization, everything is clear because one explicitly constructs the possibility to create and annihilate particles, so the above equation just follows mathematically.

But from the viewpoint of field quantization, I never introduced single-particle wave functions, I just applied the QM axioms to fields so that there is no obvious interpretation of the function which is represented by the above matrix element.
Does this make any sense?

 

[Demystifier]

There is no generally accepted answer to your question.
Nevertheless, a discussion that might be helpfull is provided in Secs. 8 and 9 of
http://xxx.lanl.gov/abs/quant-ph/0609163 [Found. Phys. 37 (2007) 1563]

 

[ledamage]

Wow, this is odd! I have found this nowhere in standard textbooks on that topic...instead I spent hours blaming myself for not getting this allegedly obvious relation...

Thanks very much!

 

[Icosahedron]

There is no position operator and no states in position space in QFT!

Read the first 7 pages of these https://www.physicsforums.com/newreply.php?do=newreply&noquote=1&p=1784740" .

 

[Demystifier]

There is no position operator and no states in position space in QFT!

Read the first 7 pages of these https://www.physicsforums.com/newreply.php?do=newreply&noquote=1&p=1784740" .

It seems that you gave a wrong link. Can you correct it?

 

[Icosahedron]

http://www2.physics.utoronto.ca/~luke/PHY2403/References_files/lecturenotes.pdf

Otherwise, Srednicki writes in his book on page 10, that keeping X and promoting t to an operator can also be done to get a relativistic particle theory.

Which leaves me puzzled.

 

[Demystifier]

http://www2.physics.utoronto.ca/~luke/PHY2403/References_files/lecturenotes.pdf

It contains some frequent wrong claims.

For example, it says that a particle cannot be localized in a region smaller than its Compton length. However, if it was true, it would imply that a massless photon could not be localized in any region smaller than its Compton length. On the other hand, the position of the photon can be measured by certain accuracy.

Next, it says that in relativistic QM there is no such thing as a 2, 1, or 0 particle state. But for free relativistic fields, such things certainly do exist. Moreover, a 1 and 0 particle states are stable even for interacting particles.

 

[Icosahedron]

On the other hand, the position of the photon can be measured by certain accuracy.

Position of a photon? But you know that a photon has no rest frame...

Next, it says that in relativistic QM there is no such thing as a 2, 1, or 0 particle state.

Right. I think every QFT text starts with this observation. What is wrong with it?

 

[Demystifier]

1. Position of a photon? But you know that a photon has no rest frame...

2. I think every QFT text starts with this observation. What is wrong with it?

1. So what? Take for example a classical photon. It has no rest frame too. Yet, at any time it has a definite position. The position, of course, changes with time, but it does not mean that the position does not exist.

2. For example, what is the state |0>, if not the state with exactly zero number of particles? And no, not every QFT text starts with this observation.

 

[ledamage]

For example, it says that a particle cannot be localized in a region smaller than its Compton length.

This is a point related to my question. What does particle localization mean in QFT? I am able to superpose some one-particle momentum eigenstates. But to determine the position of the "quantum object" thus obtained I need a position operator which does not exist. I cannot see more than a formal analogy to second quantization where it is just a mathematical implication that the Fourier transform of the superposition weight function is the one-particle wave function where an ordinary QM position operator can act on.

 

[strangerep]

What does particle localization mean in QFT?

It depends first on how you define "particle". I.e., which group you start with for which
you find the unitary irreducible representations. The Poincare algebra doesn't come
with a position generator, so some people advocate constructing one (Newton-Wigner)
from the enveloping ring. Other people are unimpressed with such "schtick". :-)

The x parameters in orthodox QFT (and Minkowski space in general) just serve as
a representation space for the Poincare group. The vacuum state has deterministic
energy-momentum of 0, and therefore cannot be a localised state in the usual sense.
Some people have suggested that this fact is at the core of the Reeh-Schlieder
paradox (which "proves" that fields over "there" can be recovered entirely from
fields "here" -- where "here" and "there" mean spacelike-separated regions).

Here's some fairly recent papers that debate these still-controversial points.

1) Fleming, "Reeh-Schlieder Meets Newton-Wigner", available via
http://philsci-archive.pitt.edu/archive/00000649/00/RS_meets_NW,_PDF.pdf

Abstract:
The Reeh-Schlieder theorem asserts the vacuum and certain other states to be spacelike
superentangled relative to local fields. This motivates an inquiry into the physical status
of various concepts of localization. It is argued that a covariant generalization of Newton-
Wigner localization is a physically illuminating concept. When analyzed in terms of
nonlocally covariant quantum fields, creating and annihilating quanta in Newton-Wigner
localized states, the vacuum is seen to not possesses the spacelike superentanglement that
the Reeh-Schlieder theorem displays relative to local fields, and to be locally empty as
well as globally empty. Newton-Wigner localization is then shown to be physically
interpretable in terms of a covariant generalization of the center of energy, the two
localizations being identical if the system has no internal angular momentum. Finally,
some of the counterintuitive features of Newton-Wigner localization are shown to have
close analogues in classical special relativity.


2) Halvorson, "Reeh-Schlieder defeats Newton-Wigner: On alternative localization
schemes in relativistic quantum field theory".
Available as quant-ph/0007060.

Abstract:
Many of the “counterintuitive” features of relativistic quantum
field theory have their formal root in the Reeh-Schlieder theorem,
which in particular entails that local operations applied to the vacuum
state can produce any state of the entire field. It is of great interest
then that I.E. Segal and, more recently, G. Fleming (in a paper entitled
“Reeh-Schlieder meets Newton-Wigner”) have proposed an alternative
“Newton-Wigner” localization scheme that avoids the Reeh-Schlieder
theorem. In this paper, I reconstruct the Newton-Wigner localization
scheme and clarify the limited extent to which it avoids the counterin-
tuitive consequences of the Reeh-Schlieder theorem. I also argue that
there is no coherent interpretation of the Newton-Wigner localization
scheme that renders it free from act-outcome correlations at spacelike
separation.


3) DeBievre, "Where's that quantum"? Available as math-ph/0607044

Abstract:
The nature and properties of the vacuum as well as the meaning
and localization properties of one or many particle states have at-
tracted a fair amount of attention and stirred up sometimes heated
debate in relativistic quantum field theory over the years. I will review
some of the literature on the subject and will then show that these is-
sues arise just as well in non-relativistic theories of extended systems,
such as free bose fields. I will argue they should as such not have given
rise either to surprise or to controversy. They are in fact the result
of the misinterpretation of the vacuum as “empty space” and of a too
stringent interpretation of field quanta as point particles. I will in par-
ticular present a generalization of an apparently little known theorem
of Knight on the non-localizability of field quanta, Licht’s character-
ization of localized excitations of the vacuum, and explain how the
physical consequences of the Reeh-Schlieder theorem on the cyclic-
ity and separability of the vacuum for local observables are already
perfectly familiar from non-relativistic systems of coupled oscillators.

 

[ledamage]

Thanks very much for the detailed reply! I will have a look at some of these papers. But I'm still confused why in standard textbooks on QFT, say, a Gaussian superposition of momentum eigenstates is called localized (blurred around a certain point x) in space. Which reasoning is behind this?

 

[strangerep]

[...] But I'm still confused why in standard textbooks on QFT,

Did you have specific textbook quote in mind?

say, a Gaussian superposition of momentum eigenstates is called localized
(blurred around a certain point x) in space. Which reasoning is behind this?

IMHO, I think that's just what people would like to think. I.e., we'd like to
think that the x's we use when constructing orthodox QFT do indeed correspond
meaningfully to points in physical space. But one must delve into the subtleties
of position operators, understand the Reeh-Schlieder theorem, etc, (and maybe
also Haag's theorem), before it becomes apparent that something is deeply
puzzling about this whole subject.

Of course, such delving is clearly not necessary to calculate the stunningly
accurate predictions of QFT, hence most people don't worry very much about it.

 

[ledamage]

Did you have specific textbook quote in mind?

Srednicki for example, Chapter 5 (LSZ formula), Eqs. (5.6) and (5.7). He speaks of creating a particle localized in momentum and position space. (A draft of the book is available at http://www.physics.ucsb.edu/~mark/qft.html" .)

Okay, thanks to you all for your answers! I think I know now what to read to get on...

 

[Fredrik]

Strangerep, thanks for posting those references. I'll definitely take a closer look at the last one and maybe the other two as well some time soon.

You seem to know this stuff pretty well already, so I'd like to ask you specifically about something that I have believed to be true until now. (Right now I'm really confused. I have e.g. never heard of the Reeh-Schlieder theorem before). I've been saying things like this:

The state of a photon is in general a superposition of states with different momenta. Let's ignore other degrees of freedom and express this as

\int d^3p f(\vec p)a^\dagger(\vec p)|0\rangle

where a^\dagger(\vec p) is the creation operator that creates a one-particle state with momentum p when it acts on the vaccum. The Fourier transform of f can be interpreted as an ordinary wave function.

Is this completely false? I guess I always thought that since the f above is pretty much the same in relativistic QM as in non-relativistic QM, its Fourier transform should also be pretty much the same as in non-relativistic QM.

 

[Demystifier]

Is this completely false? I guess I always thought that since the f above is pretty much the same in relativistic QM as in non-relativistic QM, its Fourier transform should also be pretty much the same as in non-relativistic QM.

The right questions are: Can this Fourier transform be interpreted as the probability amplitude? In other words, is such a probability conserved? And is it defined in a relativistic covariant way?
See e.g. the Appendix in
http://xxx.lanl.gov/abs/0804.4564 [not yet published]

 

[ledamage]

The right questions are: Can this Fourier transform be interpreted as the probability amplitude?

If I get this right, this is the same as asking whether you can interpret a first-quantized field as a particle system obtained by second quantization which obviously is done without ever asking for reasons.

 

[strangerep]

Strangerep, thanks for posting those references. I'll definitely take a closer look at the last one and maybe the other two as well some time soon.
You seem to know this stuff pretty well already, so [...]

(Urgle!?) No, I'm not an expert. I'm aware that the various puzzles exist, but I
don't know how to resolve them satisfactorily. (Actually, I don't think
anyone does - hence the ongoing debate.)

 

[Fredrik]

If I get this right, this is the same as asking whether you can interpret a first-quantized field as a particle system obtained by second quantization which obviously is done without ever asking for reasons.

No, that's not the question here. The a^\dagger(\vec p)|0\rangle states are eigenstates of the momentum operator that you can construct from the quantum field. (Noether's theorem tells us how). The question is about the meaning of the Fourier transform of the coefficients when an arbitrary state is expressed as a "linear combination" of momentum eigenstates.

In non-relativistic QM, that Fourier transform is the wave function, and as you know the square of its absolute value is a probability density describing where in space the particle is likely to be found. I've been assuming that the interpretation as a probability density holds in relativistic QM too, but I'm still not sure how correct that is. (I still haven't had time to really think about Demystifier's answer. It looks like a very good one but I need to think some more about what it means).

 

[ledamage]

No, that's not the question here. The a^\dagger(\vec p)|0\rangle states are eigenstates of the momentum operator that you can construct from the quantum field. (Noether's theorem tells us how). The question is about the meaning of the Fourier transform of the coefficients when an arbitrary state is expressed as a "linear combination" of momentum eigenstates.

But this is exactly what I meant, isn't it? You treat a first-quantized field like a second quantized theory by interpreting

\psi(x_1, \hdots, x_n) := \langle 0 | \hat{\psi}(x_1) \cdots \hat{\psi}(x_n) | \alpha \rangle

as a many-particle wave function, where |\alpha\rangle is some superposition of momentum eigenstates. But the point is that the matrix element \langle 0 | \hat{\psi}(x_1) \cdots \hat{\psi}(x_n) | \alpha \rangle itself gives no key to this interpretation since the field operator is obtained by first quantization. And in fact, as has been pointed out here, there are reasonable doubts to do so.

 

[strangerep]

You treat a first-quantized field like a second quantized theory by interpreting

\psi(x_1, \hdots, x_n) := \langle 0 | \hat{\psi}(x_1) \cdots \hat{\psi}(x_n) | \alpha \rangle

as a many-particle wave function, where |\alpha\rangle is some superposition of momentum eigenstates. [...]

In such areas, it helps to go back and think about how the multi-particle Fock space
is constructed. One starts with a (single-particle) Hilbert space H_1 carrying
a +ve energy representation of the Poincare group. By various abstract nonsense
(the spectral theorem from Functional Analysis) we can then assert that H_1
is spanned by momentum eigenstates, subject to the mass-shell condition
\delta^{(4)}(p^2 - m^2).

If we try to Fourier-transform to position space (i.e., to obtain an alternate basis
of position eigenstates), it ceases to be physically meaningful if we don't include
the \delta^{(4)}(p^2 - m^2) factor in the Fourier transform.
Remember now that a product in momentum space Fourier-transforms into a
convolution in position space, so you don't get very nice position eigenstates.
(IMHO, this is another reason why we hit paradoxes like Reeh-Schlieder.)

Now, what is a 2-particle Hilbert space? We take the tensor product of two
copies of H_1, i.e., H_2 := H_1 \otimes H_1.
Remember: I'm still thinking of these in momentum basis, or rather, two tensored
copies of momentum space, each with their separate \delta^{(4)}(p^2 - m^2)
constraints. Now think about the horrors when you try to Fourier-transform
this tensor product to a 2-position basis. Convolutions everywhere. (And even
that's only if you can somehow make the Fourier transform itself make sense.)

And yet, we try to kid ourselves that the tensor product of 2 separately-transformed
H_1 spaces can be handled in the simple way shown at the start of
this post. Really, it becomes mathematical nonsense very quickly. You can't
work with infinite-dimensional spaces and their duals sensibly without
paying careful attention to topological matters, unboundedness of operators,
and the like.

(That probably didn't help, I know. :-)

 

[jostpuur]

If you assume that the position and momentum space wave functions \psi(x) and \hat{\psi}(p) are related by the Fourier transforms, define the single particle state as

<br /> |\psi\rangle = \int d^3p\; \hat{\psi}(p) a_p^{\dagger}|0\rangle,<br />

derive the time evolution

<br /> |\psi(t)\rangle = \int d^3p\; \hat{\psi}(p) e^{-iE_p t} a_p^{\dagger}|0\rangle<br />

from the Shrödinger's equation (after fixing the Hamilton's operator from whatever QFT principles you want to postulate initially...) (here E_p=\sqrt{|p|^2+m^2}) and then recognize the evolving momentum space wave function

<br /> \hat{\psi}(t,p) = \hat{\psi}(p) e^{-iE_p t},<br />

it follows that in the position space representation the time evolution is given by the old-fashioned relativistic Schrödinger's equation

<br /> \psi(t,x) = e^{-it\sqrt{-\nabla^2 + m^2}}\psi(x),<br />

which satisfies the normalization conserving condition

<br /> \int d^3x\; |\psi(t,x)|^2 = \int d^3x\; |\psi(x)|^2 = 1<br />

at all times. To me this looks all good, and I must say I'm slightly confused about why this (assuming I've understood the mainstream view correctly) is widely considered incorrect. The QFT books always start doing some tricks with the 1/(2E_p) factor, and that's where I usually get lost.

 

[strangerep]

[...] define the single particle state as

<br /> |\psi\rangle = \int d^3p\; \hat{\psi}(p) a_p^{\dagger}|0\rangle,<br />

Aren't you in trouble already, even at this early stage? You don't
have the E_p factor in the denominator, so maybe
this expression is not Lorentz-covariant?

 

[jostpuur]

Aren't you in trouble already, even at this early stage? You don't
have the E_p factor in the denominator, so maybe
this expression is not Lorentz-covariant?

Hello. Immediately after posting the previous post, I realized there was several places where I could have still placed energy-factors so that I would have still understood what I'm doing, but I decided not to try any quick edit fixes. About your question... I'm not sure if that's trouble. It's more like a convention question. I think there are several equivalent ways to do this stuff, and the stuff gets confusing because it is usually not clear what conventions are being used.

For example. Suppose I had defined |p\rangle = \sqrt{2E_p} a_p^{\dagger}|0\rangle. Then I could have defined the state like

<br /> |\psi\rangle = \int \frac{d^3p}{\sqrt{2E_p}} \hat{\psi}(p)|p\rangle,<br />

and actually nothing changed! Of course I could also try to define the state like

<br /> \int d^3p\; \hat{\psi}(p)|p\rangle<br />

or

<br /> \int \frac{d^3p}{\sqrt{2E_p}} \hat{\psi}(p)a_p^{\dagger}|0\rangle<br />

and these expressions would not be equivalent.

Besides the raising operator, different factors could also be absorbed into the definition of the wave function \hat{\psi} too, but I better not say anything more now, because I'm not fully sure what all these conventions would mean right now.

 

[strangerep]

[...] I realized there was several places where I could have still placed energy-factors so that I would have still understood what I'm doing, but I decided not to try any quick edit fixes. About your question... I'm not sure if that's trouble. It's more like a convention question. I think there are several equivalent ways to do this stuff, and the stuff gets confusing because it is usually not clear what conventions are being used.

The crucial part is to be clear first about what should be
Lorentz invariant/covariant under any convention.

E.g., P&S use a Lorentz-invariant normalization -- see their eq 2.36, i.e.,

<br /> \langle p|q\rangle ~=~ 2E_p \, (2\pi)^3 \, \delta^{(3)}(p-q) ~,<br />

(which corresponds to your earlier |p\rangle = \sqrt{2E_p} a_p^{\dagger}|0\rangle),
and the discussion surrounding it about how to make a 3D delta fn invariant.

 

[Demystifier]

Jostpuur, strangerep is right that your expressions are not Lorentz invariant. See also the Appendix in my paper with correct Lorentz-invariant measures.

 

[jostpuur]

(edit: Ouch. Using \vec{p}' wasn't so good idea, because the prime gets messed into the arrow. Try to manage with it...)

My point in the post #26 was that the mere appearance of the energy-factor doesn't tell if things are going right or wrong, because that factor can be made appear or disappear with change of conventions. The post wasn't fully logical though, but a slightly heuristic instead.

I must admit that I'm not fully sure about the meaning of "covariant". I've understood that it is related to some form of expression remaining the same in some situations, but what's the difference between calling the expression invariant? Anyway, I'll have a careful and critical look at these integrals here:

Using on-shell four momentums it is possible to define \Lambda:\mathbb{R}^3\to\mathbb{R}^3 that transforms three momentum \Lambda(\vec{p})=\vec{p}&#039;. With a boost in 3-direction we have

<br /> p^3 = \frac{(p&#039;)^3 + uE_{\vec{p}&#039;}}{\sqrt{1-u^2}}<br />

<br /> \frac{dp^3}{d(p&#039;)^3} = \frac{1 + u(p&#039;)^3/E_{\vec{p}&#039;}}{\sqrt{1-u^2}} = \frac{1}{E_{\vec{p}&#039;}}\frac{E_{\vec{p}&#039;} + u(p&#039;)^3}{\sqrt{1-u^2}} = \frac{E_{\vec{p}}}{E_{\vec{p}&#039;}}<br />

Some arguments of rotational symmetry reveal that with arbitrary boost we have

<br /> \textrm{det}(D\Lambda^{-1}(\vec{p}&#039;)) = \frac{E_{\vec{p}(\vec{p}&#039;)}}{E_{\vec{p}&#039;}},\quad \vec{p}(\vec{p}&#039;) =\Lambda^{-1}(\vec{p}&#039;).<br />

This means, that if we have some function f:\mathbb{R}^3\to\mathbb{R}, and an integral

<br /> \int d^3p\; f(\vec{p}),<br />

we can perform a change of variables, and write the same integral as an integral over variable \vec{p}&#039;. The integral is then

<br /> \int d^3p&#039;\;\textrm{det}(D\Lambda^{-1}(\vec{p}&#039;)) f(\vec{p}&#039;) = \int d^3p&#039;\; \frac{E_{\vec{p}(\vec{p}&#039;)}}{E_{\vec{p}&#039;}} f(\vec{p}&#039;).<br />

This shows that on the other hand an expression

<br /> \int \frac{d^3p}{E_{\vec{p}}} f(\vec{p})<br />

has the particular covariant form, because under the change of variables it becomes

<br /> \int \frac{d^3p&#039;}{E_{\vec{p}&#039;}} f(\vec{p}&#039;).<br />

So far this is easy, because it is merely calculating. Things get more difficult when you say that the equation

<br /> |\psi\rangle = \int d^3p\;\hat{\psi}(\vec{p})a_{\vec{p}}^{\dagger}|0\rangle<br />

is not properly covariant or invariant. Do you know what you meant by this yourselves? We are never going to be interested in the change of variable \vec{p}&#039;=\Lambda(\vec{p}) in this integral. It would have no meaning. If you have a wave function \hat{\psi}(t,\vec{p}) in the space time, then it is possible to solve a boosted wave function \hat{\psi}&#039;(t&#039;,\vec{p}&#039;)=\hat{\psi}(t,\vec{p}), but there does not exist any boost for a wave function defined at some instant. For this reason, there does not exist boost for the state vector. There is no transformation |\psi\rangle\mapsto \Lambda|\psi\rangle. Because there does not exist Lorentz boosts for the left or the right side of the equation, it does not make sense to ask if the equation is covariant or invariant. All there exists is one change of variable, which in this context has no clear meaning. The integrals where this change of variable is related to boosts occur in different kind of situations, and I believe that in those the energy-factor plays important role.

My point in the post #24 was that the way how this convention gives the conserving probabilities looks very good. Because of this, it is difficult for me to believe that it could be simply wrong. Instead, if there exists some other expression to write the single particle state, which for sure is right, my first thought is that these expressions must be equivalent in some sense then.

 

[Demystifier]

Jostpuur, you have not specified the commutation relations among the creation and destruction operators, so it is impossible to say whether your alternative representations of the same thing are all invariant or all non-invariant. See the second equation of my Appendix (Eq. (44) in v2 of arXiv:0804.4564).

But this is not the real problem. Even if you have fixed these normalizations such that the wave function is invariant, the problematic expression is
\int d^3x |\psi({\bf x},t)|^2
This expression is not Lorentz invariant. Compare with Eq. (49) in v2 of arXiv:0804.4564, which is Lorentz invariant.

 

[strangerep]

I must admit that I'm not fully sure about the meaning of "covariant". I've understood that it is related to some form of expression remaining the same in some situations,

That's "invariant".

"Lorentz-covariant expression" means that the expression transforms as a tensor under
Lorentz transformations. The crucially important feature about this is that if a covariant
expression is 0 in one frame of reference, it is 0 in every frame of reference.
(That's the mathematical representation of physical statements like "the laws of physics are
the same for all inertial observers".)

[...] Things get more difficult when you say that the equation

<br /> |\psi\rangle = \int d^3p\;\hat{\psi}(\vec{p})a_{\vec{p}}^{\dagger}|0\rangle<br />

is not properly covariant or invariant. Do you know what you meant by this yourselves?

Well, I'm pretty sure I know what I meant. :-) And I'd bet a large amount of money
that Demystifier does too.

Here, one is dealing with Fock space (I'll assume bosonic, so that canonical commutation
relations apply to the annihilation/creation operators). It is an essential part of the original
construction of any such Fock space that it should "carry" an irreducible representation of
the full Poincare group. In less high-falutin' language, this means that (eg), given an abstract
Lorentz transformation \Lambda (eg a boost in some direction), we can find
a concrete representation U_\Lambda, (constructed somehow from the basic
annihilation and creation operators), such that
U_\lambda |\psi\rangle = |\Lambda\,\psi\rangle, and such that we can
satisfy the various commutation relations of the Poincare group. (The "irreducible" adjective
earlier means here that any operator on the Fock space can be constructed from
the basic annihilation and creation operators.)

Further, under a Lorentz boost that takes \vec{p} \to \vec{p&#039;} := \Lambda \vec{p},
we have

<br /> a_{\vec{p&#039;}} ~=~ U_\Lambda ~ a_{\vec{p}}^{\dagger} ~ U_\Lambda^\dagger<br /> ~=~ a_{\Lambda \vec{p}}^{\dagger}<br />

We are never going to be interested in the change of variable \vec{p}&#039;=\Lambda(\vec{p}) in this integral.It would have no meaning.

Sure we might, sure it does...

We could perform a rotation, or we could transform between the lab frame
and the center-of-momentum frame, or heaps of other things to make our life easier when
calculating stuff. We could also transform between Cartesian p_1, p_2, p_3 to
polar p_r, p_\theta, p_\phi (which sometimes makes integrals easier to solve.

For this reason, there does not exist boost for the state vector. There is no transformation |\psi\rangle\mapsto \Lambda|\psi\rangle. Because there does not exist Lorentz boosts for the left or the right side of the equation,

Yes, there does. Yes, there is. Yes, there are.

it does not make sense to ask if the equation is covariant or invariant.

And yes, it does.

All there exists is one change of variable, which in this context has no clear meaning.

The equation

<br /> |\psi\rangle = \int d^3p\;\hat{\psi}(\vec{p})a_{\vec{p}}^{\dagger}|0\rangle<br />

expresses that the state vector |\psi\rangle is a linear combination of
the basis state vectors a_{\vec{p}}^{\dagger}|0\rangle. Under the
Lorentz transform U_\Lambda, we have:

<br /> U_\Lambda |\psi\rangle <br /> ~=~ U_\Lambda \int d^3p\;\hat{\psi}(\vec{p})a_{\vec{p}}^{\dagger}|0\rangle<br /> ~=~ U_\Lambda \int d^3p\;\hat{\psi}(\vec{p})a_{\vec{p}}^{\dagger}<br /> U_\Lambda^{\dagger} U_\Lambda |0\rangle<br /> ~=~ U_\Lambda \int d^3p\;\hat{\psi}(\vec{p})a_{\vec{p}}^{\dagger}<br /> U_\Lambda^{\dagger} |0\rangle<br />

where I have used the assumption that U_\Lambda |0\rangle = |0\rangle,
i.e., that the vacuum is Lorentz-invariant.

[continued in next post...]

 

[strangerep]

Continuing my previous too-long post...

I wrote

<br /> U_\Lambda |\psi\rangle <br /> ~=~ U_\Lambda \int d^3p\;\hat{\psi}(\vec{p})a_{\vec{p}}^{\dagger}|0\rangle<br /> ~=~ U_\Lambda \int d^3p\;\hat{\psi}(\vec{p})a_{\vec{p}}^{\dagger}<br /> U_\Lambda^{\dagger} U_\Lambda |0\rangle<br /> ~=~ U_\Lambda \int d^3p\;\hat{\psi}(\vec{p})a_{\vec{p}}^{\dagger}<br /> U_\Lambda^{\dagger} |0\rangle<br />

where I have used the assumption that U_\Lambda |0\rangle = |0\rangle,
i.e., that the vacuum is Lorentz-invariant.

Now, the U_\Lambda can only be moved legally through to
integration operation to act on the integrand if the measure d^3p
is Lorentz-invariant. But it's not, so we need to go back and start instead with the different
Lorentz-invariant measure d^3p/E_p. In that case, we could move
the U_\Lambda through to the left side of a_{\vec{p}}^{\dagger}
and get the original state vector, re-expressed in terms of p' basis instead of the
original p basis. (Yes, I know I should probably make that clearer, but I need to go.)

Back to Jostpuur's

My point in the post #24 was that the way how this convention gives the
conserving probabilities looks very good. Because of this, it is difficult for
me to believe that it could be simply wrong.

The tricky parts happen when one tries to combine the probability concepts
sensibly with special relativity.

 

[Horbacz]

Quantum mechanics uses partial differential equations to model how probability densities evolve over time. This produces `waves' of probability densities not unlike the waves on the surface of the ocean. The higher the wave at a given point the more likely the particle is to be located at that point. Everything that exists has a wave aspect. One cannot model what the particle does between observations of the particle. One can only model how the probability of observing it changes over time. These probabilities have a wave like character. Everything also has a particle aspect. Every particle interacts with other particles as if it were located at a single point in space.
It is these two aspects of a particle that causes physicists to speak of the ``collapse'' of the wave function. Prior to observing a particle it could be located at anywhere the wave function is not zero. After observing it the particle is known to be within a much smaller region of space. The wave function has collapsed. Of course there cannot be a physical wave function that collapses because this would violate special relativity. However the wave function does at times seem to be a physical entity because wave functions from two particles can interfere with each other just as physical waves do.

 

[Haelfix]

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.

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).

 

[reilly]

Continuing my previous too-long post...

I wrote ?Now, the U_\Lambda can only be moved legally through to
integration operation to act on the integrand if the measure d^3p
is Lorentz-invariant. But it's not, so we need to go back and start instead with the different
Lorentz-invariant measure d^3p/E_p. In that case, we could move
the U_\Lambda through to the left side of a_{\vec{p}}^{\dagger}
and get the original state vector, re-expressed in terms of p' basis instead of the
original p basis. (Yes, I know I should probably make that clearer, but I need to go.)

Back to Jostpuur's

The tricky parts happen when one tries to combine the probability concepts
sensibly with special relativity.

In my opinion, strangerep and demystifier have written excellent posts in this thread. 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. Thus you'll have something like UaU*, which will give the Lorentz transformed destruction operator. The detailed form will have a SQRT(E/E') factor, where E' is the transformed energy -- change of measure. This whole subject is extensively discussed in Weinberg's book.

Regards,
Reilly Atkinson

 

[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

<br /> \psi&#039;(t&#039;,x&#039;)=\psi(t,x).<br />

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&#039;(x&#039;). 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
<br /> [J_{\mu\nu} , J_{\rho\sigma}]<br /> ~=~ \eta_{\mu\rho} J_{\nu\sigma} - \eta_{\mu\sigma} J_{\nu\rho}<br /> - \eta_{\nu\rho} J_{\mu\sigma} + \eta_{\nu\sigma} J_{\mu\rho}<br />
(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
<br /> [J_{\mu\nu}, P_{\rho}] ~=~ \eta_{\mu\rho} P_{\nu} - \eta_{\nu\rho} P_{\mu}<br />
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
<br /> N ~:=~ \int_k \dots a_k^\dagger \, a_k<br />
and the momentum operator is like
<br /> P_i ~:=~ \int_k \dots \, k_i \, a_k^\dagger \, a_k<br />
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

<br /> \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}},<br />

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

<br /> \psi&#039;(t&#039;,\boldsymbol{x}&#039;) = \psi(t,\boldsymbol{x}) = \psi\big(\frac{t&#039; + u(x&#039;)^3}{\sqrt{1-u^2}}, (x&#039;)^1, (x&#039;)^2,\frac{(x&#039;)^3 + ut&#039;}{\sqrt{1-u^2}}\big) = \psi((\Lambda^{-1})^0{}_{\mu} (x&#039;)^{\mu},\ldots, (\Lambda^{-1})^3{}_{\mu} (x&#039;)^{\mu}).<br />

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

<br /> \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}).<br />

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

<br /> \hat{\psi}&#039;(t&#039;,\boldsymbol{p}&#039;) = \int d^3x&#039;\;\psi&#039;(t&#039;,\boldsymbol{x}&#039;)e^{-i\boldsymbol{x}&#039;\cdot\boldsymbol{p}&#039;} = \int d^3x&#039;\;\psi\big(\frac{t&#039; + u(x&#039;)^3}{\sqrt{1-u^2}}, (x&#039;)^1, (x&#039;)^2, \frac{(x&#039;)^3 + ut&#039;}{\sqrt{1-u^2}}\big) e^{-i\boldsymbol{x}&#039;\cdot\boldsymbol{p}&#039;}<br />

<br /> =\int d^3x&#039;\;\Big(\int\frac{d^3p}{(2\pi)^3}\hat{\psi}\big(\frac{t&#039; + u(x&#039;)^3}{\sqrt{1-u^2}},\boldsymbol{p}\big)e^{i((x&#039;)^1,(x&#039;)^2,((x&#039;)^3 + ut&#039;)/\sqrt{1-u^2})\cdot\boldsymbol{p}}\Big)e^{-i\boldsymbol{x}&#039;\cdot\boldsymbol{p}&#039;}<br />

<br /> =\int\frac{d^3x&#039;\;d^3p}{(2\pi)^3}\hat{\psi}\big(\frac{t&#039; + u(x&#039;)^3}{\sqrt{1-u^2}},\boldsymbol{p}\big) e^{-i(\Lambda^{-1})^i{}_{\mu}(x&#039;)^{\mu} p_i - i\boldsymbol{x}&#039;\cdot\boldsymbol{p}&#039;} = \cdots<br />

<br /> \hat{\psi}\big(\frac{t&#039; + u(x&#039;)^3}{\sqrt{1-u^2}},\boldsymbol{p}\big) = e^{-i((t&#039; + u(x&#039;)^3)/\sqrt{1-u^2})\sqrt{|\boldsymbol{p}|^2 + m^2}}\hat{\psi}(0,\boldsymbol{p}) = e^{-i(\Lambda^{-1})^0{}_{\mu}(x&#039;)^{\mu}p_0}\hat{\psi}(0,\boldsymbol{p})<br />

<br /> \cdots = \int\frac{d^3x&#039;\;d^3p}{(2\pi)^3} \hat{\psi}(0,\boldsymbol{p}) e^{-i(\Lambda^{-1})^{\nu}{}_{\mu}(x&#039;)^{\mu}p_{\nu} - i\boldsymbol{x}&#039;\cdot\boldsymbol{p}} = \cdots<br />

At this point a change of variable

<br /> \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}<br />

comes handy.

<br /> \cdots = \int\frac{d^3x&#039;\;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&#039;)^{\mu}\bar{p}^{\alpha} - i\boldsymbol{x}&#039;\cdot\boldsymbol{p}&#039;} = \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&#039;)^0\bar{p}_0} \int\frac{d^3x&#039;}{(2\pi)^3} e^{i\boldsymbol{x}&#039;\cdot(\bar{\boldsymbol{p}} - \boldsymbol{p}&#039;)}<br />
<br /> =\frac{E_{\Lambda^{-1}(\boldsymbol{p}&#039;)}}{E_{\boldsymbol{p}&#039;}} \hat{\psi}(0,\Lambda^{-1}(\boldsymbol{p}&#039;)) e^{-it&#039;E_{\boldsymbol{p}&#039;}}.<br />

With t&#039;=0 that means

<br /> E_{\boldsymbol{p}&#039;}\hat{\psi}&#039;(0,\boldsymbol{p}&#039;) = E_{\Lambda^{-1}(\boldsymbol{p}&#039;)}\hat{\psi}(0,\Lambda^{-1}(\boldsymbol{p})) = E_{\boldsymbol{p}}\hat{\psi}(0,\boldsymbol{p}).<br />

If we now define

<br /> \phi(\boldsymbol{p}) = E_{\boldsymbol{p}}\hat{\psi}(\boldsymbol{p}),<br />

we can write the spatial wave function like

<br /> \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}},<br />

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

<br /> \psi&#039;(\boldsymbol{x}&#039;)=\psi&#039;(0,\boldsymbol{x}&#039;) = \int\frac{d^3p&#039;}{(2\pi)^3}\hat{\psi}&#039;(0,\boldsymbol{p}&#039;)e^{i\boldsymbol{x}&#039;\cdot\boldsymbol{p}&#039;} = \int\frac{d^3p&#039;}{(2\pi)^3}\frac{1}{E_{\boldsymbol{p}&#039;}} \phi&#039;(\boldsymbol{p}&#039;)e^{i\boldsymbol{x&#039;}\cdot\boldsymbol{p}&#039;}<br />

could as well be obtained by substituting \phi&#039;(\boldsymbol{p}&#039;)=\phi(\boldsymbol{p}) and performing the change of integration variable \boldsymbol{p}\to\boldsymbol{p}&#039; 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:

<br /> 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&#039;}{(2\pi)^3}\frac{1}{E_{\boldsymbol{p}&#039;}} \underbrace{\phi(\Lambda^{-1}(\boldsymbol{p}&#039;))}_{=\phi&#039;(\boldsymbol{p}&#039;)} |\boldsymbol{p}&#039;\rangle<br />

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

<br /> \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}},<br />

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

<br /> \psi&#039;(t&#039;,\boldsymbol{x}&#039;) = \psi(t,\boldsymbol{x}) = \psi\big(\frac{t&#039; + u(x&#039;)^3}{\sqrt{1-u^2}}, (x&#039;)^1, (x&#039;)^2,\frac{(x&#039;)^3 + ut&#039;}{\sqrt{1-u^2}}\big) = \psi((\Lambda^{-1})^0{}_{\mu} (x&#039;)^{\mu},\ldots, (\Lambda^{-1})^3{}_{\mu} (x&#039;)^{\mu}).<br /> (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

<br /> \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}).<br />

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&#039;(t&#039;,x&#039;)=\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?