# A Fock space and Poincaré invariance

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1. Jun 27, 2016

### Gedankenspiel

Hi all,

is Fock space Poincaré invariant? As far as I can see, the scalar product in Fock space involves the scalar products in its N-particle subspaces, which, in turn, are the integrals of the properly (anti-)symmetrized wave functions over space.

This works well in a Galilei-invariant theory because a Galilei transformation just tilts the time axis and leaves the space axis untouched, so space is the same in all inertial systems. But a Lorentz transformation also tilts the space axis (in 2D for simplicity). This means that the spatial integrals in one inertial system extend over points of space-time with different time coordinates in another inertial system. So the dynamics seems to play an important role in guaranteeing the normalization of the state. Is this accounted for in the construction of Hamiltonians in QFT?

What's more, the state has to be normalized for any inertial observer: for any possible space axis the sum over particle number of the spatial integrals has to be one. This seems to be an extremely restrictive condition to me. Can this be guaranteed? Can Fock space be a state space for interacting QFT at all?

Cheers,

Gedankenspiel

2. Jun 28, 2016

### vanhees71

By construction any relativistic QFT is defined by a unitary representation of the covering group of the proper orthochronous Poincare group. In addition it's assumed to have a Hamiltonian that's bounded from below (i.e., there exists a stable ground state) and that the interactions are local. This ensures that the S-matrix is Poincare covariant and the validity of the linked-cluster property. For details, see Weinberg, Quantum Theory of Fields, vol. I.

3. Jun 28, 2016

### A. Neumaier

There are different Fock spaces, two (one bosonic and one fermionic) for every 1-particle space. The Fock space is Poincare invariant iff the single-particle space one starts with is Poincare invariant.

4. Jun 28, 2016

### A. Neumaier

No, by Haag's theorem. But it can be approximated by Fock spaces with cutoffs. This is the background for renormalization theory.

5. Jun 28, 2016

### Gedankenspiel

This equivalence makes perfect sense to me. But I cannot imagine a way in which a single particle state space can be Poincaré invariant: This would mean that there is a complex function on Minkowski space whose squared modulus integrated over ANY possible space axis yields one. I did not attempt to prove it but this seems an extremely restrictive condition to me which no function might be able to fulfil.

However, if Fock space is the state space of non-interacting QFT, there must be a way out. How can this come about?

6. Jun 28, 2016

### A. Neumaier

The Hilbert space of a single free Klein-Gordon (or Dirac) particle of mass $m$ is Poincare invariant, and the corresponding bosonic (or fermionic) Fock space describes noninteracting fields.

Last edited: Jun 29, 2016
7. Jun 28, 2016

### DarMM

Your close to the answer here. The single particle space for a single relaitivistic boson is (in momentum space for ease):

$L^{2}(\mathbb{R}^{3},d\nu)$

However the measure $d\nu$ is not $d^{3}p$, but $\frac{d^{3}p}{2E_{p}}$. Try a Lorentz transformation and you'll see it is preserved under transformations.

8. Jun 30, 2016

### Gedankenspiel

Let's stay in position space for now, in 1+1 dimensions.

If I have a single free Klein-Gordon particle then the squared modulus of the wave function in the t-x-system integrated over the x-axis must give 1 for any time coordinate t (because the wave function is a wave packet, not a plane wave, otherwise it would be no valid state: the particle must be ANYWHERE with certainty).

Now making a Lorentz transformation into the t'-x'-system. In order to be a valid state for this observer as well, the integral over the x'-axis must be 1, too. But in fact, if I did not miscalculate, the integral gives $\frac{1}{\gamma (1-v)}\neq 1$ with $\gamma = \frac{1}{\sqrt{1-v^2}}$ (with m=0 for simplicity, but a massive particle is no better).

So I am led to the conclusion that a one-particle space is not Lorentz invariant and so isn't the whole Fock space, even for free particles. This makes me wonder what the state space for QFT is.

Or is there any flaw in the argument?

9. Jun 30, 2016

### DarMM

Yes, you must transform the measure, which in position space will involve an inverse differential operator.

Also there isn't really anything to be gained in understanding by moving back to position space, as the reason is the same:

The measure is composed of two parts, the standard measure $dtdx$ and a weighting factor $L^{-1}$ and their Lorentz transformations cancel out.

10. Jul 1, 2016

### vanhees71

Well, all this only holds, if you assume that you could interpret relativistic wave equations as in non-relativistic Schrödinger wave mechanics. This is, to a certain extent, possible for non-interacting particles, but it doesn't work for interacting particles. The reason is very easy to understand nowadays: A single-particle picture doesn't make sense in the relativistic realm since at relativistic energies it is always probable to create and/or destroy particles in scattering processes. Thus you need a formalism for reactions with non-conserved particle numbers, and that's QFT. What's conserved in relativistic physics are not particle numbers but the various conserved charges, particularly those, to which the gauge fields couple.

Another more formal argument you have given yourself. It's hard to define, within the single-particle picture, a Lorentzinvariant probability measure. The reason is that the conserved charge-like quantities (from invariance under global phase transformaions, i.e., U(1) symmetries) are usually not positive definite and thus cannot be used to define a probability measure. This implies that it is impossible to define a probability measure in terms of a positive definite charge density that is conserved, and this implies that the the normalization integral is not Lorentz invariant. So also the math is telling you, why you should use a QFT and not a single-particle wave-mechanics formalism to build a quantum theory of relativistic particles. A great deal of the structure of the very successful relativistic QFT (aka Standard Model) can be analyzed and understood by constructing the unitary representations of the proper orthochronous Lorentz group. For an introduction have a look at my QFT manuscript:

http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf

or for the full story

S. Weinberg, The Quantum Theory of Fields, vol. 1, Cambridge University Press

Last edited: Jul 1, 2016
11. Jul 2, 2016

### Gedankenspiel

Thanks for all your answers! But I still lack some understanding here.

Here is my current understanding:

1. State space of non-interacting QFT is a Fock space.

2. Under Lorentz transformation the N-particle subspaces transform separately, as I have learned from one of the answers above. So, a one-particle state in one inertial system is just that in another.

3. For a one-particle state I have a wave function in position space whose integral over the x-axis (in 1+1 for simplicity) is 1 for any time t. (Otherwise it wouldn't be the wave function of a one-particle state. Moreover, the argument with the non-positivity of charge-like quantities does not seem to apply here as total probability is, though conserved, not charge-like. Its conservation is not due to a continuous symmetry)

4. The same must hold true in the transformed system, e.g. for t'= 0. This means that as soon as I have the wave function $\psi(t',x')$, the equation $\int dx' |\psi(t'=0,x')|^2 = 1$ must hold. I cannot see why any factor from a measure transformation is supposed to enter here. This equation does not even make reference to the unprimed system!

5. Having a normalized wave packet $\psi(t=0,x)$ I know the wave function at all times because the theory is free.

6. I can express $(t,x)$ in terms of $(t',x')$, so I have $\psi(t',x')$.

7. Calculating the above "normalization" it turns out to be not 1 but a velocity-dependent factor as shown in #8.

Conclusion: Fock space is not Lorentz-invariant even in a non-interacting theory.

It may well be that momentum space would be easier to use, sure. But things should look right seen from all angles, shouldn't they? I am just trying to find out exactly where and if this line of reasoning goes wrong.

12. Jul 2, 2016

### strangerep

It's hard to comment on that if you don't show your calculation...

One possibility is that maybe you're not doing the calculation in such a way as to remain in an irreducible representation? But that's best done in momentum space where one can easily express the constraint $E^2-p^2=M^2$. That gets quite messy in position space.

Also remember that the 1-particle state spaces are constructed explicitly such that they carry a (causal) field rep of the Poincare group. So if you find that not to be the case, then you're working with the space incorrectly.

13. Jul 3, 2016

### A. Neumaier

The probability interpretation of the 1-particle wave function is inherently nonrelativistic, hence not directly related to the covariant representation used in quantum field theory, but given by an observer-dependent so-called Foldy-Wouthuysen transformation.

This reconciles the apparent contradiction. Your steps 3-7 apply to the Foldy-Wouthuysen form of the wave function, not to the covariant one.

14. Jul 3, 2016

### DarMM

It's step 4. The correct measure isn't $dx$, it's $L^{-1}dx$ where $L$ is the differential operator which is the Fourier transform of $2E_{p}$.

15. Jul 4, 2016

### Gedankenspiel

@strangerep: Maybe the calculation is unnecessary with Arnolds post. I'll post it if the FW transformation doesn't answer my questions.

@DarMM: I doubt this. But even in this case the factor I get has a difficult dependence on v which the sole $L^{-1}$ couldn't account for.

Moreover, for different masses this factor is different. So, if L is not mass dependent as well, it won't work. But L cannot be mass dependent because the measure in space-time cannot know about the masses of the particles I want to describe in it.

@A. Neumaier: That's interesting! I heard about that transformation but never knew much about it. I'll check it out and come back if my problem is not resolved.
But maybe you know literature (easily accessible, if possible) that deals explicitly with Fock space and Lorentz transformations and addresses my questions?
Thanks a lot!

16. Jul 4, 2016

### A. Neumaier

The lorentz invariance of the free field is treated at the beginning of every textbook about quantum field theory. Your questions have nothing to do with Fock space per se, as the problems in the Fock space are solved once they are solved in the single-particle case.

17. Jul 4, 2016

### DarMM

Why?

I mean, it basically has to involve $L$ rather than a flat measure, as this is the position-space implementation of confinement to the mass-shell.

It does, just do the Lorentz transformation.

L is mass dependent. It is the Fourier transformation of $2E_{p}$ after all.

Why?

The measure I've given is the correct measure, see p.13 of these notes by Arthur Jaffe:
http://www.arthurjaffe.com/Assets/pdf/IntroQFT.pdf

Last edited: Jul 5, 2016
18. Jul 5, 2016

### Gedankenspiel

The Lorentz invariance of the free field indeed is, but for the question I am after I couldn't find an answer although it is very basic: if in a free QFT I have a one-particle state described by a wave packet in position space, $\psi(t,x)$, how then does the wave packet look like in another inertial frame, i.e., $\psi(t',x')$?

I still have not digested your statement that the probability interpretation of the 1-particle wave function is inherently nonrelativistic. Fock space (and the one-particle space as a subspace) is the state space of relativistic QFT. If the states in it are not to be interpreted probabilistically, what do they mean then?

19. Jul 5, 2016

### Gedankenspiel

20. Jul 5, 2016

### A. Neumaier

$\psi(x)$ with 4D $x$ transforms to $\psi(\Lambda x)$ under the Lorentz transformation $\Lambda$. This should be in every textbook!
For a normalized state $\psi$ and any operator $A$, the number $\bar A=\psi^*A\psi$ is the frame-independent ensemble mean of $A$. This is enough to interpret the theory. But position probabilities need a particular frame since the integral of probabilities over space at a fixed time must integrate to 1 in the Lebesgue measure on each space slice. This makes the probabilities frame-dependent.