# I Wilsonian viewpoint and wave function reality

1. Mar 26, 2016

### atyy

In the Wilsonian viewpoint, quantum electrodynamics is an effective theory, where the low energy predictions are obtained by coarse graining eg. some form of lattice QED where the lattice is taken very finely.

In the Copenhagen interpretation, we are agnostic as to whether the wave function is real, or for that matter whether observables are real. Only experimental results are real classical events, and theory only gives the probabilities of experimental results.

Does the Wilsonian viewpoint require taking the lattice to be real or classical, in the same way that Minkowski spacetime is considered real or classical in special relativistic quantum field theory?

2. Mar 27, 2016

### vanhees71

I don't think so since the lattice regularization is only one possible regularization of QFT. You can as well use a momentum cutoff or a smooth cutoff function. So why should an arbitrary space-time lattice have any more "reality" than any other cutoff? Particularly QED is a Dyson-renormalizable theory, which means that for observables at low energies the high-energy scales don't play lot of a role. That's why you can take the cutoff to infinity at the prize of having to introduce a renormalization scale, upon which the renormalized coupling depends.

3. Mar 27, 2016

### A. Neumaier

Removing the cutoff (in the lattice case taking the continuum limit) is essential to get the correct Poincare symmetry. The approximate theories are only construction tools, not the real thing.

4. Mar 27, 2016

### atyy

I am not asking specifically about the lattice cutoff. I am asking about the idea of coarse graining. What is being coarse grained? Is the thing that is being coarse grained real?

5. Mar 27, 2016

### A. Neumaier

The stuff QFT is about: expectations of fields at a point and correlation functions, expectations of suitably (time- or normally) ordered products of fields at several points.

The sufficiently coarse-grained version is observable; nobody talks about what is real.

But there is no reason to suppose that this implies a lack of reality. Talk about reality is needed only when someone questions it.

Last edited: Mar 27, 2016
6. Mar 28, 2016

### vanhees71

Real is what is observable! So the answer is that the coarse-grained quantities that are really measured are real, what else?

7. Mar 28, 2016

### atyy

But what is being coarse grained?

For example, in http://arxiv.org/abs/1502.05385 the renormalization gives a flow in the space of wave function and Hamiltonians. Are the flows of the wave function and Hamiltonians just tricks, and is it always more fundamental to conceive the flow in the space of correlation functions, as A. Neumaier says?

Last edited: Mar 28, 2016
8. Mar 28, 2016

### vanhees71

What's being coarse-grained are the microscopic degrees of freedom, we even might not know with our contemporary experimental abilities. That's the great point of the renormalization group in Wilson's interpretation of it: It doesn't matter what the "underlying" microscopic degrees of freedom really are, but we can use QFT as an effective description of what's observable or better said resolvable with our detectors. If you look at, say, a proton at very low energies it's pretty well described as a heavy point particle or as a Coulomb field concerning the electrons surrounding it building an atom. If you start scattering electrons on it, you'll find that it is in fact an extended object and you can characterize it with a form factor, relating to a charge distribution of an extended object. If you enhance the energy even further you start to resolve the constituent valence quarks and even the sea quarks and gluons etc. Who knows, whether this is the final answer to what a proton really is, but we can describe at the so far accessible energies (or resolutions) the observations with regard of what we call a proton. I'd not dare to say that we understand protons (or any other hadron) fully, but only within some "blurred" picture limited by the resolution of our probes and detectors, and that observables are real. Anything else we might think about "what's coarse-grained" is speculation as long as it isn't observable and thus it's good to stick to the rule that "real is what's observable", and that might change when we get refined measurement devices and that's why the picture about what a proton is changed from, say, Rutherford to today a great deal, and I guess it will change also great deal in the future :-).

9. Mar 28, 2016

### A. Neumaier

Are quarks real? Measured are only the cross sections of leptons and hadrons....

10. Mar 28, 2016

### vanhees71

Quarks are real. The only question is what you call quarks. Quarks are observable, e.g., in the deep inelastic scattering, demonstrating Bjorken scaling and its violation, which is well explained by QCD (DGLAP equation etc.). For sure what's not directly observable are the quanta of the quark fields occuring in the QCD Lagrangian, because these "current quarks" do not occur as "asymptotic free states" in our detectors. This holds also for their properties like the current-quark masses, which are not directly observable but inferred from theory as parameters entering the standard-model Lagrangian (in terms of Yukawa Couplings of the quarks to the Higgs field).

11. Mar 28, 2016

### A. Neumaier

Coarse-graining always means restricting the algebra of observables to a smaller one consisting of fewer and/or more averaged quantities. This is visible in the most general form of the projection operator formalism that is the basis of all reduced descriptions derived with sufficient care. Instead of a big algeba of observables evolving in the Heisenberg picture on the fixed state of the universe, one only considers an algebra of relevant observables, and restricts the state of the universe to this effective algebra.

In the tensor network paper you linked to, the state of the ''universe of interest'' is the ground state of a spin system, and the effective algebra is the algebra of linear operators preserving a subspace of the Hilbert space, the reduced Hilbert space selected by the tensor network structure. The state of the universe restricted to this algebra is a state of this reduced Hilbert space - in general a mixed state of the reduced system but approimated by a pure state (to be determined by a variational process).

In Wilson's view, one has a parameterized family of algebras, labelled by a scale parameter $\Lambda$ inducing a labelling of all observables $A(\Lambda)$. Each of these can be obtained by some explicit expression in the observables $A(\Lambda')$ for some or all $\Lambda'>\Lambda$, which defines the embedding of the algebras in each other. Wilson renormalization is the fact that the temporal dynamics of the $A(\Lambda)$ can be approximately described by an effective Hamiltonian $H(\Lambda)$ for which one can derive a renormalization group equation.

12. Mar 28, 2016

### A. Neumaier

On the surface, only deep elastic scattering is observable, and explained by the unobservable quarks.

13. Mar 28, 2016

### vanhees71

True, but then you can also say the same for particles that exist as asymptotic free states like electrons. We never see "electrons" in this sense but only their manifestations in interacting with macroscopic matter, like in a discharge tube by the glow of the rest gas ionized by them or as dots on an old-fashioned TV screen or as pixel signals stored on a hard disk at CERN etc.

14. Mar 28, 2016

### A. Neumaier

Indeed. That's why the question of what exists is a nontrivial one. Some argue for the existence of virtual photons by saying that we see their manifestations. So one needs more definite criteria for what is real.

My proposal for defining ''being real'' is ''having a Heisenberg state'' - see my new Insight Article. Having a state means having expectations and correlation functions, just as required above.

15. Mar 28, 2016

### vanhees71

The Insight Article is great. Now we have it to link to whenever somebody asks about "virtual particles" :-)).

One should also note, and that many professionals are not aware of, that the only clear definition of a resonances mass and width is given as the pole of the S-matrix element of a specific process. Even an apparently "simple" resonance as the $\rho$ meson is only well defined when this is kept in mind, and that's why in the Review of Particle Physics it's very well defined what is used to determine its mass and width in the tables, namely via $\mathrm{e}^+ + \mathrm{e}^- \rightarrow \pi^+ + \pi^-$ and (consistently) $\tau \rightarrow 2\pi+\nu_{\tau}$. Already in the Dalitz decays of the pseudo-scalar mesons $\eta$ and $\eta'$ lead to differences if the proper dynamics is not taken into account (not to talk about the Dalitz decays of baryon resonances, where the $\rho$ meson appears in the vector-meson resonance model for the corresponding transition-form factors, which turns out to be not too bad a model). But that's a bit off-topic here.

I've not found an explanation/definition of what you mean by "Heisenberg state" in your Insight article. It may also be good to reformulate the definition of "state" to: "A quantum state is represented by a trace-class positive semidefinite self-adjoint operator with trace 1." That holds for both pure and mixed states. The pure states are exactly represented by statstical operators that are projectors, i.e., obeying $\hat{\rho}^2=\hat{\rho}$. Another equivalent definition is that a pure state is represented by rays in Hilbert space, but this somewhat cumbersome definition is avoided by using statistical operators for both pure and mixed states.

16. Mar 28, 2016

### A. Neumaier

This and a version of the remainder should be posted in the other thread, not here.

There it doesn't matter. In the present context, it is just a state in the Heisenberg picture, so that time correlations make sense.
This would make it worse - not in the sense of incorrect but in the sense of adding technicalities that only detract from the real message.

Semidefinite already implies Hermitian, trace 1 already implies trace-class, and Hermitian trace-class operators are automatically self-adjint. Thus there is no need to introduce these technical terms. Pure states are only those projectors $\rho$ that project to a ray. Of course the trace is in this case the dimension of the target space, so your definition is ok. But these technicalities are also not needed for the insight article since I never do anything with the states.

17. Mar 28, 2016

### Ilja

How do you know that Poincare symmetry is the real thing? Because we observe it in the large distance limit? So what, all those lattice theories have no problem to recover Poincare symmetry in the large distance limit.

I would say if Poincare symmetry would be the real thing, one would be able to construct models without infinities which have full Poincare symmetry, and one would not need such non-Poincare-symmetric "construction tools".

18. Mar 28, 2016

### A. Neumaier

It is typical in mathematics that nice objects are first constructed in a messy way.
The real numbers have very nice properties but to construct them one needs artifacts that make the numbers appear to be complicated sets (Dedekind cuts, euqivalence classes of Cauchy sequences, etc.).
The exponential function has many nice properties, but to construct it one needs limits of simpler functions that do not have this property.

Another reason why Poincare symmetry is the real thing is that it gives rise (via Noether's theorem) to the conservation laws on which all basic physics relies. Drop symmetries - and you have nothing left to guide your theory building.

19. Mar 28, 2016