# From observable to operators in QFT

#### bhobba

Mentor
In plain quantum mechanics. The energy and momentum are in the hamiltonians of the wave function.
That's incorrect.

QM is a theory about observations. It only has properties like energy or momentum when its observed to have those properties. To determine the statistical outcome of those observations you need both the observable and the state then you use the Born rule. A QFT field is as real or not real as any observable in QM. It's not real in the sense of having that value - it is real in the sense it allows us to determine things that are very real - observations.

Thanks
Bill

#### lucas_

That's incorrect.

QM is a theory about observations. It only has properties like energy or momentum when its observed to have those properties. To determine the statistical outcome of those observations you need both the observable and the state then you use the Born rule. A QFT field is as real or not real as any observable in QM. It's not real in the sense of having that value - it is real in the sense it allows us to determine things that are very real - observations.

Thanks
Bill

"I think the only way to understand QFT is to accept that quantum fields are NOT some "physical objects" that can have "states" and "observables". Quantum fields are just purely mathematical constructs (abstract operators in the Fock space) which appear to be useful for building relativistic Hamiltonians. I suggest you to re-read Weinberg's vol. 1. This book is excellent in everything except its title. QFT is not about dynamics of fields. QFT is a theory about systems with varying numbers of particles. Quantum fields play only a technical role there.

Eugene."

In case you know Eugene. How was his QFT different from the mainstream? Im using his context as I learnt it yesterday from him. Do you also believe the above? But he said opposite to you.

#### bhobba

Mentor
"I think the only way to understand QFT is to accept that quantum fields are NOT some "physical objects" that can have "states" and "observables". Quantum fields are just purely mathematical constructs (abstract operators in the Fock space) which appear to be useful for building relativistic Hamiltonians. I suggest you to re-read Weinberg's vol. 1. This book is excellent in everything except its title. QFT is not about dynamics of fields. QFT is a theory about systems with varying numbers of particles. Quantum fields play only a technical role there.
That's his view - not mine. I managed to find the following that gives mine pretty well:
http://www.quora.com/Are-quantum-fields-real-or-merely-a-mathematical-tool-used-to-describe-elementary-particles [Broken]

However it makes zero difference to the theory itself.

Thanks
Bill

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#### lucas_

In quantum mechanics, you need to get the operators from your pocket in terms of a calculator to compute the eigenstate (eigenvalues) of the observable.
But in quantum field theory, the operators are labels in points of spacetime, hence the operators are just floating in the vacuum (before measurements) instead of being located inside your calculator.

This is an accurate way to imagine it right guys (others beside Bhobba)? What you think?

#### strangerep

In quantum mechanics, you need to get the operators from your pocket in terms of a calculator to compute the eigenstate (eigenvalues) of the observable.
But in quantum field theory, the operators are labels in points of spacetime, hence the operators are just floating in the vacuum (before measurements) instead of being located inside your calculator.

This is an accurate way to imagine it right guys (others beside Bhobba)? What you think?
What do I think? Well,... for most of the statements you make, I feel a desire to say: "rubbish" -- so I bite my tongue and (mostly) stay silent.

But I'll try to be more constructive: I think you should put QFT aside for a while and go back and study ordinary QM properly. Basic stuff: like how a Hilbert space is constructed to be a representation of a (physically relevant) dynamical symmetry group; like how the "unitary irreducible representations" of the rotation group give the observed half-integral angular momentum spectrum; and like how a tensor product of Hilbert spaces is used to model multiparticle systems.

Can you access a copy of Ballentine's QM text? If you can cope with the level of math therein, I guarantee that such study would be much more profitable than persisting with all the ill-informed heuristic flubbing around that's been happening in this thread.

------------------------------
BTW, I know Eugene reasonably well, and I was here when those older threads were in progress. The underlying points of difference between Eugene's view and the view of some others are on a rather advanced level -- concerned with difficult convergence issues in advanced QFT, and how useful QFTs may be constructed. I don't recommend that any QFT beginner get embroiled in that stuff, but rather master a basic textbook level of QFT first before confronting the more archane aspects.

#### atyy

In quantum mechanics, you need to get the operators from your pocket in terms of a calculator to compute the eigenstate (eigenvalues) of the observable.
But in quantum field theory, the operators are labels in points of spacetime, hence the operators are just floating in the vacuum (before measurements) instead of being located inside your calculator.

This is an accurate way to imagine it right guys (others beside Bhobba)? What you think?
The construction of the non-relativistic quantum mechanical momentum and position operators from the quantum field operators is given in David Tong's QFT notes http://www.damtp.cam.ac.uk/user/tong/qft.html, section 2.8.1.

#### QuasiParticle

Apparently the momentum of one particle cannot be expressed in the second quantization formalism.
I'm a little confused as to what this means. Surely in second quantization one can use the momentum representation, where the momenta are analogous to the positions in the position representation.

#### atyy

I'm a little confused as to what this means. Surely in second quantization one can use the momentum representation, where the momenta are analogous to the positions in the position representation.
Me too. It looks like the total momentum operator is the usual momentum operator of one particle - see Tong's notes linked two posts above.

#### vanhees71

Gold Member
I do not understand what you mean. The momentum operator is given by the corresponding momentum density which is given by the (Belinfante) energy-momentum tensor. For the free Klein-Gordon field it's given by
$$\vec{\pi}=-:\dot{\phi} \vec{\nabla} \phi^{\dagger}+\dot{\phi}^{\dagger} \vec{\nabla} \phi:.$$
The colons mean normal ordering.

#### atyy

I do not understand what you mean.
Neither do I, I was quoting a reference given in post #30.

The question I was trying to answer is: if we use the second-quantized langauge and write non-relativistic quantum mechanics of many identical particles as a quantum field theory, how are the usual momentum and position operators of single-particle quantum mechanics written in the second quantized form?

As I understand the answer is given in section 2.8.1 of Tong's notes linked in post #56.

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#### vanhees71

Gold Member
Yes, that's the non-relativistic case. Note that this is a bit incomplete. One has to carefully check that the suggested operators fulfill the operator algebra, i.e., the commutation relations following from their group-theoretical meaning according to Noether's theorem. For non-relativistic QT this is the Lie algebra of a central extension of the covering group of the Galilei group, and for relativistic QT that of the covering group of the Poincare group, SL(2,C).

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