Fields as operator valued distributions

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  • #1
naima
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Main Question or Discussion Point

As fields ##\phi ## are ill defined at precise time and position i read that fields have to be smeared. So we have test functions f in bounded regions in space time. We have a Hilbert space and ##\phi (f) ## is an operator which acts on H.
Maybe we can retrieved the usual wave function when it acts on the vacuum of H?
So we start from a test function defined on a region of spacetime. Does this distribution evolves in time? Have i to write ##\phi (t, f) ## ?
If f has a support in a space time domain D,what about the wave function outside of D?
I can understand that f is null behind the walls of the lab, before and after the apparatus is switched off. What is the relation to the state vector of the measured observable ?

thank you.
 

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  • #2
A. Neumaier
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The state $$\Phi(\psi_1)\otimes \Phi(\psi_2)\otimes\cdots\otimes \Phi(\psi_N)|vac\rangle$$ is (for nonrelativistic fields) just the tensor product of N one-particle states $$\psi_k (k=1,\ldots,N),$$ hence it is an N-particle wave function.
 
  • #3
naima
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The state $$\Phi(\psi_1)\otimes \Phi(\psi_2)\otimes\cdots\otimes \Phi(\psi_N)|vac\rangle$$ is (for nonrelativistic fields) just the tensor product of N one-particle states $$\psi_k (k=1,\ldots,N),$$ hence it is an N-particle wave function.
if ##\Phi(\psi)|vac\rangle## is the ##\psi ## wave function how can we compute ##\psi (x)##?
How does it evolve in time?
 
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A. Neumaier
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if ##\Phi(\psi)|vac\rangle## is the ##\psi ## wave function how can we compute ##\psi (x)##?
How does it evolve in time?
For any ##\psi## there is an associated wave function. Like everywhere in quantum mechanics, you cannot compute it but need to make assumptions on it that define the state at a particular time. Change of time is as always given by the Schroedinger equation for the whole state. if there are interactions, the evolved state will be entangled, hence no longer of the above simple tensor product form.

For relativistic fields, ##\psi## is a function of space-time, and the tensor product state has 4N coordinates. Now time evolution is just shifting all times by some fixed time interval.
 
  • #5
naima
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Thank you Neumeier

My "problem" is about time.
I suppose that a field is something which evolves with time according to an Hamiltonian.
So at t = 0 it associates to a test function f a vector in H. This test function is defined on a compact region of the space time which can be in the far future of t = 0.
in fact we have a state ## \Psi (t=0 , f(x,t')) \rangle ##
the field erases the notion of the test function time (which is bounded) and makes the result evolves with no upper limit in time.
Another test function will give another wave function.
We have probability rules for state vectors but as they come from test functions have we rules for test functions?
 
  • #6
A. Neumaier
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I suppose that a field is something which evolves with time according to an Hamiltonian.
So at t = 0 it associates to a test function f a vector in H. This test function is defined on a compact region of the space time which can be in the far future of t = 0.
in fact we have a state ## \Psi (t=0 , f(x,t')) \rangle ##
the field erases the notion of the test function time (which is bounded) and makes the result evolves with no upper limit in time.
Another test function will give another wave function.
We have probability rules for state vectors but as they come from test functions have we rules for test functions?
The notion of a test function is just a mathematical concept to turn a distribution into something rigorously defined. By the way, nothing demands that you smear with a function of compact support; in the standard examples you only need a Schwartz function since the distributions are tempered. The physical meaning of a test function is, as I had explained, that of a 1-particle wave function smeared in space and time; so for an interpretation in physical terms one should write ##\psi(x,t)## rather than ##f(x,t)## and equate the two concepts. The traditional meaning of expectations is then an expectation smeared in time, too (as in fact for any real measurements). The usual quantum-mechanical state at any time ##t_0## is obtained in the limit where ##\psi(x,t)## factors as $$\psi(x,t)=\psi(x)\delta(t-t_0)$$ which is realizable as a limit of truly smeared states where the smearing in time gets more and more tiny.

This works perfectly in the nonrelativistic case and in 2-dimensional relativistic quantum field theories, but such limiting states are believed not to exist for 4-dimensional relativistic quantum fields - which makes ''moments of time'' intrinsically uncertain.
 
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naima
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Sorry I had in mind that the support of a test function had to be compact in the wightman formalism. I do not remember where this idea came from.
 
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I find where i read that.
It was PCT, Spin and statistics and all tht.
Wightman writes.
"For example , in the case of the electric field ##\mathscr{E}(x,t)## is not a well defined operator, while ##\int dx dt f(x,t) \mathscr{E}(x,t) = \mathscr{E}(f)## is. Here f is ANY infinitely differentiable function of compact support in space-time"

You say that the test functions have to be identified with the wave functions of the field:
##f(x,t)## eq ##\langle x |\Psi(f) \Omega \rangle##
But i do not find an axiom for something like that.
Do you want to say that the only interesting test functions are those with the possible identification?
in http://arxiv.org/abs/gr-qc/0212074
Rovelli writes
"
Consider a finitely extended open subset O of space-time
. The algebra generated by all operators ##\phi (f)##
where f has support in O
is interpreted as representing physical operations that can be performed
within O"
He does not seem to think that all these test functions are only math.
 
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  • #9
A. Neumaier
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I find where i read that.
It was PCT, Spin and statistics and all that.
Wightman writes.
"For example , in the case of the electric field ##\mathscr{E}(x,t)## is not a well defined operator, while ##\int dx dt f(x,t) \mathscr{E}(x,t) = \mathscr{E}(f)## is. Here f is ANY infinitely differentiable function of compact support in space-time"
This gives a sufficient condition only. In fact, since the possible singularities of the e/m field are known you can use much more general test function - test functions that (in momentum space) are L^2 everywhere and sufficiently nice on the light cone.

You say that the test functions have to be identified with the wave functions of the field:
##f(x,t)## eq ##\langle x |\Psi(f) \Omega \rangle##
But i do not find an axiom for something like that.
It doesn't need axioms for this. Look at the second quantization procedure in any textbook on statistical mechanics to convince you of that.

in http://arxiv.org/abs/gr-qc/0212074
Rovelli writes
"Consider a finitely extended open subset O of space-time
. The algebra generated by all operators ##\phi (f)##
where f has support in O
is interpreted as representing physical operations that can be performed
within O"
He does not seem to think that all these test functions are only math.
He defines the local algebra of observables in the region O. They are enough for consideration of everything a given physicist (with finite lifetime) can measure, but not enough for consideration of a system with unlimited lifetime. They don't say anything about the field outside O but the latter is needed for a complete description since it influences O. In local, algebraic quantum field theory (Haag) one needs the collection of all local algebras and has as wave functions the functions in the Hilbert space of an irreducible representation of the whole net of local algebras. If one uses the Fock representation, the states of this Hilbert space contain all the states I had been talking about. One gets more than the wave functions with compact support since the Hilbert space is complete, and sequences of functions with compact support can converge to certain functions that are nonzero everywhere.
 
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  • #10
naima
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Is the nuclear theorem related to what you try to tell me from the beginning?
 
  • #11
A. Neumaier
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Is the nuclear theorem related to what you try to tell me from the beginning?
No. The spaces of interest in quantum field theory don't seem to have a nuclear structure.
 
  • #12
naima
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Herr Neumaier,

Your answers are like fireworks in the dark night.
 

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