I Commuting observables for Fermion fields?

  • Thread starter maline
  • Start date
395
51
In nonrelativistic QM, we usually describe the Hilbert space by choosing a complete set of commuting observables, so that the set of states that are eigenstates of all the observables can be used as a basis. For instance, the "wavefunction" is the state as expressed in terms of "states" with definite positions for all particles (not normalizable, but that can be dealt with), which works because the positions (3 for each particle) are a complete set of commuting observables. Another classic example is the n,l,m,s basis for the states of the hydrogen atom.

For free boson fields, this concept seems to carry over well into QFT. For instance, in the case of a complex scalar field, the real and imaginary parts of the field at each point (on a spacelike hypersurface) are Hermitian and commute, so we can think of the well-defined field configurations as "basis states", and then a general state will be a functional giving a quantum amplitude to each configuration.

This "simple" setup obviously does not work for fermion fields. There, we are given anticommutation relations, which don't seem very helpful for defining basis states. Even the real and imaginary parts of the field at the same point do not commute.

What would a complete set of commuting observables look like for a fermion field, say a free Dirac field, expressed in terms of the field operators?
 

A. Neumaier

Science Advisor
Insights Author
6,851
2,797
What would a complete set of commuting observables look like for a fermion field, say a free Dirac field, expressed in terms of the field operators?
quadratic expressions in the fermion fields
 
395
51
quadratic expressions in the fermion fields
Yes, I guessed that that could be a good direction. Do you have a detailed idea for a complete set? Can the products be arranged so as to form, say, eight commuting Hermitian operators (for the real & imaginary parts of the four components) at each space point? What would be the spectra of these operators?
 

A. Neumaier

Science Advisor
Insights Author
6,851
2,797
Yes, I guessed that that could be a good direction. Do you have a detailed idea for a complete set? Can the products be arranged so as to form, say, eight commuting Hermitian operators (for the real & imaginary parts of the four components) at each space point? What would be the spectra of these operators?
Strictly speaking, this is an ill-defined problem as the fields are distribution-valued only and hence have no spectra.

But if you discretize space you get a finite collection of qubits, and you can diagonalize each one separately. Then take a continuum limit.
 
395
51
Okay, so we're talking about operators with eigenvalues zero and one? A yes/no answer to "is there a particle at this point"?
Would there be four or eight such?
 

Want to reply to this thread?

"Commuting observables for Fermion fields?" You must log in or register to reply here.

Related Threads for: Commuting observables for Fermion fields?

Replies
2
Views
2K
Replies
3
Views
471
  • Posted
Replies
6
Views
1K
  • Posted
Replies
8
Views
1K
Replies
9
Views
4K
Replies
37
Views
5K
Replies
7
Views
2K
Replies
5
Views
2K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top