Fermionic Fock space superselected?

  • Thread starter Jazzdude
  • Start date
  • Tags
    Space
In summary: I am not convinced that this argument is convincing. It is not because the experiments are not the ones involved in the univalence superselection, that they cannot be taken as a reliable proof of the validity of the univalence superselection.In summary, the conversation discusses the concept of univalence superselection in fermionic Fock states. It is argued that the state resulting from a full 2pi spatial rotation is not equivalent to the initial state due to a change in relative phase. This leads to the conclusion that the fermionic particle space separates into two superselection sectors of odd and even particle numbers. However, there is debate about the validity of this concept and its implications for physical observables. Some believe that the spin-statistics
  • #1
Jazzdude
244
49
If you construct a fermionic Fock state

|psi> = a|psi1> + b|psi2>

where |psi1> is a 1-particle state and |psi2> is a 2-particle state, and you apply a full 2pi spatial rotation to it, then each particle contributes a factor of -1 to the amplitude because of the half-integer spin SO(3) representation it comes with.

R(2pi) |psi> = -a|psi1> + (-1)(-1)b|psi2> = -a|psi1> + b|psi2>

Now this state is not equivalent to the one we started with because the relative phase has changed. But because the 2pi rotation is an exact symmetry for the physical state the usual argument is that this coherent superposition should not be possible. That's the argument used in fermion-boson (univalent) superselection. The usual solution is to say that the different spin states are in different superselection sectors, and there are no physically constructible observables that can determine the relative phase, so that we can as well mix them incoherently.

So is it commonly accepted that the fermionic particle space separates into two superselection sectors of odd and even particle numbers? If not, why not?

In case I've made a silly mistake in my argument please forgive me. I didn't get a lot of sleep recently.

Thanks,
Jazz
 
Physics news on Phys.org
  • #3
Jazzdude said:
So is it commonly accepted that the fermionic particle space separates into two superselection sectors of odd and even particle numbers?
Yes, it's common.

But I like to think of it in a different, perhaps more fundamental way. Is there a QFT interaction which can create such a superposition from an initial state which is not in such a superposition? If you have some experience with practical QFT, you can find out that interaction Hamiltonian leading to such a production would not be represented by a hermitian operator, while we need hermiticity for the sake of unitarity (probability conservation).
 
  • #4
Demystifier said:
Yes, it's common.

But I like to think of it in a different, perhaps more fundamental way. Is there a QFT interaction which can create such a superposition from an initial state which is not in such a superposition? If you have some experience with practical QFT, you can find out that interaction Hamiltonian leading to such a production would not be represented by a hermitian operator, while we need hermiticity for the sake of unitarity (probability conservation).
I fail to see this. E.g. [itex]1+\psi(x)+\psi^+(x)[/itex] would create a superposition of even and odd particle number states out of a state with only even or odd particle number. However, this operator, while being hermitian, is not invariant under a 360 degree rotation.
 
  • #5
DrDu said:
I fail to see this. E.g. [itex]1+\psi(x)+\psi^+(x)[/itex] would create a superposition of even and odd particle number states out of a state with only even or odd particle number. However, this operator, while being hermitian, is not invariant under a 360 degree rotation.
Hm, but psi must have a spinor index (due to the spin-statistics theorem), while a Lagrangian should not have one. So you must contract it with something having another spinor index, which, I think, should contain another fermion field. But then you cannot create a single fermion, but only a pair of them.

But then again, the spin-statistics theorem, as well as the claim that Lagrangian should not have a spinor index, rest on the assumption of relativistic invariance. On the other hand, in recent years it is quite popular to study field theories which break relativistic invariance. So after all, maybe you right. Maybe superpositions of even and odd numbers of fermions are not impossible.
 
Last edited:
  • #6
Demystifier said:
Hm, but psi must have a spinor index (due to the spin-statistics theorem), while a Lagrangian should not have one.

That the lagrangian should not be spinorial is a good argument. However, the spin statistics theorem is not important here, as Jazzdudes argument will hold for any particle of half integer spin.

Consider the following situation: Take a long box containing exactly one electron at low temperature so that only the symmetric ground state will be populated. Now insert a wall in the middle of the box. Each box will then contain a superposition of 0 and 1 electron if such a superposition were possible at all. Now rotate one of the boxes by 360 degrees and recombine the two halves: Tataa, the electron has disappeared!
 
  • #7
DrDu said:
Consider the following situation: Take a long box containing exactly one electron at low temperature so that only the symmetric ground state will be populated. Now insert a wall in the middle of the box.
So far so good.

DrDu said:
Each box will then contain a superposition of 0 and 1 electron ...
That's wrong. Your trick with boxes is not different from a beam-splitter experiment with photons. In essence, there is no such thing as "state in the box". There is only a total state, which in this case has a form
|1>=|1L>+|1R>
which is a superposition of particle in the Left box and particle in the Right box. If you perform a measurement which determines in which box the particle is, the state will "collapse" to either |1L> or |1R>. But the vacuum state |0> plays no role here.
 
  • #8
Demystifier said:
There is only a total state, which in this case has a form
|1>=|1L>+|1R>
which is a superposition of particle in the Left box and particle in the Right box. If you perform a measurement which determines in which box the particle is, the state will "collapse" to either |1L> or |1R>. But the vacuum state |0> plays no role here.

After writing the previous post I had gnarling doubts that this thought experiment is really due to my ingenuity and in fact I saw that it is discussed in the article by Wightman cited in a previous post and also I realized that the electron would not disappear. Rather, the rotation should transform a symmetric state into an antisymmetric state what has been observed experimentally.
The discussion of Wightman I find somewhat dissatisfying:
"As far as the relevance of these experiments to the
univalence superselection rule is concerned, the standard response is that the
experiments are beautiful, but they are not the ones involved in the univalence
superselection rule where you must rotate the entire isolated system. "
 
  • #9
Thanks for the clarification! I know about univalent superselection, but I was just surprised that I've never seen any mentioning of superselection when constructing the fermionic multi particle state space. But I'm more a foundations of QT guy than a field theorist. So thanks for confirming my thoughts!

Cheers,

Jazz
 

Related to Fermionic Fock space superselected?

What is Fermionic Fock space superselected?

Fermionic Fock space superselected refers to the mathematical framework used to describe quantum systems with a large number of fermions, such as electrons. It involves creating a basis of states that can be used to represent the different possible configurations of the fermions.

What is the significance of Fermionic Fock space superselected?

Fermionic Fock space superselected is significant because it allows for the description and analysis of complex quantum systems with a large number of fermions. It also helps to explain phenomena such as superconductivity and the behavior of electrons in materials.

How is Fermionic Fock space superselected different from other mathematical frameworks in quantum mechanics?

Fermionic Fock space superselected differs from other frameworks in quantum mechanics, such as the Schrödinger equation, in that it specifically deals with systems containing fermions. It also takes into account the Pauli exclusion principle, which states that no two fermions can occupy the same quantum state simultaneously.

Can Fermionic Fock space superselected be applied to other types of particles besides fermions?

No, Fermionic Fock space superselected is specifically designed for fermions and cannot be applied to other types of particles such as bosons. This is because fermions have unique quantum properties that require a different mathematical approach.

How is Fermionic Fock space superselected used in practical applications?

Fermionic Fock space superselected is used in a variety of practical applications, including condensed matter physics, quantum chemistry, and quantum computing. It provides a powerful tool for understanding and predicting the behavior of fermionic systems in these fields.

Similar threads

Replies
1
Views
523
  • Quantum Physics
Replies
5
Views
928
Replies
12
Views
2K
  • Quantum Physics
Replies
4
Views
1K
  • Quantum Physics
Replies
2
Views
1K
  • Quantum Physics
Replies
9
Views
883
  • Quantum Physics
Replies
1
Views
931
  • Quantum Physics
Replies
1
Views
1K
Replies
4
Views
2K
  • Quantum Physics
Replies
7
Views
1K
Back
Top