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We always hear that

  • The Fock space is constructed with multiple $~a^\dagger_{\vec p}$ acting on the vacuum for different values of ${\vec p}$ (we can use alternatives notations to ${\vec p}$ because we can choose other operators eigenstates, momentum is not necessary).

  • The field operator $φ(x)$, the one that is present in every spacetime point and that is the solution of the QFT equations of motions, can be defined in terms of $~a^\dagger$ and $~a$.

So the question is: how can we construct the Fock space with Field operators $φ(x)$, bypassing the $~a^\dagger _{\vec p}$ (or whatever substitute for ${\vec p}$)? What would be the physical interpretation of this?

Qmechanic
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TrentKent6
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Yes you can. The family of all possible finite linear combinations of vectors $ \hat{\phi}(f_1)\cdots\hat{\phi}(f_n)|0\rangle$ where the smooth functions $f_k$ are compactly supported (and $n$ is arbitrary) is dense in the Fock space. So you can arbitrarily approximate every vector of the Fock space in terms of the ones above.

More strongly, it happens even if the supports of the functions are all included in a common bounded region as established in the Reeh-Schlieder theorem.

Alternatively you can smear the field operator in terms of solutions of the equations of motion $\varphi$: $$\sigma(\hat{\phi}, \varphi):=\int_\Sigma (\hat{\phi}(x) \partial_t \varphi(x)- \varphi(x) \partial_t \hat{\phi}(x) )d^3x $$ as in the LSZ formalism in place of the spacetime smearing $$\hat{\phi}(f):= \int \hat{\phi}(x) f(x) d^4x$$ obtaining the same.

The connection between the two smearing procedures is $$\sigma(\hat{\phi}, \Delta(f)) = \hat{\phi}(f),$$ where $\Delta$ is the causal propagator.

Valter Moretti
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    Just to confirm: so we can say that we can construct a Fock space with only field operators, where by "field operators" i mean the solution of the equations of motions (such as Klein-Gordon etc) ? – TrentKent6 Oct 31 '23 at 13:29
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    Yes, you can equivalently define the Fock space as th Hilbert completion of the vector space spanned by the vectors I wrote in my answer. – Valter Moretti Oct 31 '23 at 13:31
  • i don't get why you treat the "field operator" different from the "solutions of the equations of motion"; i have always learned that the solutions of the equations of motion are the field operators, including here: https://physics.stackexchange.com/questions/785756/does-the-qft-klein-gordon-equation-describe-the-state-of-the-field-or-the-field#comment1765384_785756 – TrentKent6 Oct 31 '23 at 15:15
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    No. There are the number-vaued fields satisfying the Klein-Gordon equations and the field operators, satisfying the KG equation distributionally. – Valter Moretti Oct 31 '23 at 15:27
  • can you elaborate please? this seems quite important – TrentKent6 Oct 31 '23 at 15:28
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    Well, that is the standard you should know before asking your question, in my view. I wrote what is actually written in every book of QFT. I suggest you to open another question in PSE on this issue, because replying as a comment is quite difficult . – Valter Moretti Oct 31 '23 at 15:32
  • I am sorry but i have read 2 books/resources (DAVID TONG and KLAUBER) and both make very, very, very clear that (i quote literally) "QFT wave equation solutions are operators". In addition to this, see my previous question, where the replies acted as if "the standard" was actually knowing that the field operators were these solutions (https://physics.stackexchange.com/questions/785756/does-the-qft-klein-gordon-equation-describe-the-state-of-the-field-or-the-field#comment1765384_785756) – TrentKent6 Oct 31 '23 at 15:51
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    It is just matter of jargon. I dislike that jargon because it causes misunderstandings as it is quite evident in your case. However, sorry, if you already have your answer found on that books, I do not understand the reason why you are still asking something about these issues. Finally, the use of any “ipse dixit” is useless in scientific discussions. I think I will stop replying you. – Valter Moretti Oct 31 '23 at 16:10