1

In the book on Quantum Field Theory by Peskin and Schroeder, it is explained how the field is promoted to an operator, now my question is that in Quantum Mechanics, operators act on kets, what does this field operator act on?

Qmechanic
  • 201,751
Cbb Ttt
  • 70
  • 3
    State vectors?! – Tobias Fünke Oct 15 '22 at 16:54
  • Tobias Fünke, sorry,that was an unnoticed autocorrection, it would be ket vectors instead of state vectors.I have edited the question. – Cbb Ttt Oct 15 '22 at 17:04
  • 2
    Kets are just notation. My comment was an answer, actually. Operators on a vector space act on the elements of the vector space, by definition. – Tobias Fünke Oct 15 '22 at 17:04
  • Tobias Fünke, I am actually looking for a more physical explanation, in the sense that yes operators act on elements of a vector space, which in the context of Quantum Mechanics would represent states of a particle , what would those vectors represent (physically ) in this case? – Cbb Ttt Oct 15 '22 at 17:46
  • 2
    A Hilbert space in ordinary QM is much simpler because there it does not happen that particles get created and annihilated all the time. In QFT you should look at Fock Space. – Kurt G. Oct 15 '22 at 18:13

1 Answers1

4

There is a Hilbert space also for quantum fields and the field operator acts on the vectors ("kets") of that Hilbert space. From this perspective there is no difference between quantum mechanics and QFT.

Valter Moretti
  • 71,578
  • 1
    I think the question is about the physical interpretation. What is the Hilbert (Fock?) space of a field? The collection of all possible classical field configurations? Is that a meaningful concept? – FlatterMann Oct 22 '22 at 10:29
  • Maybe you are right. Let the OP clarify if it is the case. – Valter Moretti Oct 22 '22 at 10:41
  • 4
    You are right actually, I just noticed the comments under the question. OK one should write a book to answer... – Valter Moretti Oct 22 '22 at 10:42
  • That's what I saw. I don't have a good answer to that question myself. If you do, it would be great if you could enlighten us. Thanks! – FlatterMann Oct 22 '22 at 10:43
  • 1
    QFT's Fock space is fundamentally different from QM's Hilbert space. Hilbert space is countable, while Fock space is NOT countable. See here: https://physics.stackexchange.com/questions/617834/is-the-fock-space-defined-in-the-context-of-the-quantum-field-theory-really-a/617932#617932 – MadMax Oct 24 '22 at 14:27
  • 1
    @MadMax, it is false if countable means separable. The Fock space is a special case of Hilbert space and it is separable if an only if the one-particle subspace is separable. – Valter Moretti Oct 24 '22 at 15:04
  • 'countable' does not mean much un this context. – Valter Moretti Oct 24 '22 at 15:07
  • @ValterMoretti, Let me give you one example to illustrate the difference between Fock space and Hilbert space: The spontaneous symmetry breaking can be only realized in the QFT Fock space, which entails some thermodynamic limit. On the other hand, QM Hilbert space can not accommodate that, since QM Hilbert space would allow the ground state wave function to be the sum of all the degenerate states which would restore symmetry. – MadMax Oct 24 '22 at 15:36
  • 1
    I agree with you that QFT permits more interesting phenomena than QM. For instance, as you suggest, SSB. However this is not a mathematical property of the Fock space, but of the algebra of observables. The Fock space of the massive Klein-Gordon quantum field is isomorphic as a Hilbert space to $L^2(\mathbb{R}^3, dk^3)$ which is a typical space of QM. You may say that this isomorphism is not physically meaningful. And I would agree with you. – Valter Moretti Oct 24 '22 at 16:46
  • 1
    The point is that the differences are not of mathematical nature, they concern physics. The algebra of observables, not the mathematical structure of Hilbert space. The Weyl algebra (i.e. the CCR) of QFT arise from an infinite dimensional symplectic space, whereas those of QM arise from a finite dimensional space. (maybe this point was the "countablity" you pointed out previously). This is again a feature of the observables. However, that is my personal view. – Valter Moretti Oct 24 '22 at 16:53
  • Can you please suggest any beginner friendly but mathematically rigorous book(s) which answers the questions asked by FlatterMann in the first comment? – Apoorv Potnis Oct 26 '22 at 13:21
  • 1
    Friendly I do not know, but for instance https://link.springer.com/book/10.1007/978-94-009-0491-0 – Valter Moretti Oct 26 '22 at 15:08
  • 1
    Another possible book is Local Quantum Physics by R. Haag – Valter Moretti Oct 27 '22 at 10:32
  • Thank you for your response, Prof. Moretti. – Apoorv Potnis Oct 27 '22 at 19:09