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I Where exactly does QFT differ from QM? (in their formalisms)

  1. Apr 17, 2016 #1
    Hi.

    First off, I'm not sure if this is the right sub to talk about QFT. Apologies if t isn't.

    I'm halfway through an introductory QFT course and I still don't get what it is. What is different in its formalism that makes it able to tackle problems that quantum mehanics can't deal with?

    From what I know so far I would say that QFT is a generalization of quantum mechanics in which you can "choose" a different equation of evolution, so that QM is a particular case of QFT in which the equation evolution is Schrödinger's equation (or Heisenberg, depending on the picture you choose).

    I know this is wrong, though, because another important difference is that the position in space is "demoted" from a operator to a variable, so that it is on the same foot as time. But this makes me wonder many things:

    What do we mean exactly when we write Φ(x)?
    Are they still wavefunctions or should I think of them as a different concept?
    Can these functions be expressed as a scalar product of kets <x|Φ>, as wavefunctions are? If they can what do we substitute the position eigenkets with, since we don't have a position operator anymore?

    As you see I am completely lost when I try to see the "big picture". I thought I would find an explicit formalism for QFT somewhere, that is, a presentation of some postulates and the conclusions that follow from them, as there is for quantum mechanics, but every book I've found so far has been unsatisfactory in this regard.

    I'd love to hear your thoughts on this.

    Thank you for your time.
     
  2. jcsd
  3. Apr 17, 2016 #2

    radium

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    Are you familiar with second quantization?

    QFT comes about when you have an infinite number of degrees of freedom which is required to reconcile QM with special relativity (you can create and destroy particles) but is also natural in condensed matter where you have a very large number of particles and are talking about emergent properties.

    \phi(x) is a field with a given amplitude at a point and particles are seen as excitations of the field. You can expand these field in modes of creation and annihilation operators.

    And the fields are most definitely operators, you quantize the theory by imposing the canonical commutation relations.
     
  4. Apr 17, 2016 #3
    QFT takes the field as the fundamental thing that exists rather than the particle and as such already incorporates special relativity into it. This allows creation and annihilation operators and hence the creation and annihilation of particles.
     
  5. Apr 17, 2016 #4
    Here's the way I think of it - may not be 100% correct.

    QM is essentially non-relativistic. When you integrate relativity with QM you're forced to go to QFT.

    Schroedinger's equation works as a single-particle equation. The wave function is just a regular function which (in the position representation) gives the probability of finding the particle at that position (by squaring).

    When Schroedinger's is made relativistic you get Klein-Gordon (spin 0) or Dirac's equation (spin 1/2). Neither of these equations works right with a single particle. For one thing you get negative energy solutions (which we wind up interpreting as representing the anti-particle). With Klein-Gordon you also fail to get a positive-definite probability density. These facts suggest that we need to abandon the "particle" as a central concept.

    Meanwhile experimenters found out that, for one thing, antiparticles do in fact exist; and that particles can be created and destroyed. So they also reinforced the idea that the particle-oriented view was unsatisfactory.

    The solution was so-called "second quantization". As I understand it (could be not entirely rigorous) you change the wave function to an operator - actually an infinite bunch of operators - by treating each point of it as an operator. We call that a "field". The particle is demoted to "an excitation of the field" - i.e. a place where the field is particularly intense. Thus there's no longer really a particle, therefore no longer a "position operator". The field itself is a probabilistic representation of the position, just like the wave function it was derived from.

    Now it's easy to incorporate multiple particles, creating / destroying particles, and antiparticles, as mere modifications of the field values.

    So Φ(x) is, you can informally say, the wavefunction quantized. And Quantum Field Theory comprises the rules for manipulating such objects.

    Reviewing your OP I think the above addresses all the questions there, please ask if you think I've missed something.

    [EDIT] Of course QFT can be used non-relativistically but it was motivated by relativitistic considerations. In Weinberg *, QFT is brought forward as an unavoidable consequence of the reconciliation of quantum mechanics with special relativity.

    "Second quantization" is better termed "field quantization" but the intuitive term is still in use, although deplored by pedants.

    In general, for an intuitive understanding, read physics history to see how QFT (or any other topic) was originally motivated.

    * Weinberg, S. (2005). The Quantum Theory of Fields 1. Cambridge University Press. ISBN 978-0521670531. page 15.
     
    Last edited: Apr 17, 2016
  6. Apr 17, 2016 #5

    atyy

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    As radium says, one approach to second quantization is that it is a way of the exactly rewriting the non-relativistic quantum mechanics of many identical particles as a field theory. (The name second quantization is terrible because historically it was thought of as quantizing the wave function, which is not a very helpful concept, although it is a quick way to generate correct equations.)

    So QFT is used in non-relativistic and relativistic physics.
     
  7. Apr 18, 2016 #6

    Demystifier

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    QM is a theory of states in the Hilbert space ##{\cal H}_n## with a fixed number ##n## of particles.
    QFT is a theory of states in the Hilbert space which is a direct sum ##{\cal H}=\bigoplus_{n=0}^{\infty}{\cal H}_n##.
     
  8. Apr 18, 2016 #7

    rubi

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    The difference between mechanics and field theory is that mechanics is concerned with the motion of point particles (i.e. Newtonian mechanics) and field theory is concerned with the motion of fields (i.e. electrodynamics). The prefix "quantum" just says that we're describing this motion (of point particles or fields respectively) in the framework of quantum physics.

    Algebraically, the difference between point particles is encoded in the algebra of observables. For a point particle, you have the following Poisson brackets:
    ##\{x_i,p_j\}=\delta_{ij}##
    For a field theory, the Poisson brackets are given by:
    ##\{\phi[f],\pi[g]\} = \int_{\mathbb R^3} f(x)g(x)\mathrm d x##
    Here, ##f## and ##g## are elements of some suitable test function space and ##\pi## is the conjugate momentum of ##\phi##. In the quantum theory, you replace Poisson brackets ##\{-,-\}## by commutators ##\frac{1}{i\hbar}[-,-]##. The difference between particles and fields is captured completely by these algebraic relations, both in the classical theory and in quantum physics. Everything else stays the same.

    I wouldn't say this is the difference. First of all, there is a difference between many-particle QM and quantum field theory. The former is concerned with many particles. The latter is concerned with fields. Secondly, Fock space isn't adequate for interacting QFT's anymore due to Haags theorem.
     
    Last edited: Apr 18, 2016
  9. Apr 18, 2016 #8
    Thank you all.

    I think I understand what you all mean, but it still not clicks (maybe it will in time and I am just being impatient). Please tell me if this is correct:

     
  10. Apr 18, 2016 #9

    A. Neumaier

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    No, it is meaningless.

    The state of a quantum field system assigns expectation values to all products of its (smeared) fields just as the state of an anharmonic oscillator assigns expectation values to all products of its position and momentum operators. Particles are those states (of an asymptotically free field) where the corresponding number operator (an integral of quadratics in the field) has it as eigenvector with eigenvalue 1.
     
  11. Apr 18, 2016 #10
    From this I understand that ##\langle A\vert \phi(x) \vert A\rangle## means the expectation value of the field amplitude in that state and in that position, which is more or less what I (poorly) tried to say. Is that correct?

    I assume the ones with eigenvalue 1 will be one-particle states, and the ones with eigenvalue ##n## will be n-particle states, correct?
     
  12. Apr 18, 2016 #11
    From the expert point of view QFT may be just a minor gloss on QM: particles => fields, with a few obvious little changes here and there. But for beginners (like you and me), they seem very different. QFT introduces topics such as (non-Hermitian) creation / annihilation operators, Fock states (no good for interactions anymore due to Haags), Feynman diagrams, renormalization, gauge fields, symmetry breaking, Faddeev-Popov ghosts, vacuum state, ... and so forth which you may not have seen in QM.

    They can't be addressed in a few posts: that's what your course is for.

    But one misunderstanding I can clear up: there are many other field operators besides ϕ! ϕ is a generic term for any field's position operator, the analogue of q or x. There's also a conjugate momentum operator π, analogous to p.

    In a given problem there may be many fields considered, most of which have their own ϕ operator. For example in electroweak theory here are some of the types of fields, with their field operators, encountered: lepton spinor fields (Left and Right); W and Z gauge fields - first massless, then given mass by interaction with the isospinor scalar Higgs field; electromagnetic (photon) field; hadron fields; hypercharge field.

    I can't imagine how electroweak theory would look in QM, with a particle interpretation.

    Perhaps you're getting the main point, but don't think QFT is simple - just change particles to fields and you're done. All the topics mentioned above are probably new, and they're complicated. Your QM knowledge is relevant but QFT is, from the point of view of a student, quite a different beast and you've got your work cut out for you. Good luck!
     
    Last edited: Apr 18, 2016
  13. Apr 18, 2016 #12
    I know its a different beast, but when I learned QM I was presented with some postulates and was showed their consequences. In QFT I feel like we are using some parts of the formalism of QM and dropping others, but we are never explicitly told which ones. Maybe I'm just overthinking it, I have a tendency for that.

    Thanks for your answer.
     
  14. Apr 18, 2016 #13

    A. Neumaier

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    yes, yes. But correlations such as ##\langle A\vert \phi(x) \phi(y)\vert A\rangle## are also interesting....
     
  15. Apr 18, 2016 #14

    Avodyne

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    I would say that QFT is QM applied to fields. You start with a classical field theory, and then quantize it (using the rules of canonical quantization that apply to any classical mechanical system). This quantized field theory then turns out to describe particles. If the classical field theory is Lorentz invariant, then the particles obey special relativity.

    Everything else is details.
     
  16. Apr 18, 2016 #15

    atyy

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    All the postulates of QM hold in QFT. In QM, one has Hilbert space, commutation relations, Hamiltonian, observables as Hermitian operators, Born rule and state reduction. All of these hold in QFT.
     
  17. Apr 18, 2016 #16

    dextercioby

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    There are so many formulations of QFT and so many formulations of QM, that saying how and where one theory differs from the other is almost impossible. It would be much more useful to learn why the QM (i.e. quantized particle theory) of Dirac, Schrödinger, Heisenberg and von Neumann had to be quickly abandoned in favor of quantized field theory. What exact formalism of QFT you wish to learn (using only path integrals like Bailin and Love do it, or operators like in the classic books) is then your choice (or rather your university lecturer's). The only way to reconcile Dirac's semiclassical theory (electrons quantized, electromagnetic interaction between them beingclassical) with the necessity to quantize the e-m field (first Dirac 1927) was to "invent" a field for the electron, too and from it a Lagrangian and that's how, in time, modern QED was born. Then all particles discovered (positron, neutrino, mesons, etc.) had to follow the old electron and grab a field of their own. So at the level of 1950, it was safe to assume there was only QFT.

    This is a must read for any human being studying vigorously physics at university level and hoping to be a physicist someday: http://arxiv.org/abs/hep-th/9702027
     
    Last edited: Apr 18, 2016
  18. Apr 19, 2016 #17

    Demystifier

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    Well, maybe not for all physicists, but it is definitely a must read for all theoretical physicists using QFT. It is remarkable how often many particle physicists still argue that certain QFT theory is not appropriate because it is not renormalizable.
     
  19. Apr 19, 2016 #18

    kith

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    I made a similar experience when I started to learn QFT. The structure seemed very different from ordinary QM and although everybody said that QFT supersedes it, nobody bothered to relate the two.

    Here are some ressources, I found valuable for this (I am actually no expert on QFT and haven't worked through them in detail):
    Lecture notes on QFT by David Tong (especially section 2.8.1 "Recovering Quantum Mechanics")
    "Student Friendly Quantum Field Theory" by Bob Klauber
    "Quantum Field Theory for the Gifted Amateur" by Tom Lancaster and Stephen J. Blundell
     
    Last edited: Apr 19, 2016
  20. Apr 19, 2016 #19

    kith

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    As others have said, [itex]\phi(\vec x)[/itex] is an operator in QFT. One way to picture it is by its action on the vacuum. The vacuum state is a state with zero particles which is denoted by [itex]|0\rangle[/itex]. If I act on it with the field operator [itex]\phi(\vec x)[/itex], I get a particle localized at location [itex]\vec x[/itex]:
    [tex]\phi(\vec x) |0\rangle = |\vec x\rangle[/tex]
    So the field operator isn't necessarily an observable but something similar to the creation and annihilation operators you already know from ordinary QM.

    What I have written here is by no means rigorous and there are quite a few difficulties. For example, we cannot define a position operator with the usual properties for massless particles like photons. So the very concept of localizing photons isn't well-defined.
     
  21. Apr 19, 2016 #20

    A. Neumaier

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    In free relativistic QFT, typically ##\phi(x)=a(x)+a^*(x)## (or a constant multiple of it), where ##a(x)## is an annihilation operator distribution in spacetime. (In the classical version, ##a(x)## is called the ''analytic signal'' corresponding to ##\phi(x)##.)

    In the interacting case there is no simple relation.
     
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