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I Field operator

  1. Jun 24, 2017 #1
    Hello! So I understand that in QFT and based on the second quantization, one introduce the hermitian operator ##\hat{\phi}(x)##. So, if we have a state with n particles ##|n>## we can get the configuration space representation as: ##\psi(x_1,..,x_n,t)=<0|\hat{\phi}(x_1)...\hat{\phi}(x_n)|n>## (please let me know if anything I said is wrong). Then I read that writing the action for a field, in the end we can derive ##H[\pi,\phi]##, with ##\pi## the conjugate momentum of ##\psi## and here is where I get confused. What is the meaning of ##\phi## without a hat? It means it is just a scalar field, not a field operator? So it is just as ##\phi## was previously, in the first quantization QM? And if in the end we go back to a scalar field and not an operator, what is the point of the second quantization? I.e., if we solve the Klein-Gordon equation for ##\phi(x)## we still get a plane wave, so where is the operator-like behavior of ##\phi##? And I see that both ##\phi## and ##\psi## appear in this formalism, so they can't represent both the same thing. Thank you!
     
    Last edited: Jun 24, 2017
  2. jcsd
  3. Jun 25, 2017 #2
    Firstly, a field operator [itex]\hat{\phi}(x)[/itex] is not necessarily hermitian. For example, a complex scalar field is not hermitian but the Hamiltonian is always. A more clearer statement would be [itex]\langle 0|\hat{\phi}(x_1)\hat{\phi}(x_2)...\hat{\phi}(x_n)|n\rangle[/itex] is the configuration space wavefunction [itex]\psi(x_1,x_2,...x_n)[/itex], corresponding to the n-particle Fock state.

    You're mixing up quantum states with fields. For a classical field [itex]\phi(x)[/itex], it has a classical Hamiltonian (density) which is a function of the field [itex]\phi(x)[/itex], and the corresponding conjugate momentum [itex]\pi(x)[/itex] i.e., [itex]H=H(\phi,\pi)[/itex]. There is nothing called conjugate momentum of [itex]\psi[/itex]. I think, it's a typo in your question.

    [itex]\phi(x)[/itex] is the classical field, a function of spacetime coordinates [itex]x[/itex]. In the quantized version, the scalar field [itex]\hat{\phi}(x)[/itex] is a promoted to the status of an operator (not necessarily hermitian). The quantized field Hamiltonian is however a hermitian operator and is given by [itex]\hat{H}=\hat{H}(\hat{\phi},\hat{\pi})[/itex].

    Second quantization is a bad name. Historically, the field [itex]\phi(x)[/itex] used to be regarded as the wavefunction and Klein-Gordon equation as a relativistic generalization of Schrodinger equation. This used to be called first quantization. However, relativistic quantum mechanics has severe pathologies and therefore was later superseded by quantum field theory (QFT). In QFT, [itex]\phi(x)[/itex] is no longer the wavefunction of a relativistic particle. The wavefunctions are what you mentioned i.e., [itex]\psi[/itex].

    Yes. The solution of Klein-Gordon equation can be decomposed in terms of plane-waves with amplitudes. Those amplitudes are numbers in classical field theory and promoted to creation and annihilation operators in QFT.
     
    Last edited: Jun 25, 2017
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