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I Difference between field and wave equation

  1. Jun 25, 2017 #1
    Hello! I am reading some introductory stuff on Klein-Gordon equation and I see that the author mentions sometimes that in a certain context the K-G equation "is a classical field equation, not a quantum mechanical field equation". I am not sure I understand. What is the difference between the two? Isn't the wave function a field, as it assigns to each point in space a number? I.e. can someone give me a clear explanation of what is the difference between a classical field, the wave function of a particle and a quantum field? Thank you!
  2. jcsd
  3. Jun 25, 2017 #2


    Staff: Mentor

    Please give the specific reference you are reading.
  4. Jun 25, 2017 #3


    Staff: Mentor

    What context? Please be specific.
  5. Jun 26, 2017 #4


    Staff: Mentor

    You may be referring to the following:

    The Kelin Gordon Equation with zero mass and no sources or sinks is really Maxwell's Equations in disguise which is very interesting and may be telling us something - maybe something important. But the KG equation can be derived from simple symmetry considerations so it maybe that's all that's going on.

    Classically fields reveal themselves by what they do eg exerting forces on charged particles or in the case of gravity exerting forces on mass. They sometimes can be explained, as in the case of gravity, by something more fundamental but still classical such as space-time curvature.

    In standard QM the wave-function is a field that is technically the representation of a quantum state in terms of the position basis without detailing exactly what that is because you really need the advanced concept of Rigged Hilbert Spaces to do it. If you are interested some threads have discussed it eg

    In Quantum Field Theory the field is a field of quantum operators. It turns out mathematically they are the same as the second quantization formulation of QM which is one of the formulations of QM:

    That's how 'particles' come out of QFT.

  6. Jun 26, 2017 #5
    No, the wave function is a complex field defined on the configuration space. There is only one particular exceptional case where the configuration space is the same as the usual space, namely a single point particle without any additional properties like spin or charge. If you have, instead, two point particles, the configuration space is already 6-dimensional, because the configuration is defined by two points in usual space, the positions of the two particles.

    Instead, classical wave equations are wave equations on usual three-dimensional space ##\mathbb{R}^3##. The classical example is the electromagnetic field. The Klein-Gordon equation is even simpler, given that it has only a single component. Which makes it an equation for a scalar field.

    There is also another difference. You cannot measure the wave function - what you can measure are probabilities, which can be computed given the wave function. Instead, classical fields can be measured: The electric and magnetic field can be easily measured. There may be some problems with this if the field is a potential, and what is measurable is not the potential itself, but only potential differences. Nonetheless, even in this case a classical field is much closer to something measurable.
  7. Jun 26, 2017 #6


    Staff: Mentor

    Not really, no. A scalar field and a vector field are not identical. Similar, yes, but not identical.
  8. Jun 26, 2017 #7


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    I would say it's maybe a careful book, indicating right from the beginning that the Klein-Gordon equation cannot be interpreted as a non-relativistic Schödinger equation to describe the position probability of a single particle. The reason is that the Noether current of phase invariance for the complex field is not positive semi-definite and thus cannot be interpreted as probability distribution.

    Also already for the free-particle case, the single-particle energy is not bounded from below as for the Schrödinger equation since for the plane-wave modes of the KG equation (i.e., the would-be momentum eigenstates for a particle) you have ##E=\pm \sqrt{\vec{p}^2+m^2}## while for the Schrödinger equation it's uniquely ##E=\vec{p}^2/(2m)##.

    The physical reason for all this trouble is well-known nowadays: If you have interacting particles (and only these are interesting, because they are observable), scattering with relativistic energy transfers (i.e., energies larger than the mass of the particles), you can always easily destroy and create particles, and this is what in fact happens in accelerators like the LHC, where you produce a lot of new particles when colliding two protons at very high energies (of now around 13 TeV).

    That's why relativistic QT has to describe the typical situation where the particle number is not conserved, and thus a many-body description is necessary. The most convenient way is to formulate this in terms of quantum field theory, and indeed the most successful QT ever is the Standard Model of elementary particles. The use of QFT allows to solve the above mentioned problems: Quantizing the Klein-Gordon field, you can decompose (for the free-field case) it into plane-wave modes. The trick to avoid the interpretation problem for the field modes with negative frequencies is to just write a creation operator in front of them. This leads to the prediction that the Klein-Gordon field describes two kinds of particles: The one occuring in ##\hat{\phi}(x)## with an annihilation operator in front of the positive-frequency solutions (usually called "particle states") and the one with a creation operator in front of the negative-frequency solutions (usually called "anti-particle states"). It turns out that the particles and anti-particles have precisely the same mass.

    Now the conserved Noether current from the symmetry under redifinition of the phase factors, a U(1) symmetry), becomes a physical meaning: It's related to positive charges for the particle modes and to negative charges for the anti-particle modes. If you include interactions and want the U(1) symmetry stay intact you've always some function of ##\hat{\phi}^{\dagger} \hat{\phi}##, i.e., Lagrangians like ##\mathcal{L}_{\text{int}}=-\lambda (\hat{\phi}^{\dagger} \hat{\phi})^2##. Then in the interactions the net charge stays always constant in scattering processes, and thus the net-particle number (i.e., number of particles minus number of anti-particles) stays constant, but it doesn't need to be a positive number to be interpreted as a charge (the electric charge is an example for this).

    If in addition you make the interactions local, i.e., make the Lagrangian polynomials of the field operators and their 1st derivatives with respect to the spacetime variables and demand a Hamiltonian bounded from below you necessarily get some very fundamental consequences: One is that you can work out wave equations for particles of any spin. Klein-Gordon fields describe particles with spin 0. Weyl or Dirac fields describe particles with spin 1/2, vector fields particles with spin 1, etc. etc. Then it turns out when quantizing such kinds of QFTs you must have bosons for integer-spin (i.e., canonical commutation relations for fields and canonical field momenta) and fermions for half-integer-spin (i.e., canonical anticommutation relations for fields and canonical field mometa) in order to have an energy bounded from below. This holds true for all so far observed particles (be they "elementary" or "composed" particles).

    Another general consequence is the PCT theorem, according to which any Poincare-invariant local QFT with an energy bounded from below is automatically also invariant under the PCT transformation, i.e., the transformation consisting of spatial reflections (P for "parity"), charge conjugation (C, which exchanges each particle by its anti-particle), and time-reversal transformation (T) is always a symmetry. Also this theorem has been tested and never failed so far (while the weak interaction breaks P, T, and CP invariances, it still obeys PCT invariance).
  9. Jun 28, 2017 #8


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    What's a little strange to me is that on the one hand, the Klein Gordon equation is the natural relativistic generalization of the Schrodinger equation. On the other hand, the interpretation of [itex]|\psi|^2[/itex] as a probability density, which is the heart of the interpretation of the wave function in QM, doesn't apply in any straight-forward manner to the solutions of the KG equation. The KG equation is best interpreted as a field equation for a relativistic scalar field, not as describing the evolution of a probability amplitude. (Of course, probabilities pop again when it's reinterpreted as a quantum field, rather than a classical field)
  10. Jul 1, 2017 #9


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    The key to the interpretation of ##\psi^* \psi## as a probability distribution in the case of the Schrödinger equation is that it is the density of a conserved Noether charge. The symmetry is symmetry under multiplication of ##\psi## with a constant phase factor. The application of Noether's theorem to the Schrödinger equation leads to the corresponding current
    $$\vec{j}=-\frac{\mathrm{i}}{2m} (\psi^* \vec{\nabla} \psi-\psi \vec{\nabla} \psi^*).$$
    It's easy to prove from Schrödinger's equation that this fullfills the continuity equation with ##\rho=\psi^* \psi##. Indeed we have
    $$\partial_t (\psi^* \psi)=\dot{\psi}^* \psi + \psi^* \dot{\psi}=-\mathrm{i} (\psi \hat{H} \psi^*-\psi^* \hat{H} \psi)=\frac{\mathrm{i}}{2m} (\psi \Delta \psi^*-\psi^* \Delta \psi)=-\vec{\nabla} \cdot \vec{j},$$
    which means that
    $$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \psi^* \psi=\text{const}$$
    under the time evolution governed by the Schrödinger equation. Since ##\psi^* \psi## is interpreted as the probability density for the position of the particle the constant is normalized to ##1##, and once normalized it stays normalized for all times, as it must be.

    In the KG case the same arguments hold true too, but the conserved four-current reads
    $$j^{\mu} = \mathrm{i} [\phi^* \partial^{\mu} \phi - \phi \partial^{\mu} \phi^*],$$
    i.e., the Noether charge is not positive definite, and thus you cannot interpret ##j^0## as probability distribution for the position of the particle.
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