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Can someone give me a clear explanation of what is the difference between a classical field, a wave function of a particle and a quantum field? I haven't find a clear explanation. For example for Klein-Gordon equation, the solution $\phi(x)$ is a plane wave, but $\phi(x)$ can be interpreted in any of the 3 ways I mentioned above and I am not sure what is the difference between them. (for example I am not sure why the wave function is not a field, as it assigns to any point in space a value, so it seems to behave like a field). Thank you!

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    $\begingroup$ It's been a while, but it should work something like this: Classical fields are sections over spacetime. Wavefunctions are probability amplitudes over configuration space. Quantum fields are local operators over spacetime acting on wavefunctionals over the space of classical field configurations. $\endgroup$ – Christoph Jun 25 '17 at 18:59
  • $\begingroup$ The whole motivation for Dirac to search for his namesake equation was that Klein-Gordon solutions don't admit the probability amplitude interpretation. $\endgroup$ – OON Jun 25 '17 at 19:18
  • $\begingroup$ Related. $\endgroup$ – Cosmas Zachos Oct 28 '19 at 21:59
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Ignore spin, polarization, and even Lorentz issues, absorb all superfluous constants, and consider time and one space dimension.

  • A one-particle complex wavefunction $\psi(x,t)=\langle x|\psi \rangle$ is a spacetime function serving as a probability density amplitude. For the quantum oscillator in coordinate space, it obeys the suitable Schroedinger equation, $(i2\partial_t +\partial^2_x -x^2)\psi =0$. (Well, in a twisted sense, it is some sort of "probability amplitude classical field", as you suggest.)

  • A real classical field $\phi(x,t)$ is a spacetime function. As you review in classical mechanics texts, a free one represents a superposition of normal modes of an infinity of coupled oscillators on a line (lattice, as their spacing vanishes). Here, however, x represents the equilibrium location of each oscillator, instead, and not the displacement from equilibrium (as it did for the classical parent of the above single oscillator). The Euler-Lagrange equation it satisfies is $(\partial_t^2-\partial_x^2 + m^2)\phi=0$. The decoupled normal modes are more apparent in the Fourier transform, $\tilde{\phi}(k,t)=\int dx ~e^{-ikx} \phi(x)$, so $\tilde{\phi}(k,t)=e^{it\sqrt{k^2+m^2}} c_k + e^{-it\sqrt{k^2+m^2}} c^*_k$, for numerical coefficients ck.

  • A quantum field $\Phi(x,t)$ is a noncommutative operator, linked to the above expression, except with suitable (k-dependent) normalized creation and annihilation operators $a_k^\dagger, a_k$ supplanting the above c-numbers ck and ck* of the classical field. (x remains a classical parameter!) These normal modes are decoupled from each other (the string Lagrangian has been diagonalized for them), and commute among themselves; they evidently obey the same linear equation as the classical field, since the operator coefficients do not affect it: the wave operators act on the c-number parameters. That is, each normal mode $\tilde{\phi}_k$ of the above string was “first quantized” as a classical oscillator, but this organized assembly of an infinity of identical such amounts to “second quantization”; also see WP. Since it packages an infinity of oscillators (excitations, identical particles), each configuration state vector is a particular assembly of creation operators acting on the Fock vacuum, while transition amps are vacuum expectation values of strings of such fields. In the first volume of the text by Bjorken and Drell, you may go through the solution of dynamical equations as wavefunctions, and in the second, the very same equations and solutions as fields. As long as you stick to the proper interpretation and use...

  • If your life depended on it, you could construct some type of a wavefunction out of a field operator, $\langle x| \Phi (x',t)|0\rangle$, now an almost localized c-number function, but careful treading is necessary to avoid a range of confusions.

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