Many text books emphasize that the quantum field is not wavefunction. But because of the similarity in the format, I could not stop from wondering whether they are actually the same thing.

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    $\begingroup$ No, a quantum field is a superposition of operators ;-) $\endgroup$ – Vladimir Kalitvianski Dec 13 '14 at 19:33
  • $\begingroup$ @VladimirKalitvianski, your answer may be valid. But this is not what I expected. Is there a more physical explanation on this argument? $\endgroup$ – Fine Observer Dec 13 '14 at 20:26
  • $\begingroup$ In my opinion, it is a matter of definition. We define quantum fields, find out their equations and use them to calculate the occupation number evolutions. $\endgroup$ – Vladimir Kalitvianski Dec 13 '14 at 20:30
  • $\begingroup$ This question might be helpful: physics.stackexchange.com/questions/16091/… $\endgroup$ – jinawee Dec 13 '14 at 20:37
  • $\begingroup$ I read several links in SE on this matter, such as "is a quantum field observable?", and tried hard to understand this issue in my own way. The underlying state is point in Fock space, the superposition of 1 particle, 2 particles, ... states. By acting on one of the states, the quantum field $\phi (x,t)$ attaches a space-time meaning to the underlying state. $\endgroup$ – Fine Observer Dec 14 '14 at 21:06

A wavefunction is (typically understood to be) a complex valued function on configuration space; a wave function assigns a complex number (probability amplitude) to each point in configuration space. For a system of $N$ particles, the system's wavefunction is $3N$ dimensional.

A quantum field, on the other hand, is an operator valued function on physical space; a quantum field assigns an operator to each event in spacetime.

These are clearly fundamentally different.

One can make a connection between a single particle wavefunction and a quantum field in the following way.

$|0\rangle$ is the vacuum (no particle) state of the system, $\phi^*(x,t)$ is the operator which creates a particle with definite location $x$ at time $t$, and $\Psi(x,t)$ is a 1 particle wavefunction.

Then, a 1 particle state described described by the wavefunction $\Psi$ is given by

$$\int dx \;\Psi(x,t)\;\phi^*(x,t)|0\rangle$$

That is, the state is a superposition over all 1 particle states, $\phi^*(x,t)|0\rangle$, weighted by $\Psi(x,t)$.

For a 2 particle state, the generalization is

$$\int dx_1 dx_2 \;\Psi(x_1, x_2,t)\;\phi^*(x_1,t)\phi^*(x_2,t)|0\rangle $$


A quick answer is that simple wave functions can not describe the processes of annihilation and creation represented in fields. Fields support multi-particles states in a way that wave functions can not.


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