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In quantum electrodynamics, the classical Hamiltonian is obtained from the classical electromagnetic Lagrangian. Then the classical electric and magnetic fields are promoted to operators, as is the classical 4-vector potential $A_{\mu}$. The appropriate commutation relations are expected between the fields and their conjugate momenta.

Now, my question is, do the principles of quantum electrodynamics follow as a consequence of the fact that the charged particle producing the field is a quantum particle which must follow the principles of quantum mechanics?

Let me give a specific example. Consider a slow moving(for simplicity) free electron moving with a constant velocity initially.

Now, classically, the magnetic field at a point $P$ would be given by a function $\vec{B} = \vec{f}(\vec{r},\vec{x},\vec{p})$, where $\vec{r}$ is the position vector of the point at which the field is being 'measured' and $x$ and $p$ are the position and momenta of the charged particle evaluated at the retarded time.

Now, supposing I apply the principles of quantum mechanics to this electron and promote the above mentioned expression for the magnetic field at point $P$ to an operator by the usual quantum mechanical prescription. Would this prescription yield the correct values for the measured magnetic field at point $P$? Why? or Why not?

The bottom line of my entire question is whether the quantum field theory of an electron is a direct consequence of the fact that the particle producing the field is a quantum particle (and not a classical one) or does it involve much more than that?

EDIT: Thank you for your responses. I would also like to know if the above mentioned prescription for obtaining the magnetic field would yield accurate results for slow moving electrons(non-relativistic)?

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3 Answers

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The bottom line of my entire question is whether the quantum field theory of an electron is a direct consequence of the fact that the particle producing the field is a quantum particle (and not a classical one) or does it involve much more than that?

It involves "much more than that":

If I understand correctly, you're taking the classical expression for, say, the Coulomb field resulting from a source charge, or a magnetic field resulting from a current element and then saying that, since the position/momentum of the sources are quantized, they become operators and in this way the field becomes an operator since it's a function of those positions/momenta.

In QED, it's possible to describe a freely propagating field quantum (e.g. a photon). Freely propagating means that once it's been produced, its existence is now independent of any source. I don't see how this is possible in the scheme where you just quantize the source. Any time dependence of the source would always be immediately transferred to the electromagnetic field.

In QED, you quantize the electromagnetic field and the electron/positron fields independently. Neither is in any sense more fundamental than the other. One can act as a source for the other only after you've introduced an interaction term in the theory. So the source/field relationship isn't the basis of the quantization.

Also one of the key features of quantum field theory which distinguishes it from quantum mechanics is that it offers a mechanism to create and destroy particles. This would not be possible with a prescription such as the one you describe. Even your electron description is still a single particle one.

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Thank you for the informative post. In addition, I would like to know if the procedure outlined by me to calculate the magnetic field of the electron at the point P, would give accurate results for a slow moving(v<<c) electron? –  guru Jun 24 '13 at 16:59
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In Quantum mechanics , you have operators $X(t)$, where $t$ is a parameter, and $X$ is the operator.

What happens in Quantum Field theory ?

If we take a real scalar field, you have operators $\Phi(x,t)$, where $x$ and $t$ are parameters, and $\Phi$ is the operator.

In Quantum Electrodynamics, the photonic field is represented by operators $A_{\mu}(x,t)$, where $x$ and $t$ are parameters, and $A_{\mu}$ is the operator. The electron/positron field is represented by operators $\Psi(x,t)$, where $x$ and $t$ are parameters, and $\Psi$ is the operator.

So, in Quantum Field Theory, $x$ is not an operator, it is a (space) parameter. So, you have not the right to "mix" the 2 formalisms.

Especially, you have not the right to think about something like $\vec B=f(\vec X,t)$, where $\vec B$ and $\vec X$ would be operators representing magnetic field and position, because that will be a incoherent mixing of the 2 formalisms of Quantum Mechanics and Quantum Field Theory.

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Sure you can. –  Michael Brown Jun 24 '13 at 8:51
    
@MichaelBrown : Very Interesting paper, like this one, but the idea in these papers, if I correctly understand, is that a $1$st quantized formalism would be equivalent to a $2$nd quantized formalism. But there is no mixing of the 2 formalisms. –  Trimok Jun 24 '13 at 16:22
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Now, my question is, do the principles of quantum electrodynamics follow as a consequence of the fact that the charged particle producing the field is a quantum particle which must follow the principles of quantum mechanics?

First, a semantic issue. A principle is a starting point so, if an alleged principle follows from something else, can it in fact be called a principle?

Anyhow, what are the principles of QED or, more generally, QFT?

Certainly, one principle is the identification of fundamental "particles" as the quanta of quantized field modes. Each mode of a (free) field obeys a quantum harmonic oscillator equation and thus, each mode has associated ladder operators that create and destroy quanta, i.e., particles.

So, an electron (or positron) with definite momentum $k$ is identified as a quantum of the $k$th mode of Dirac "field".

And, a photon with definite momentum $k$ is identified as a quantum of the $k$th mode of the vector potential "field".

But, importantly, the Dirac field is not the source of the vector potential field in this picture. In fact, the vector potential is seen as a gauge field that is required by the principle of local gauge invariance.

So, it seems to be the case that the answer to your question must be no.

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