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Consider the electron field in quantum field theory (or the electromagnetic field if easier). My understanding is that what we call an electron is the result of applying the creation operator to the vacuum state $\left| 0 \right>$ of the electron field, i.e. electrons are quanta of the electron field (or photons are quanta of the electromagnetic field).

What does it mean to apply the creation operator TWICE to the vacuum state? Is this an electron with twice the energy, or two electrons at the same point in spacetime, or what?

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  • $\begingroup$ $(a^\dagger)^2=0$ is one of the basic properties of the fermionic operators. $\endgroup$ – Vadim Mar 12 at 8:44
  • $\begingroup$ Note that there is not a single creation operator. There an infinite number of creation operators that create modes with different energies and momenta. I assume you mean applying the same creation operator to the vacuum twice. $\endgroup$ – John Rennie Mar 12 at 9:57
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Start with the electromagnetic field. It is conceptually easier I think.

Each creation operator is associated with a mode of the field. A mode is a pattern of field whose quantum behavior is like that of a harmonic oscillator. In the case of a bosonic field such as the electromagnetic field, applying a given creation operator twice increases the excitation of that mode by 2. That means you get two photons, if you want to think of it in terms of photons. Their location is not at a point, but throughout the region of spacetime occupied by the given mode. It could for example be a plane wave, or some other shape. Both photons are there throughout the region. They are just another way of saying that the mode in question is now in its second excited state, two above the ground state.

To get a single photon of twice the energy, you would use the creation operator of a mode whose frequency is twice that of the one you considered first. It is a different mode and a different creation operator.

In the case of a fermionic field such as Dirac field, the square of the creation operator for any given mode gives zero. In this case the mode does not have a second excited state. It just has a ground state (no electrons) and a first excited state (one electron). This is the form that the Pauli exclusion principle takes when you do field theory. To get two electrons you would need to consider two different modes. For example: $$ \hat{a}_3^\dagger \hat{a}_5^\dagger | 0,0,0,0,0,0, \ldots \rangle = | 0,0,1,0,1,0, \ldots \rangle $$

Finally, none of these operators actually create anything, of course. That is just a loose way of speaking. The operators are mathematical abstractions. It is the fields themselves which interact and evolve in a physical sense; the operators give us a mathematical language to express this.

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  • $\begingroup$ How do you "create" two electrons? (+1) $\endgroup$ – Deschele Schilder Mar 11 at 22:55
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    $\begingroup$ @DescheleSchilder I added a sentence to show this. $\endgroup$ – Andrew Steane Mar 12 at 8:18
  • $\begingroup$ Thank you Andrew. What exactly do you mean by a 'pattern of field', or 'the region of spacetime occupied by the given mode [...] for example a plane wave'? What values does the field take here? $\endgroup$ – acephalous Mar 18 at 20:09
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    $\begingroup$ @chaos For a plane wave there is a region of spacetime filled by the wave. It is as wide as the wave is wide and as long as the wave is long. But you can also have other shapes such as a Gaussian amplitude cross-section, like a typical laser beam, and in fact all shapes are possible. $\endgroup$ – Andrew Steane Mar 18 at 21:26
  • $\begingroup$ Thanks. I guess my confusion comes down to this: what values does 'pattern of field' take here? A real number? What exactly is this 'wave'? Sorry for my ignorance in this area. $\endgroup$ – acephalous Mar 20 at 9:48

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