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Let's consider a free scalar quantum field, at "a single point in time".

Let $p$ be a momentum state.

I could write this: $a^\dagger_p|0\rangle$ , and as far as I understand, this denotes a field with a single particle in a particular momentum state.

Now this can be expanded into position states. If I understand correctly, it is a superposition of single particle states at every position. But, the "amount" of particle at each position is infinitesimal (ie. the prefactors of eigenvectors in the position basis are infinitessimal. Equivalently there is 'infinitessimal chance' to find a particle at any position if there was an observation).

My problem comes when I try to consider the field at any point. It is like a harmonic oscillator, and we can calculate it by combining the ladder states in the same way. But in this case, at any point, we have a single inifinitessimal piece of 1 particle state, and all the rest (all the other contributions of the infinite points in the integral) are of the ground state. So what does the field look like? An infinitesimal deviation from the harmonic oscillator ground state? That seems awkward. Is that valid? Or is my understanding wrong? Or are single particle states not really valid on their own?

user183966
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    AFAIK the field theoretical "particles" used to calculate Feynman diagrams for comparison with data cannot describe a real particle in space. since these are plane waves, i.e. over all space. One has to use the concept of wave packet see here http://www.quantumfieldtheory.info/H_operator_and_noneigen_states.pdf , – anna v Dec 14 '19 at 05:42
  • Since this is quantum field theory, you have to remember the field is, well, quantum. What you're asking is like asking "if a harmonic oscillator is in the state $a^\dagger |0 \rangle$, what is the position?" There isn't a definite position, the whole point is that it's a superposition of different positions. – knzhou Dec 14 '19 at 08:34
  • Similarly, this particular state is a superposition of different quantum field eigenstates, i.e. there is one expansion coefficient for every possible space-dependent function. These coefficients are very clunky to work with, so we don't. – knzhou Dec 14 '19 at 08:35
  • @annav thank you, that document is very interesting – user183966 Dec 14 '19 at 09:47
  • @knzhou thank you. Is what I am asking similar to "What's the position"? As best I understand, at any position, in a free scalar quantum field, the fields value has a distribution, like the harmonic oscillator. The spectrum of the field is the ladder eigenvalues. So all the info about the field can be represented by a general but definite distribution (but not a single definite value) at every point. I appreciate the coefficients are clunky to use. I am trying specifically to improve my fundamental theoretical understanding rather than calculating ability. – user183966 Dec 14 '19 at 10:02
  • @knzhou sorry, is that correct - there is an expansion coefficient for every possible space-dependent function? I mean yes I understand there is a value for every $\langle space function | state \rangle$ but that is over-specified isn't it? All the information by the coefficients of a smaller basis for example plane wave states? Or have I misunderstood something fundamentally? – user183966 Dec 14 '19 at 10:05
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    Are you asking about localization of fields? – Cosmas Zachos Dec 14 '19 at 12:03
  • @CosmasZachos thank you! I think I may be. The question you linked is very interesting and I think relevant. So, perhaps it is not single particle states that do not make sense, but rather position states $a_x^\dagger |0\rangle$ , and I should be thinking about them as a density to integrate over maybe. I will clear up some related things first then maybe come back to this if needed. – user183966 Dec 14 '19 at 12:48

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