Can one define a “particle” as space-localized object in quantum field theory?

In Peskin and Schroeder, while discussing creation and annihilation operators for a Klein-Gordon field (p.22), the authors say, as we all know the creation operator $a_p^{\dagger}$ acts on vacuum to produce a state with momentum $p$ having energy equal to $\sqrt{p^2+m^2}$. Then they state: "It is quite natural to call these excitations particles, since they are discrete entities that have the proper relativistic energy-momentum relation. (By a particle we do not mean something that must be localized in space; the creation operator creates particles in momentum eigenstates)"

This clashes with my previous conception of a particle to be something localized in space; a point particle located at point $(x,y,z)$ for example, or an extended massive object occupying certain region in space. Form above should I conclude that no such concept of space-localized point-particle exists in quantum field theory? I am not familiar with the process of visualizing something in coordinate space which is known to be localized in momentum space. Please help.

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This might answer your question arxiv.org/abs/gr-qc/0409054 – alexarvanitakis May 22 '13 at 9:13
This derivation is effectively done in the momentum basis---which is just a valid and correct as the position basis---but because these excitations have well defined momentum Heisenberg tells us that viewed in that basis they do not have well defined position. To get the thing that you are used to you need to pick a basis in which both position and momentum are roughly localized; that's the wave-packed formalism that Trimok talks about, but it is mathematically difficult and we mostly calculate in momentum space. – dmckee May 22 '13 at 14:10

Your notion of "my previous conception of a particle to be something localized in space" is a classical (non- quantum) conception.

So, in Quantum Mechanics, it it a false conception. And it is the same thing in Quantum Field Theory.

In Quantum Field Theory (as in Quantum Mechanics), you are working with operators. The most natural way - because of the relation with the quantum harmonic oscillator - is to use the impulsion-time representation for these operators : $a(\vec k,t) = a(\vec k) \, e^{i k_0t} , \, a^+(\vec k,t) = a^+(\vec k) \, e^{-i k_0t}$, with $k_0^2 = \vec k^2 + m^2$

But you may use the position representation also :

$$A(\vec x, t) = \int d^4k \,\delta(k^2 - m^2) \, (a(\vec k,t) \, e^{i \vec k.\vec x}+ a^+(\vec k,t) \, e^{-i \vec k.\vec x})$$

$$A(\vec x, t) = \int d^4k \,\delta(k^2 - m^2) \, (a(\vec k) \, e^{i k.x}+ a^+( \vec k) \, e^{-i k.x})$$

So you must be very careful when speaking about "particles". In Quantum field theories, we are working with operators whose eigenvalues are fields (as in Quantum mechanics, the eigenvalues of the Position operator are the only possible measurable positions)

Of corse, instead of speaking about operators, you can speak about states, for instance, the state $|\vec k> = a^+(\vec k)|0>$. But it is a 1-particle state, it is not an operator, and it is not a multi-particle state.

You can also build a "wave-packet" state by considering some particular linear combination of 1-particle states, for instance :

$$|Wave Packet> = \int d^4 k \,\delta(k^2 - m^2) \, f(\vec k) \, |\vec k>$$

And, if you choose correctly the function $f$, this state could be localized in a precise region of space-time.

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I would not go as far as calling a localized object "false" from a quantum mechanical point of view. It is simply not the mathematically easiest way to handle things but a still perfectly valid. – Neuneck May 22 '13 at 15:24

A particle (say a scalar boson) is something that has the right mass, the right spin and the right properties. If the operator $a_{p}^{\dagger}$ creates a state of definite $p$ acting on the vacuum $|0\rangle$ and such that the Einstein relation between energy and momentum is satisfied, then it is more or less a natural thing to say that the state $a^{\dagger}_{p}|0\rangle$ represents an excitation of the Klein Gordon field, and that mathematical object represents a particle.

Even in non-relativistic quantum mechanics you can't have perfectly defined position states, at say, $\vec{r}_0$$\psi(\vec{r})_{\vec{r_0}} \propto\delta^{3}(\vec{r}-\vec{r}_0)$ because this is not an allowed function (not square integrable) and you can't apply Born's rule and the Heisenberg uncertainty principle doesn't apply either.

The particle as a localized object with a definite trajectory (a twice-differentiable function of the coordinates with respect to time) it is not in the realm of quantum mechanics. The mathematical language of quantum mechanics, and its postulates, forbid us to speak of trajectories in the classical (non quantum mechanical) sense.

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Of course you can have perfectly defined position states. At least in the same way that you can have perfectly defined momentum states. $\psi(r) \propto \int dk \psi(k) exp(i k r)$ – jjj May 22 '13 at 10:32
@jjj I guess Nivalth's objection is that neither position nor momentum eigenstates are normalizable, they don't live in $L^2(\mathbb{R}^n)$, but rather in the "rigged" version, where you add distributional functions into the mix. – twistor59 May 22 '13 at 12:06