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In the quantization of the Klein-Gordon field (for example) we interpret the quanta of the fields with definite momentum and energy as particles but are they also localized in space? Shouldn't a physical particle in quantum mechanics be represented by a wave-packet in the peaked about some point is space? Is the scenario same in quantum field theory? If yes, how can we write such a state mathematically? In a nutshell, are the field quanta point particle?

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To represent the states of a single particle in the $3$-space you should exploit the so called Newton-Wigner position representation.

The point is that the so obtained wavefunction, for a KG field, is a complex function defined in the rest space of the reference frame you are adopting and it has a strongly non-local behavior under the action of Lorentz transformation. This is different from what happens for quantum fields that locally transform under Lorentz transformations.

Position-defined states are again formally represented by Dirac deltas in that representation. If $\psi_{x_0}= \delta(x-x_0)$ is the wavefuntion of a KG particle in NW representation spatially localized at $x_0$ for $t=0$, the field configuration $\phi_{x_0}(t,x)$ associated to it is a solution of KG equation such that, for $t=0$, is localized around $x_0$ just up to the Compton length of the particle.

ADDENDUM

Let us focus on the momentum representation space for a real scalar KG particle with mass $m>0$. There are several, unitarily equivalent, possibilities to define that (one-particle) Hilbert space $\cal H$ used to build up the Fock space of the field. The most elementary one is thath referring to the standard measure $d^3p$ (instead of the covariant one $d^3p/E$).

In this representation ${\cal H}= L^2({\mathbb R}^3, d^3p)$ and states are represented by normalized wavefunctions $\psi =\psi(\vec{p})$ belonging to that functional space.

The momentum operator is therefore defined as (I adopt here the signature $-+++$ and $c=\hbar=1$): $$\left(P^i \psi\right)(\vec{p}):= p^i \psi(\vec{p})\:, \quad \left(P^0 \psi\right)(\vec{p}):= E(\vec{p})\psi(\vec{p})=\sqrt{\vec{p}^2 +m^2}\psi(\vec{p})\:,\quad i=1,2,3\:,$$ for those wavefunctions such that the right-hand sides still are elements of $L^2(\mathbb R^3, d^3p)$. These operators are automatically self-adjoint and the $\mathbb R^3$ in $L^2(\mathbb R^3, d^3p)$ turns out to coincide with the joint spectrum of $P_1,P_2,P_3$ as usual.

Let us come to the NW position operators and the corresponding representation of $\cal H$. In the Hilbert space $L^2(\mathbb R^3, d^3p)$, the three components of the three Newton-Wigner position operators are simply defined as the unique self-adjoint extensions of the symmetric operators:
$$\left(X^i\psi\right)(\vec{p}) := - i\frac{\partial}{\partial p^i}\psi(\vec{p})\qquad \psi \in C_0^\infty(\mathbb R^3)\:.$$ As usual, the (Newton-Wigner) position representation is the representation of the formalism in an unitarily equivalent Hilbert space where the $X^i$s are multiplicative. So the $L^2$ space is the one constructed upon joint spectrum of these operators: $L^2(\mathbb R^3, d^3x)$. In practice, also introducing time evolution, the wavefunctions in this space are related with those in the momentum representation by the usual Fourier-Plancherel transform. For $\psi \in L^1(\mathbb R^3, d^3p)\cap L^2(\mathbb R^3, d^3p)$: $$\phi_{NW}(t,\vec{x}):= \int_{\mathbb R^3} \frac{e^{i \vec{p}\cdot \vec{x}}}{(2\pi)^{3/2}} e^{-i tE(\vec{p})}\psi(\vec{p}) d^3p\:.$$ Conversely, the field representation is: $$\phi(t,\vec{x}):= \int_{\mathbb R^3} \frac{e^{i \vec{p}\cdot \vec{x}}}{(2\pi)^{3/2}\sqrt{2E(\vec{p})}} e^{-i tE(\vec{p})}\psi(\vec{p}) d^3p\:.$$ When acting on $\psi$ with the usual unitary action of the Lorentz group (henceforth $x= (t,\vec{x})$ and $p=(E(\vec{p}),\vec{p})$ are $4$-vectors and I define $E(p):= E(\vec{p})$), i.e.: $$\left(U_\Lambda\psi\right)(\vec{p}) = \sqrt{\frac{E(p)}{E(\Lambda^{-1} p )}}\psi\left( \vec{\Lambda^{-1} p}\right) \:,$$ the field representation transforms locally $$\phi \to \phi'\:,\quad \phi'(x):=\phi(\Lambda^{-1}x)\:,$$ whereas the transformation law of the Newton-Wigner representation, $\phi_{NW}$, is given in terms of a quite complicated integral formula and the action turns out to be highly non-local, as the transformed $\phi'_{NW}(x)$ also depends on the values of the functions $\phi_{NW}(x)$ at events different for $x$ itself.

Finally, consider a particle localized at $\vec{x}_0$ for $t=0$. Therefore, at least formally, we can write: $$\phi_{NW}(0,\vec{x}) = \delta(\vec{x}-\vec{x}_0)\:.$$ The covariant representation instead reads: $$\phi(0,\vec{x}):= \int_{\mathbb R^3} \frac{e^{i \vec{p}\cdot (\vec{x}-\vec{x}_0)}}{(2\pi)^{3}\sqrt{2E(\vec{p})}} d^3p\:.$$ It is possible to prove that the shape of that function is concentrated in a set including $\vec{x}_0$ with linear extension of the order of $1/m$, namely the Compton length of the particle.

There is a wide, quite technical, literature on these topics. Some results can be extended to QFT in curved spacetime for static spacetimes. An elementary discussion in that context can be found in Fulling's textbook on QFT in curved spacetime. A general discussion though confined to Minkowski spacetime, but even including fields with spin, appears in chapter 20 of Barut-Raczka's textbook on the theory of group representations. Several attempts to encompass a physically sensible notion of time observable exists in the literature especially making use of POVM.

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@ V.Moretti-How do you define a position space wavefunctions in field theory? In field theory, $$\hat \phi(x)|0\rangle\simeq \int d^3p \frac{1}{\sqrt{2E_p}}e^{-ip\cdot x}|p\rangle$$. Right? How do you define position-space wavefunction here? Position is supposed to be a parameter in field theory, not an operator. Therefore, can form a basis such that you can construct position space wave function corresponding to an abstract ket vector in the Fock space. Isn't it? –  Roopam Feb 12 at 5:27
    
@RoopamSinha I have inserted a long addendum to answer! It is false that position is supposed to be only a parameter in QFT. At least for one-particle states it can be defined also as an observable. –  V. Moretti Feb 12 at 8:36
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