The action-functionals describing the motion $\mathbf{x}:[a,b]\to \mathbb{R}^3$ of a free particle of mass $m$ and the evolution $\varphi:[a,b]\times \Omega\to \mathbb{R}$ of a free scalar field of mass $m$ are, respectively, \begin{align}S[\mathbf{x}]:=&\int_a^b\frac{m}{2}\frac{\mathrm{d} \mathbf{x}}{\mathrm{d}t}^2\mathrm{d}t.\tag{1} \\S[\varphi]:=&\int_a^b\left(\int_{\Omega}\frac{m}{2}\mathrm{d}\varphi^2\,\mathrm{d}\mathbf{x}\right)\mathrm{d}t.\tag{2}\end{align} Where $\Omega\subset \mathbb{R}^3$ is a bounded region of space.

This leads me to think, based on the fact that free scalar quantum fields can be thought of as position space wavefunctions and vice versa, that if we make the substitution $$\varphi(\mathbf{x},t):=\delta^3(\mathbf{x}-\mathbf{x}(t))$$ and perform the integral over $\Omega$ in $(2)$, we should recover $(1)$. Is this true? If not, is it still possible to convert lagrangians to lagrangian densities?

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    $\begingroup$ Comment to the question (v3): The square of a distribution is mathematically ill-defined, cf. e.g. this Phys.SE post and links therein. $\endgroup$
    – Qmechanic
    Dec 29 '14 at 13:54
  • $\begingroup$ Huh - that's interesting. I guess I'm misled by quantum mechanics, where definitions like $\Psi(\mathbf{x})=\delta^3(\mathbf{x})$ are tossed around too much $\endgroup$
    – Dave
    Dec 29 '14 at 13:57
  • $\begingroup$ that is fine, what's not fine is to take the square of that $\endgroup$
    – Phoenix87
    Dec 29 '14 at 13:58
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    $\begingroup$ The free scalar quantum field cannot be thought of directly as a wavefunction. In the same way that a wave-function defines superpositions over position states, the wave-functional describes superpositions of field configurations. However, it is true that when you restrict the field to be extremely localized, these become the same thing. $\endgroup$
    – jwimberley
    Dec 29 '14 at 14:11
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    $\begingroup$ Before starting the discussion - (2) is not the action of a free massive scalar field. Depending on what you mean by $d\phi$ this is either mathematically wrong notation, or an action for a massless field. $\endgroup$
    – Fedxa
    Feb 18 '17 at 23:35

Particles in $n+1$ dimensions ($n$ spatial and one temporal) can be mathematically thought of as fields in $0+1$ dimensions (no spatial, one temporal) where position of the particle $n+1$ dimension corresponds to the value of the field in $0+1$ dimensions. This is the essence of the similarity of these actions.

For example, a particle in a harmonic oscillator potential can be mathematically thought of as a massive field in $0+1$ dimensions.

To answer your question:

The field $\varphi(\mathbf x,t)=\delta^3(\mathbf x-\mathbf x(t))$ is not a solution to equation of motion of massless (or massive) scalar field. If you put field equal to a delta function and its time derivative equal to zero as initial conditions (i.e. $\varphi(\mathbf x,0)=\delta^3(\mathbf x-\mathbf x_0)$ and $\dot \varphi(\mathbf x, 0)=0$), the most scalar fields will spread, i.e. they will not remain equal to a delta function.


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