Interpreting SM Lagrangian "differential particle" terms

  • #1
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The Standard Model Lagrangian contains terms like these:

##-\partial_\mu \phi^+ \partial_\mu\phi^-##

##-\frac{1}{2}\partial_\nu Z^0_\mu\partial_\nu Z^0_\mu##

##-igc_w\partial_\nu Z^0(W^+_\mu W^-_\nu-W^+_\nu W^-_\mu)##

How should one interpret the "derivative particle fields" like ##\partial_\mu\phi^+## physically? What are they or what do they represent?

Or, more generally, how should one interpret the differentiation, especially when applied to what represents physical objects?
 

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  • #2
Orodruin
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It is not differential particles. The standard model is a quantum field theory and contains fields. The derivatives are the derivatives of those fields.
 
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It is not differential particles. The standard model is a quantum field theory and contains fields. The derivatives are the derivatives of those fields.
My mistake. But does that mean the fields interact with their own rates of change or something? Is there an easy way to make physical sense of it?
 
  • #4
Orodruin
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In the momentum representation, the derivatives become the 4-momenta corresponding to the fields in the Feynman diagrams.
 
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  • #5
Hepth
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Basically yes, if you want a way to think about it then imagine that whatever field your working with is unbound, and so has some plane-like wavefunction. Then these derivatives are basically creating interaction terms that have a proportionality to the momentum of the field.

Since it plane-like, (so like e^(x.p)) itll give you the momentum times the field itself. That way you still have creation/annihilation operators in the field, but get the momentum out of it. All of these operations will happen once you start integrating over the terms.
 
  • #6
king vitamin
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Those terms give an energy penalty to large fluctuations in space and time. Compare them to the action for a one-dimensional string:
$$
\mathcal{S} = \int dt \, dx \left[ \frac{\rho}{2} \left(\frac{\partial u(x,t)}{\partial t}\right)^2 - \frac{T}{2} \left(\frac{\partial u(x,t)}{\partial x}\right)^2 \right]
$$
where ##u(x,t)## is the amplitude, ##\rho## is the density per unit length, and ##T## is the string tension. The Euler-Lagrange equation for this action is just the wave equation, and the two terms (which are combined in a relativistic notation in your post) correspond to kinetic and potential energy respectively. Various complications occur from quantizing these theories, but the idea that large variations in space and time cost more energy is still true.
 
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  • #7
DarMM
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In the momentum representation, the derivatives become the 4-momenta corresponding to the fields in the Feynman diagrams.
Exactly, in a particle picture how strongly the particles couple to each other/the vertex for their interaction depends not just on a fixed constant like ##e## the electric charge, but also on their four momenta.
 

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