- #1
TriTertButoxy
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Something is totally not making sense. In a complex scalar field theory, I have two field degrees of freedom, which I parametrize in polar field coordinates: [itex]\phi = \rho e^{i\theta}/\sqrt{2}[/itex], where [itex]\rho[/itex] and [itex]\theta[/itex] are real-valued; and its Lagrangian takes the form:
[tex]\mathcal{L} = \frac{1}{2}\partial_\mu\rho\partial^\mu\rho-V(\rho)+\frac{1}{2}\rho^2(\partial_\mu\theta)^2.[/tex]
I have written this Lagrangian so that it is has a global U(1) symmetry (under [itex]\rho\rightarrow\rho[/itex] and [itex]\theta\rightarrow\theta+\alpha[/itex] with [itex]\alpha[/itex] being a spacetime-independent parameter). And I have a conserved charge, allowing me to get positively- and negatively- charged particles.
All is good so far; but now when I gauge the U(1), something funny happens. Suddenly [itex]\theta[/itex] becomes an "unphysical degree of freedom." If the potential [itex]V(\phi)[/itex] is such there is no spontaneous symmetry breaking, i.e. no vacuum expectation value for [itex]\rho[/itex], the Lagrangian reads:
[tex]\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}{ \partial }_\mu\rho{\partial}^\mu\rho-V(\rho)+\frac{1}{2}\rho^2({\partial}_\mu\theta)^2+e\rho A_\mu\partial^\mu\theta+e^2A^2\rho^2.[/tex]
And now, the gauge transformation rule for the fields are:
[tex]A_\mu\rightarrow A_\mu-\frac{1}{e}{\partial_\mu}\alpha[/tex][tex]\rho\rightarrow\rho[/tex][tex]\theta\rightarrow\theta+\alpha,[/tex]where now the gauge parameter [itex]\alpha[/itex] is spacetime-dependent.
The first term is the gauge-invariant kinetic term for the gauge field, and the next two terms involve the gauge invariant field [itex]\rho[/itex]. The last three terms are individually not gauge-invariant, but added together, are gauge-invariant.
I shall now make a particular choice of gauge such that [itex]\theta=0[/itex], by appropriately choosing [itex]\alpha[/itex]. This is analogous to the unitary gauge when spontaneous symmetry breaking does occur (but again, my potential is designed so this doesn't happen). With this choice of gauge my lagrangian reads:
[tex]\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}{ \partial }_\mu\rho{\partial}^\mu\rho-V(\rho)+e^2A^2\rho^2.[/tex]
And all the dependence on [itex]\theta[/itex] is gone. Given that no symmetry breaking has occurred, the gauge field remains massless, and only has two physical degrees of freedom (the transverse polarizations) and since [itex]\rho[/itex] is a real-valued field, it represents only one field degree of freedom. Yet, this theory is supposed to describe the interaction of charged particles with the gauge field. How is this describing the interaction between charged particles when I have only one real scalar field? What the heck is going on?
Bottom line: How can real-valued [itex]\rho[/itex] accommodate particle and antiparticle?
[tex]\mathcal{L} = \frac{1}{2}\partial_\mu\rho\partial^\mu\rho-V(\rho)+\frac{1}{2}\rho^2(\partial_\mu\theta)^2.[/tex]
I have written this Lagrangian so that it is has a global U(1) symmetry (under [itex]\rho\rightarrow\rho[/itex] and [itex]\theta\rightarrow\theta+\alpha[/itex] with [itex]\alpha[/itex] being a spacetime-independent parameter). And I have a conserved charge, allowing me to get positively- and negatively- charged particles.
All is good so far; but now when I gauge the U(1), something funny happens. Suddenly [itex]\theta[/itex] becomes an "unphysical degree of freedom." If the potential [itex]V(\phi)[/itex] is such there is no spontaneous symmetry breaking, i.e. no vacuum expectation value for [itex]\rho[/itex], the Lagrangian reads:
[tex]\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}{ \partial }_\mu\rho{\partial}^\mu\rho-V(\rho)+\frac{1}{2}\rho^2({\partial}_\mu\theta)^2+e\rho A_\mu\partial^\mu\theta+e^2A^2\rho^2.[/tex]
And now, the gauge transformation rule for the fields are:
[tex]A_\mu\rightarrow A_\mu-\frac{1}{e}{\partial_\mu}\alpha[/tex][tex]\rho\rightarrow\rho[/tex][tex]\theta\rightarrow\theta+\alpha,[/tex]where now the gauge parameter [itex]\alpha[/itex] is spacetime-dependent.
The first term is the gauge-invariant kinetic term for the gauge field, and the next two terms involve the gauge invariant field [itex]\rho[/itex]. The last three terms are individually not gauge-invariant, but added together, are gauge-invariant.
I shall now make a particular choice of gauge such that [itex]\theta=0[/itex], by appropriately choosing [itex]\alpha[/itex]. This is analogous to the unitary gauge when spontaneous symmetry breaking does occur (but again, my potential is designed so this doesn't happen). With this choice of gauge my lagrangian reads:
[tex]\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\frac{1}{2}{ \partial }_\mu\rho{\partial}^\mu\rho-V(\rho)+e^2A^2\rho^2.[/tex]
And all the dependence on [itex]\theta[/itex] is gone. Given that no symmetry breaking has occurred, the gauge field remains massless, and only has two physical degrees of freedom (the transverse polarizations) and since [itex]\rho[/itex] is a real-valued field, it represents only one field degree of freedom. Yet, this theory is supposed to describe the interaction of charged particles with the gauge field. How is this describing the interaction between charged particles when I have only one real scalar field? What the heck is going on?
Bottom line: How can real-valued [itex]\rho[/itex] accommodate particle and antiparticle?
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