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In the standard classical Maxwell theory, we use the following arguments to claim that there are only two propagating degrees of freesomfreedom

  • $A_\mu$ has 4 components

  • $A_0$ is non-dynamical (-1)

  • $\mathcal{L}_\text{Maxwell}$ enjoys gauge symmetry and should be removed (-1)

So in total, we have $4 - 1 - 1 = 2$ dofs.

However, in a quantization procedure as discussed in David Tong's Quantum Field Theory lecture note (section 6.2.2 LorentzLorenz Gauge), the Lagrangian is modified to $$ \mathcal{L} = - \frac{1}{4} F_{\mu \nu}F^{\mu \nu} - \frac{1}{2} (\partial_\mu A^\mu)^2 \ . $$$$ \mathcal{L} = - \frac{1}{4} F_{\mu \nu}F^{\mu \nu} - \frac{1}{2\xi} (\partial_\mu A^\mu)^2 \ . $$

  • $A_\mu$ has 4 components.

  • $A_0$ is dynamical again

  • there is no gauge symmetry left.

Question: at this stage, can we claim that there are only 2 propagating dofs? (I feel we cannot.)

Of course, subsequently we introduced the Gupta-Bleuler condition $$ \partial^\mu A_\mu^{(+)} |\text{phy}\rangle = 0 $$ to make the observation (or rather, to decide) that only the two transverse polarisation are "physical". I feel, only at this stage, that we are in position to fix #dof to 2.

Please correct me if my understanding is incorrect (the counting of dofs has been fuzzy in my head, although I know what words to say to get the right number).

In the standard classical Maxwell theory, we use the following arguments to claim that there are only two propagating degrees of freesom

  • $A_\mu$ has 4 components

  • $A_0$ is non-dynamical (-1)

  • $\mathcal{L}_\text{Maxwell}$ enjoys gauge symmetry and should be removed (-1)

So in total, we have $4 - 1 - 1 = 2$ dofs.

However, in a quantization procedure as discussed in David Tong's Quantum Field Theory lecture note (section 6.2.2 Lorentz Gauge), the Lagrangian is modified to $$ \mathcal{L} = - \frac{1}{4} F_{\mu \nu}F^{\mu \nu} - \frac{1}{2} (\partial_\mu A^\mu)^2 \ . $$

  • $A_\mu$ has 4 components.

  • $A_0$ is dynamical again

  • there is no gauge symmetry left.

Question: at this stage, can we claim that there are only 2 propagating dofs? (I feel we cannot.)

Of course, subsequently we introduced the Gupta-Bleuler condition $$ \partial^\mu A_\mu^{(+)} |\text{phy}\rangle = 0 $$ to make the observation (or rather, to decide) that only the two transverse polarisation are "physical". I feel, only at this stage, that we are in position to fix #dof to 2.

Please correct me if my understanding is incorrect (the counting of dofs has been fuzzy in my head, although I know what words to say to get the right number).

In the standard classical Maxwell theory, we use the following arguments to claim that there are only two propagating degrees of freedom

  • $A_\mu$ has 4 components

  • $A_0$ is non-dynamical (-1)

  • $\mathcal{L}_\text{Maxwell}$ enjoys gauge symmetry and should be removed (-1)

So in total, we have $4 - 1 - 1 = 2$ dofs.

However, in a quantization procedure as discussed in David Tong's Quantum Field Theory lecture note (section 6.2.2 Lorenz Gauge), the Lagrangian is modified to $$ \mathcal{L} = - \frac{1}{4} F_{\mu \nu}F^{\mu \nu} - \frac{1}{2\xi} (\partial_\mu A^\mu)^2 \ . $$

  • $A_\mu$ has 4 components.

  • $A_0$ is dynamical again

  • there is no gauge symmetry left.

Question: at this stage, can we claim that there are only 2 propagating dofs? (I feel we cannot.)

Of course, subsequently we introduced the Gupta-Bleuler condition $$ \partial^\mu A_\mu^{(+)} |\text{phy}\rangle = 0 $$ to make the observation (or rather, to decide) that only the two transverse polarisation are "physical". I feel, only at this stage, that we are in position to fix #dof to 2.

Please correct me if my understanding is incorrect (the counting of dofs has been fuzzy in my head, although I know what words to say to get the right number).

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Lelouch
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$R_\xi$ gauge and degrees of freedom counting

In the standard classical Maxwell theory, we use the following arguments to claim that there are only two propagating degrees of freesom

  • $A_\mu$ has 4 components

  • $A_0$ is non-dynamical (-1)

  • $\mathcal{L}_\text{Maxwell}$ enjoys gauge symmetry and should be removed (-1)

So in total, we have $4 - 1 - 1 = 2$ dofs.

However, in a quantization procedure as discussed in David Tong's Quantum Field Theory lecture note (section 6.2.2 Lorentz Gauge), the Lagrangian is modified to $$ \mathcal{L} = - \frac{1}{4} F_{\mu \nu}F^{\mu \nu} - \frac{1}{2} (\partial_\mu A^\mu)^2 \ . $$

  • $A_\mu$ has 4 components.

  • $A_0$ is dynamical again

  • there is no gauge symmetry left.

Question: at this stage, can we claim that there are only 2 propagating dofs? (I feel we cannot.)

Of course, subsequently we introduced the Gupta-Bleuler condition $$ \partial^\mu A_\mu^{(+)} |\text{phy}\rangle = 0 $$ to make the observation (or rather, to decide) that only the two transverse polarisation are "physical". I feel, only at this stage, that we are in position to fix #dof to 2.

Please correct me if my understanding is incorrect (the counting of dofs has been fuzzy in my head, although I know what words to say to get the right number).