- #1
johne1618
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The Dirac electron in the Higgs vacuum field v and an electromagnetic field with vector potential A_\mu is described by the following equation:
i \gamma^\mu \partial_\mu \psi = g v \psi + e \gamma_\mu A^\mu \psi
where g is the coupling constant to the Higgs field and e is the coupling constant to the electromagnetic field.
Let us assume that we are in the rest frame of the electron so that:
\partial_x=\partial_y=\partial_z=0
Let us also assume that there is only an electrostatic potential A_0=\phi so that:
A_x = A_y = A_z = 0
So the simplified Dirac equation is now:
i \gamma^0 \partial_t \psi = g v \psi + e \gamma_0 \phi \psi
Let us choose the Weyl or Chiral basis so that:
\gamma^0 = \begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix}
where I is the 2\times2 unit matrix.
In this representation:
\psi=\begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix}
where \psi_L and \psi_R are left-handed and right-handed two-component Weyl spinors.
Subtituting into the simplified Dirac equation above we get:
i \begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix} \begin{pmatrix} \partial \psi_L / \partial t \\ \partial \psi_R / \partial t \end{pmatrix} = g v \begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix} + e \phi \begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix} \begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix}
This equation separates into two equations of two-component Weyl spinors:
i \partial \psi_R / \partial t = g v \psi_L + e \phi \psi_R
i \partial \psi_L / \partial t = g v \psi_R + e \phi \psi_L
Now let us add these two equations together to obtain:
i \frac{\partial}{\partial t} (\psi_L + \psi_R) = (g v + e \phi)(\psi_L + \psi_R)
My question is this:
Does the state \psi_L + \psi_R describe an electron with an effective mass given by gv + e \phi?
Does the presence of an electrostatic field increase the electron's mass over and above the mass induced by the Higgs vacuum field alone?
i \gamma^\mu \partial_\mu \psi = g v \psi + e \gamma_\mu A^\mu \psi
where g is the coupling constant to the Higgs field and e is the coupling constant to the electromagnetic field.
Let us assume that we are in the rest frame of the electron so that:
\partial_x=\partial_y=\partial_z=0
Let us also assume that there is only an electrostatic potential A_0=\phi so that:
A_x = A_y = A_z = 0
So the simplified Dirac equation is now:
i \gamma^0 \partial_t \psi = g v \psi + e \gamma_0 \phi \psi
Let us choose the Weyl or Chiral basis so that:
\gamma^0 = \begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix}
where I is the 2\times2 unit matrix.
In this representation:
\psi=\begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix}
where \psi_L and \psi_R are left-handed and right-handed two-component Weyl spinors.
Subtituting into the simplified Dirac equation above we get:
i \begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix} \begin{pmatrix} \partial \psi_L / \partial t \\ \partial \psi_R / \partial t \end{pmatrix} = g v \begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix} + e \phi \begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix} \begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix}
This equation separates into two equations of two-component Weyl spinors:
i \partial \psi_R / \partial t = g v \psi_L + e \phi \psi_R
i \partial \psi_L / \partial t = g v \psi_R + e \phi \psi_L
Now let us add these two equations together to obtain:
i \frac{\partial}{\partial t} (\psi_L + \psi_R) = (g v + e \phi)(\psi_L + \psi_R)
My question is this:
Does the state \psi_L + \psi_R describe an electron with an effective mass given by gv + e \phi?
Does the presence of an electrostatic field increase the electron's mass over and above the mass induced by the Higgs vacuum field alone?
