- #1

thetafilippo

- 10

- 0

I derive the quadratic form of Dirac equation as follows

$$\lbrace[i\not \partial-e\not A]^2-m^2\rbrace\psi=\lbrace\left( i\partial-e A\right)^2 + \frac{1}{2i} \sigma^{\mu\nu}F_{\mu \nu}-m^2\rbrace\psi=0$$

And I need to find the form of the spin dependent term to get the final expression

$$g \frac{e}{2} \frac{\sigma^{\mu\nu}}{2}F_{\mu \nu}=-g\frac{e}{2}\left(i\vec{\alpha}\cdot\mathbf{E}+\vec{\Sigma}\cdot\mathbf{B}\right)$$

But I don't get this expression.I'm using the Dirac representation with these quantities

$$\vec{\alpha}=\begin{pmatrix}

0 & \vec{\sigma}\\

\vec{\sigma} & 0

\end{pmatrix} \ \ \ \ \ \vec{\Sigma}=\begin{pmatrix}

\vec{\sigma}& 0\\

0&\vec{\sigma}

\end{pmatrix}$$

Where

$$\vec{\sigma}=(\sigma_x,\sigma_y,\sigma_z)$$

is the Pauli matrix vector.

I constructed the electromagnetic tensor term by term, using the definition

$$F_{\mu\nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}$$

with the metric tensor

$$g^{\mu\nu}=\textrm{diag}(+1,-1,-1,-1)$$

and I get

$$F_{\mu\nu}=\begin{pmatrix}

0 & E_x&E_y&E_z\\

-E_x&0&B_z & -B_y\\

-E_y&-B_z&0&B_x\\

-E_z&B_y&-B_x&0

\end{pmatrix}$$

I evaluate the $$\sigma^{\mu\nu}$$ matrix starting from its definition in terms of gamma matrices $$\sigma^{\mu\nu}=\frac{i}{2}\left[\gamma^\mu,\gamma^\nu\right]$$

$$\sigma^{00}=\frac{i}{2}[\gamma^0,\gamma^0]=0$$

$$\sigma^{0i}=\frac{i}{2}[\gamma^0,\gamma^i]=[\gamma^0,\gamma^0\alpha_i]=\alpha_i-\gamma^0\alpha_i\gamma^0=-2\alpha_i$$

$$\sigma^{ij}=\frac{i}{2}[\gamma^i,\gamma^j]=[\gamma^0\alpha_i,\gamma^0\alpha_j]=\frac{i}{2}\gamma^0(\alpha_i\gamma^0\alpha_j-\alpha_j\gamma^0\alpha_i)=\frac{i}{2}

\begin{pmatrix}

-[\sigma_i,\sigma_j] &0\\

0&-[\sigma_i,\sigma_j]

\end{pmatrix}=\epsilon_{ijk}\begin{pmatrix}

\sigma_k &0\\

0&\sigma_k

\end{pmatrix}=\epsilon_{ijk}\Sigma_k$$

And the remaining terms follow by the antisymmetry property $$\sigma^{\mu\nu}=-\sigma^{\nu\mu}$$

$$\sigma^{\mu\nu}=\begin{pmatrix}

0 & 2\alpha_x & 2\alpha_y & 2\alpha_z\\

-2\alpha_x&0&\Sigma_z & -\Sigma_y\\

-2\alpha_x&-\Sigma_z&0&\Sigma_x\\

-2\alpha_x&\Sigma_y&-\Sigma_x&0

\end{pmatrix}$$

Now, my questions are:

"Why these calculations do not yield the correct result?"

"What I should do to obtain the correct result? What I'm missing?"

$$\frac{\sigma^{\mu\nu}}{2}F_{\mu \nu}=-\left(i\vec{\alpha}\cdot\mathbf{E}+\vec{\sigma}\cdot\mathbf{B}\right)$$

$$\lbrace[i\not \partial-e\not A]^2-m^2\rbrace\psi=\lbrace\left( i\partial-e A\right)^2 + \frac{1}{2i} \sigma^{\mu\nu}F_{\mu \nu}-m^2\rbrace\psi=0$$

And I need to find the form of the spin dependent term to get the final expression

$$g \frac{e}{2} \frac{\sigma^{\mu\nu}}{2}F_{\mu \nu}=-g\frac{e}{2}\left(i\vec{\alpha}\cdot\mathbf{E}+\vec{\Sigma}\cdot\mathbf{B}\right)$$

But I don't get this expression.I'm using the Dirac representation with these quantities

$$\vec{\alpha}=\begin{pmatrix}

0 & \vec{\sigma}\\

\vec{\sigma} & 0

\end{pmatrix} \ \ \ \ \ \vec{\Sigma}=\begin{pmatrix}

\vec{\sigma}& 0\\

0&\vec{\sigma}

\end{pmatrix}$$

Where

$$\vec{\sigma}=(\sigma_x,\sigma_y,\sigma_z)$$

is the Pauli matrix vector.

I constructed the electromagnetic tensor term by term, using the definition

$$F_{\mu\nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}$$

with the metric tensor

$$g^{\mu\nu}=\textrm{diag}(+1,-1,-1,-1)$$

and I get

$$F_{\mu\nu}=\begin{pmatrix}

0 & E_x&E_y&E_z\\

-E_x&0&B_z & -B_y\\

-E_y&-B_z&0&B_x\\

-E_z&B_y&-B_x&0

\end{pmatrix}$$

I evaluate the $$\sigma^{\mu\nu}$$ matrix starting from its definition in terms of gamma matrices $$\sigma^{\mu\nu}=\frac{i}{2}\left[\gamma^\mu,\gamma^\nu\right]$$

$$\sigma^{00}=\frac{i}{2}[\gamma^0,\gamma^0]=0$$

$$\sigma^{0i}=\frac{i}{2}[\gamma^0,\gamma^i]=[\gamma^0,\gamma^0\alpha_i]=\alpha_i-\gamma^0\alpha_i\gamma^0=-2\alpha_i$$

$$\sigma^{ij}=\frac{i}{2}[\gamma^i,\gamma^j]=[\gamma^0\alpha_i,\gamma^0\alpha_j]=\frac{i}{2}\gamma^0(\alpha_i\gamma^0\alpha_j-\alpha_j\gamma^0\alpha_i)=\frac{i}{2}

\begin{pmatrix}

-[\sigma_i,\sigma_j] &0\\

0&-[\sigma_i,\sigma_j]

\end{pmatrix}=\epsilon_{ijk}\begin{pmatrix}

\sigma_k &0\\

0&\sigma_k

\end{pmatrix}=\epsilon_{ijk}\Sigma_k$$

And the remaining terms follow by the antisymmetry property $$\sigma^{\mu\nu}=-\sigma^{\nu\mu}$$

$$\sigma^{\mu\nu}=\begin{pmatrix}

0 & 2\alpha_x & 2\alpha_y & 2\alpha_z\\

-2\alpha_x&0&\Sigma_z & -\Sigma_y\\

-2\alpha_x&-\Sigma_z&0&\Sigma_x\\

-2\alpha_x&\Sigma_y&-\Sigma_x&0

\end{pmatrix}$$

Now, my questions are:

"Why these calculations do not yield the correct result?"

"What I should do to obtain the correct result? What I'm missing?"

$$\frac{\sigma^{\mu\nu}}{2}F_{\mu \nu}=-\left(i\vec{\alpha}\cdot\mathbf{E}+\vec{\sigma}\cdot\mathbf{B}\right)$$

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