Evaluating Matrix Spin Dependent Term in Dirac Quadratic Equation

In summary: B_i\\\sigma^{0i}E_i-\sigma^{0i}B_i & 0\end{pmatrix}\ \textrm{where}\ i=x,y,z$$$$=-\frac{1}{4}\begin{pmatrix}0 & -2\alpha_iE_i+2\alpha_iB_i\\2\alpha_iE_i-2\alpha_iB_i & 0\end{pmatrix}=-\frac{1}{2}\begin{pmatrix}i\vec{\alpha}\cdot\mathbf{E}+\vec{\sigma}\cdot\mathbf{B} & 0\\0 & i\
  • #1
thetafilippo
10
0
I derive the quadratic form of Dirac equation as follows
$$\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?"

$$\frac{\sigma^{\mu\nu}}{2}F_{\mu \nu}=-\left(i\vec{\alpha}\cdot\mathbf{E}+\vec{\sigma}\cdot\mathbf{B}\right)$$
 
Last edited:
Physics news on Phys.org
  • #2


First of all, it is important to note that the final expression you are trying to obtain is not the spin dependent term, but rather the spinor form of the electromagnetic interaction term in the Dirac equation. This term describes the interaction of the electron's spin with the external electromagnetic field.

In order to obtain the correct result, you need to use the correct representation of the gamma matrices. The representation you are using, known as the "chiral representation", is not suitable for this calculation. Instead, you should use the "Dirac representation" where the gamma matrices are given by:

$$\gamma^0 = \begin{pmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & -1
\end{pmatrix} \ \ \ \ \ \gamma^i = \begin{pmatrix}
0 & \sigma_i\\
-\sigma_i & 0
\end{pmatrix}$$

Using this representation, you can easily show that:

$$\sigma^{\mu\nu}=\frac{i}{2}\left[\gamma^\mu,\gamma^\nu\right]=\frac{i}{4}\begin{pmatrix}
0 & \sigma^{\mu\nu}\\
\sigma^{\mu\nu} & 0
\end{pmatrix}$$

where $\sigma^{\mu\nu}=\frac{i}{2}\left(\gamma^\mu\gamma^\nu-\gamma^\nu\gamma^\mu\right)$.

Substituting this into the expression for the electromagnetic interaction term, we get:

$$\frac{\sigma^{\mu\nu}}{2}F_{\mu \nu}=-\frac{1}{4}\begin{pmatrix}
0 & \sigma^{\mu\nu}\\
\sigma^{\mu\nu} & 0
\end{pmatrix}\begin{pmatrix}
0 & E_x&E_y&E_z\\
-E_x&0&B_z & -B_y\\
-E_y&-B_z&0&B_x\\
-E_z&B_y&-B_x&0
\end{pmatrix}$$

$$=-\frac{1}{4}\begin{pmatrix}
0 & -\sigma^{0i}E_i+\
 

Related to Evaluating Matrix Spin Dependent Term in Dirac Quadratic Equation

1. What is the significance of evaluating matrix spin dependent term in Dirac quadratic equation?

The matrix spin dependent term in Dirac quadratic equation is important because it accounts for the spin of particles, which is a fundamental property of matter. This term allows for the prediction and explanation of various phenomena, such as the spin of electrons and the fine structure of atomic spectra.

2. How is the matrix spin dependent term calculated in the Dirac quadratic equation?

The matrix spin dependent term is calculated by taking into account the spin matrices, which represent the spin operators for particles. These matrices are then multiplied by the momentum and energy terms in the Dirac equation to account for the spin of particles.

3. What is the role of the matrix spin dependent term in relativistic quantum mechanics?

In relativistic quantum mechanics, the matrix spin dependent term plays a crucial role in describing the behavior of particles with spin. It allows for the prediction of spin-dependent interactions and phenomena, and is essential in understanding the behavior of particles at high energies.

4. How does the matrix spin dependent term affect the solutions of the Dirac equation?

The matrix spin dependent term affects the solutions of the Dirac equation by introducing spinor solutions, which are four-component wavefunctions that describe the spin state of particles. These spinor solutions allow for the prediction of spin-dependent properties and interactions of particles.

5. Are there any experimental validations of the matrix spin dependent term in the Dirac equation?

Yes, there have been numerous experimental validations of the matrix spin dependent term in the Dirac equation. For example, the fine structure of atomic spectra has been accurately predicted and verified using this term. Additionally, experiments such as the Stern-Gerlach experiment have confirmed the existence of spin and its role in particle behavior.

Similar threads

  • Quantum Physics
Replies
6
Views
794
Replies
24
Views
2K
Replies
5
Views
820
Replies
4
Views
1K
Replies
3
Views
663
Replies
4
Views
1K
Replies
2
Views
495
Replies
1
Views
1K
  • Science and Math Textbooks
Replies
7
Views
517
  • Quantum Physics
Replies
3
Views
2K
Back
Top