Energy and momentum of a particle can be expressed by equation $$E^2=p_1^2c^2+p_2^2c^2+p_3^2c^2+m^2c^4\hspace{40pt}(1)$$ Equation (1) can be divided into $E$ on both sides. We obtain $$E=\frac{v_1}{c}p_1\,c+\frac{v_2}{c}p_2\,c+\frac{v_3}{c}p_3\,c+\frac{v_4}{c}m\,c^2\hspace{40pt}(2)$$ where $v^2=v_1^2+v_2^2+v_3^2$, and $v_4=\sqrt{c^2-v^2}$; The Dirac equation has the form $$i\hbar\frac{\partial\psi}{\partial t}=(\alpha_1\hat p_1c+\alpha_2\hat p_2c+\alpha_3\hat p_3c+\alpha_4m\,c^2)\psi\hspace{30pt}(3)$$ where $\alpha_i$ is matrix $(i=1,2,3,4)$. From the principle of correspondence between (2) and (3) is $\alpha_i\rightarrow v_i/c$. In quantum mechanics, it is shown that the relativistic velocity operator $v_v=dx_v/dt$; $(v=1,2,3)$ is given by $\hat{v}_v=c\,\alpha_v$, ie is a matrix operator. Then relativistic velocity operator $v_4=\sqrt {c^2-v^2}$ is the matrix $\hat v_4=c\alpha_4$. Is it right? Does the equation (2) be the basis of the Dirac equation?

  • $\begingroup$ Why don't you square your equation (2) and check if it returns your equation (1). Frankly, I don't understand how you divided (1) into (2). $\endgroup$ – Horus Apr 21 '16 at 6:31
  • $\begingroup$ Let $v_4=\sqrt{c^2-v^2}$ then $$\frac{p_1c}{E}=\frac{mv_1c(c/v_4)}{mc^2(c/v_4)}=\frac{v_1}{c}$$ and so on. Let $E_0=mc^2$ then $$\frac{E_0}{E}=\frac{v_4}{c}$$; $\endgroup$ – Alexander Klimets Apr 21 '16 at 9:21
  • $\begingroup$ Well your math seems to check out. However a quick search on the Dirac equation would already tell you that the $\alpha$ matrices have the Pauli matrices for elements, independent of velocity. Also simply squaring your equation (2), it is already evident that it does not return the original equation (1), not unless $v_i v_j = 0$ where i does not equal j. I have an idea on why this is so though I think I am wrong. $\endgroup$ – Horus Apr 21 '16 at 13:39
  • $\begingroup$ From the textbook: $$\frac{dx_{\nu}}{dt}=\frac{\partial x_{\nu}}{\partial t}+[H, x_{\nu}]$$ where $$H=c\alpha_{\nu}p_{\nu}+mc^2\alpha_4$$ Since the operator $x_{\nu}$ does not depend on time, it will be $dx_{\nu}/dt=[H,x_{\nu}]$. We get $$\frac{dx_{\nu}}{dt}=[{c\alpha_{\mu}p_{\nu}+mc^2\alpha_4},x_{\nu}]$$ The matrix $\alpha_{\mu}$ commutes with $x_{\nu}$, so that the matrix $\alpha_{\mu}$ can be factored out. Finally we have $$v_{\nu}=dx_{\nu}/dt=c\alpha_{\mu}[p_{\mu},x_{\nu}]=c\alpha_{\mu}\delta_{\mu\nu}=c\alpha_{\nu}$$ $\endgroup$ – Alexander Klimets Apr 21 '16 at 18:42

The Dirac equation (according to my understanding), is an alternate form of the Klein-Gordon equation. The Dirac equation can be derived from the following equation. $$ Z_{μ\ }Z^{μ}-m^{2}=0 $$ Were,$$ Z_{μ\ }= \left(E\ \ \ P\right)$$ and, $$Z^{μ}= \begin{pmatrix} E\\-P \end{pmatrix} $$

Where "E" is the operator of energy, and "P" is the operator of momentum. The equation:

$$ Z_{μ\ }Z^{μ}-m^{2}=0 $$

Is the 4-vector representation of the equation:

$$ E^{2}-P^{2}-m^{2}=0 $$

Which is just a rearranged form of the Mass-Energy equation, with natural-unit simplification namely,

$$E^{2}=\ P^{2}+m^{2\ }$$

Therefore, the Dirac equation is just the Schrodinger equation, with a relativistic Hamiltonian. I guess, if you start with a relativistic Hamiltonian, and rearrange it in 4-vector notation, and then substitute in Quantum operators for energy and momentum, you will eventually get some form of the Dirac equation!

The connection between the equation, $$ Z_{μ\ }Z^{μ}-m^{2}=0 $$ and the Dirac equation, can be found in this video: https://www.youtube.com/watch?v=jjG2Y_dMsbI

I hope this helps!

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