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I need to elaborate the equation ,and need to know what is the physical significance and how matrices will manipulate in the equation $$ \hat{H} = (\hat{\tau_3}+i\hat{\tau_2})\frac{\hat{p}^2}{2m_0}+ \hat{\tau_3}m_0 c^2 = \left| \begin{array}{ccc} 1 & 1 \\ -1 & -1 \\ \end{array}\right| \frac{\hat{p}^2}{2m_0} + \left| \begin{array}{ccc} 1 & 0 \\ 0 & -1 \\ \end{array}\right| m_0 c^2 $$

Where $$\tau_1 , \tau_2,\tau_3 $$ are Pauli matrices and Hamiltonian comes from "Schrodinger form of the free Klein_Gordon equation And also why did we added Pauli matrices in the free Hamiltonian ?

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perhaps you should elaborate on what system that this particular Hamiltonian. –  Prathyush Oct 9 '12 at 19:20

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Your second equation is somehow mistyped: it should have only one $\frac{p^2}{2m}$ factor in the first term.

Also, the notation $m_0$ devoid of physical meaning. There is not "rest mass" (different from any other "mass") - what used to be called by some authors (e.g. early in the 20th century) the "rest mass" is now understood to be the only mass that make sense to assign to a (massive) particle. Therefore the only correct notation is $m$, not $m_0$.

I am not sure what is your question. Perhaps showing that that Hamiltonian is somehow equivalent to the Klein-Gordon's Hamiltonian? Or something else? If that is what you are asking, then the derivation is straight-forward. Let's have a slightly more general starting expression: $H= a \frac{p^2}{2m} + b m c^2$, where $a$ and $b$ are $2\times2$ matrices whose properties are to be determined. Square $H$ to get: $H^2 = a^2 \left(\frac{p^2}{2m}\right)^2 + \{a,b\} \frac{p^2}{2m} mc^2 + b^2 m^2c^4$. If we can find $a$ and $b$ such to have $a^2=0$ $\{a,b\}=2\cdot{\bf 1}$ (here $\{A,B\}:\equiv A B + B A$) and $b^2={\bf 1}$ (and ${\bf 1}$ is the unit matrix), then we would get the original dispersion equation for a free relativistic particle: $H^2 = p^2c^2 + m^2c^4$. One such a choice is $a= \tau_3+i\tau_2$ and $b=\tau_3$, but there are many other choices, too.

Working with the components:

$H \pmatrix{\chi \cr \xi} = \left[\left(\frac{p^2}{2m}+mc^2\right) \pmatrix{1 &0 \\ 0 & -1} + \frac{p^2}{2m} \pmatrix{0 & 1 \\ -1 & 0}\right]\pmatrix{\chi \cr \xi}$

gives:

$H \chi = \left(\frac{p^2}{2m}+mc^2\right) \chi + \frac{p^2}{2m} \xi$

and

$H \xi = - \left(\frac{p^2}{2m}+mc^2\right) \xi - \frac{p^2}{2m} \chi$

Notice that $[H, p^2] =0$ so one can apply $H$ one more time to the left side of any of these two equations:

$H^2 \chi = \left(\frac{p^2}{2m}+mc^2\right) H \chi + \frac{p^2}{2m} H \xi = \left(\frac{p^2}{2m}+mc^2\right)^2 \chi + \left(\frac{p^2}{2m}+mc^2\right)\frac{p^2}{2m}\xi - \frac{p^2}{2m}\left(\frac{p^2}{2m}+mc^2\right)\xi - \left(\frac{p^2}{2m}\right)^2\chi = \left[\left(\frac{p^2}{2m}+mc^2\right)^2- \left(\frac{p^2}{2m}\right)^2\right]\chi = \left[p^2 c^2 + m^2c^4\right]\chi$.

Etc.

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+1 for the point about the redundancy of $m_0$ –  Benjamin Hodgson Oct 9 '12 at 19:52
    
Thanks .I got it now. –  Unlimited Dreamer Oct 10 '12 at 3:36

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