- #1

peguerosdc

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- TL;DR Summary
- How to motivate positive and negative energy solutions as a superposition of two solutions?

Hi!

I am studying Dirac's equation and I already have understood the derivation. Following Griffiths, from factoring Einstein's energy relation with the gamma matrices:

##

(\gamma^\mu p_\mu + m)(\gamma^\mu p_\mu - m) = 0

##

You take any of the two factors, apply quantization and you arrive to Dirac's equation. Griffiths takes for example the second one:

##

(\gamma^\mu p_\mu - m) \varphi = 0

##

Comparing this with Srednicki's, they look the same just that he renames ## \gamma^\mu p_\mu = i\not\!\partial ##, inverts signs and arrives to this expression for Dirac's equation (equation 37.23):

##

(-i\not\!\partial + m) \varphi = 0

##

Now, Srednicki motivates finding two solutions (one with +E and one with -E) by showing that the wave function must obey the Klein-Gordon equation, so he proposes the solution as (eq 37.27):

##

\varphi (x) = u(\mathbf p) e^{ipx} + v(\mathbf p) e^{-ipx}

##

where u is the positive energy solution and v the negative energy solution.

So far, so good. Now, where I need help is understanding the next step. He says:

Thanks!

I am studying Dirac's equation and I already have understood the derivation. Following Griffiths, from factoring Einstein's energy relation with the gamma matrices:

##

(\gamma^\mu p_\mu + m)(\gamma^\mu p_\mu - m) = 0

##

You take any of the two factors, apply quantization and you arrive to Dirac's equation. Griffiths takes for example the second one:

##

(\gamma^\mu p_\mu - m) \varphi = 0

##

Comparing this with Srednicki's, they look the same just that he renames ## \gamma^\mu p_\mu = i\not\!\partial ##, inverts signs and arrives to this expression for Dirac's equation (equation 37.23):

##

(-i\not\!\partial + m) \varphi = 0

##

Now, Srednicki motivates finding two solutions (one with +E and one with -E) by showing that the wave function must obey the Klein-Gordon equation, so he proposes the solution as (eq 37.27):

##

\varphi (x) = u(\mathbf p) e^{ipx} + v(\mathbf p) e^{-ipx}

##

where u is the positive energy solution and v the negative energy solution.

So far, so good. Now, where I need help is understanding the next step. He says:

I don't understand why plugging eq. 37.27 into 37.23 yields equation 37.28. It looks like we are taking one different factor for each solution (that's why I introduced my question with Griffiths' derivation), but I don't understand the reasoning behind it.Plugging eq. (37.27) into the eq. (37.23) we get eq (37.28):

$$

(\not\!p + m)u(\mathbf p) e^{ipx} + (-\not\!p + m)v(\mathbf p) e^{-ipx} = 0

$$

Thus we require (37.29):

$$

(\not\! p + m)u(\mathbf p) = 0

\qquad

(-\not\!p + m)v(\mathbf p) = 0

$$

Thanks!