I Solution to the Dirac equation

[klabautermann]

Hello!

I have a question regarding the construction of solutions to the Diracequation for generell \vec{p}. In my lecturenotes (and also in Itzykson/Zuber) it is stated that it is easier than boosting the restframe-solutions, to construct them by using (\gamma^{\mu}p_{\mu}+m)(\gamma^{\nu}p_{\nu}-m)=0 But how does that help me? Why do I get the appropriate solution if I operate on the restfram-solution with the Diracoperator: u^{\alpha}(p)=\frac{1}{N}(\gamma^{\mu}p_{\mu}+m)u^{\alpha}(m,\vec{0})
Where <br /> u^{1}(m,\vec{0})=\left(\begin{array}{c}1\\0\\0\\0\end{array}\right) and
u^{2}(m,\vec{0})=\left(\begin{array}{c}0\\1\\0\\0\end{array}\right)

Thanks for your help!

 

[samalkhaiat]

Let w(p) be an arbitrary 4-component spinor. Now, using the dispersion relation (p\!\!\!/ - m)(p\!\!\!/ + m) = 0 , you can easily show that u(p) = (p\!\!\!/ + m)w(p) , is a solution to the Dirac equation (p\!\!\!/ - m)u(p) = 0 . Now take w(p) = \frac{1}{\sqrt{2m(E+m)}} u^{(\alpha)}(m,\vec{0}) \equiv \frac{1}{\sqrt{2m(E+m)}} \begin{pmatrix} \chi^{(\alpha)} \\ 0_{2} \end{pmatrix} , where \chi^{(1)} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \chi^{(2)} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} and 0_{2} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. So, you have the following solutions u^{(\alpha)}(p) = \frac{1}{\sqrt{2m(E+m)}} \left( p\!\!\!/ + m \right) \begin{pmatrix} \chi^{(\alpha)} \\ 0_{2} \end{pmatrix} . \ \ \ \ (1) In the Dirac representation, you have p\!\!\!/ + m = E \gamma^{0} - \vec{p} \cdot \vec{\gamma} + m I_{4} = \begin{pmatrix} (E+m)I_{2} &amp; - \vec{p} \cdot \vec{\sigma} \\ \vec{p} \cdot \vec{\sigma} &amp; - (E+m)I_{2} \end{pmatrix} . Substituting this in (1) and doing the matrix multiplication, we get u^{(\alpha)}(p) = \frac{1}{\sqrt{2m(E+m)}} \begin{pmatrix} (E+m)\chi^{(\alpha)} \\ (\vec{p} \cdot \vec{\sigma}) \chi^{(\alpha)} \end{pmatrix} .

 

[klabautermann]

Thank you for your reply. But why can you assume that by mutliplying the restframe spinor by \frac{1}{\sqrt{2m(E+m)}} gives you a spinor w(p) with arbitrary p?

 

[samalkhaiat]

you assume that by mutliplying the restframe spinor by \frac{1}{\sqrt{2m(E+m)}} gives you a spinor w(p) with arbitrary p?

No, I did not assume such thing because it is not correct: Multiplying u_{\alpha}(m,\vec{0}) by a constant does not turn it into a spinor w(p) with arbitrary p, because \psi and c\psi represent the same spinor. Okay, let me repeat what I did, and please pay attention to my logic.

I said: let w(p) be any (completely arbitrary) 4-component spinor. This statement means that we are free to choose w(p) to be any spinor we like.

Then, I used the dispersion relation and concluded that u(p) = (p\!\!\!/ + m) w(p) , \ \ \ \ \ \ \ \ \ \ \ (1) solves the Dirac’s equation (p\!\!\!/ - m)u(p) = 0 . Now, because w (in Eq(1)) is arbitrary, we can choose it to be the rest-frame spinor u_{\beta}(m,\vec{0}) = \begin{pmatrix} \chi_{\beta} \\ 0 \end{pmatrix}. After all, at this time, u_{\beta}(m,\vec{0}) is the only spinor we have in our pocket. So in Eq(1), instead of w(p), I substituted the rest-frame spinor N u_{\beta}(m,\vec{0}) and obtained the solutions u_{\beta}(p) = N \begin{pmatrix} (E + m)\chi_{\beta} \\ (\vec{p} \cdot \vec{\sigma}) \chi_{\beta} \end{pmatrix} . \ \ \ \ \ (2)

Now in Eq(2), the 4-momentum p does not have to be the rest-frame 4-momentum (m , \vec{0}), and N is some constant that we can choose to make our equations look nice. For example, if we insist on the normalization \bar{u}_{\alpha}(p) u_{\beta}(p) = \delta_{\alpha \beta}, we find (and I leave you to prove it) that N = \frac{e^{i\eta}}{\sqrt{2m(E+m)}} .

 

[klabautermann]

I am perfectly aware of what the word 'arbitrary' means und what a normalization constant is. Anyway, thanks for your time.