# I Solution to the Dirac equation

1. Mar 21, 2017

### 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 $$u^{1}(m,\vec{0})=\left(\begin{array}{c}1\\0\\0\\0\end{array}\right)$$ and
$$u^{2}(m,\vec{0})=\left(\begin{array}{c}0\\1\\0\\0\end{array}\right)$$

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2. Mar 21, 2017

### 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 $$p\!\!\!/ + m = E \gamma^{0} - \vec{p} \cdot \vec{\gamma} + m I_{4} = \begin{pmatrix} (E+m)I_{2} & - \vec{p} \cdot \vec{\sigma} \\ \vec{p} \cdot \vec{\sigma} & - (E+m)I_{2} \end{pmatrix} .$$ Substituting this in (1) and doing the matrix multiplication, we get $$u^{(\alpha)}(p) = \frac{1}{\sqrt{2m(E+m)}} \begin{pmatrix} (E+m)\chi^{(\alpha)} \\ (\vec{p} \cdot \vec{\sigma}) \chi^{(\alpha)} \end{pmatrix} .$$

3. Mar 22, 2017

### 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?

4. Mar 22, 2017

### samalkhaiat

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 $$N = \frac{e^{i\eta}}{\sqrt{2m(E+m)}} .$$

5. Mar 23, 2017

### klabautermann

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