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I Solution to the Dirac equation

  1. Mar 21, 2017 #1
    Hello!

    I have a question regarding the construction of solutions to the Diracequation for generell [itex] \vec{p} [/itex]. 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 [tex] (\gamma^{\mu}p_{\mu}+m)(\gamma^{\nu}p_{\nu}-m)=0 [/tex] But how does that help me? Why do I get the appropriate solution if I operate on the restfram-solution with the Diracoperator: [tex] u^{\alpha}(p)=\frac{1}{N}(\gamma^{\mu}p_{\mu}+m)u^{\alpha}(m,\vec{0})[/tex]
    Where [tex]
    u^{1}(m,\vec{0})=\left(\begin{array}{c}1\\0\\0\\0\end{array}\right) [/tex] and
    [tex] u^{2}(m,\vec{0})=\left(\begin{array}{c}0\\1\\0\\0\end{array}\right) [/tex]

    Thanks for your help!
     

    Attached Files:

  2. jcsd
  3. Mar 21, 2017 #2

    samalkhaiat

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    Let [itex]w(p)[/itex] be an arbitrary 4-component spinor. Now, using the dispersion relation [itex](p\!\!\!/ - m)(p\!\!\!/ + m) = 0[/itex] , you can easily show that [tex]u(p) = (p\!\!\!/ + m)w(p) ,[/tex] is a solution to the Dirac equation [itex](p\!\!\!/ - m)u(p) = 0[/itex] . Now take [tex]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} ,[/tex] where [itex]\chi^{(1)} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}[/itex], [itex]\chi^{(2)} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}[/itex] and [itex]0_{2} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}[/itex]. So, you have the following solutions [tex]u^{(\alpha)}(p) = \frac{1}{\sqrt{2m(E+m)}} \left( p\!\!\!/ + m \right) \begin{pmatrix} \chi^{(\alpha)} \\ 0_{2} \end{pmatrix} . \ \ \ \ (1)[/tex] In the Dirac representation, you have [tex]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} .[/tex] Substituting this in (1) and doing the matrix multiplication, we get [tex]u^{(\alpha)}(p) = \frac{1}{\sqrt{2m(E+m)}} \begin{pmatrix} (E+m)\chi^{(\alpha)} \\ (\vec{p} \cdot \vec{\sigma}) \chi^{(\alpha)} \end{pmatrix} .[/tex]
     
  4. Mar 22, 2017 #3
    Thank you for your reply. But why can you assume that by mutliplying the restframe spinor by [tex] \frac{1}{\sqrt{2m(E+m)}} [/tex] gives you a spinor w(p) with arbitrary p?
     
  5. Mar 22, 2017 #4

    samalkhaiat

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

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

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

    Now in Eq(2), the 4-momentum [itex]p[/itex] does not have to be the rest-frame 4-momentum [itex](m , \vec{0})[/itex], and [itex]N[/itex] is some constant that we can choose to make our equations look nice. For example, if we insist on the normalization [itex]\bar{u}_{\alpha}(p) u_{\beta}(p) = \delta_{\alpha \beta}[/itex], we find (and I leave you to prove it) that [tex]N = \frac{e^{i\eta}}{\sqrt{2m(E+m)}} .[/tex]
     
  6. Mar 23, 2017 #5
    I am perfectly aware of what the word 'arbitrary' means und what a normalization constant is. Anyway, thanks for your time.
     
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