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Chiral currents of a Dirac plane wave

by Slereah
Tags: chiral, currents, dirac, plane, wave
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Slereah
#1
Mar1-12, 02:04 PM
P: 7
I am currently trying to check the formula for the chiral current of the Dirac equation for a plane wave solution (found here ), that is,

[itex]j_{R}^\mu = \psi_R^\dagger \sigma^\mu \psi_R[/itex]

[itex]j_{L}^\mu = \psi_L^\dagger \sigma^\mu \psi_L[/itex]

With

[itex] \psi_R = I( \cosh(\frac{\theta}{2}) + \sigma^i n_i \sinh(\frac{\theta}{2})) \xi e^{ix^\mu p_\mu} [/itex]

[itex]\psi_R = I( \cosh(\frac{\theta}{2}) - \sigma^i n_i \sinh(\frac{\theta}{2})) \xi e^{ix^\mu p_\mu}[/itex]

χ being the rest frame spinor, θ the rapidity and n the direction of the momentum.

So for the right-handed current, I get

[itex]\psi_R^\dagger \sigma^\mu \psi_R = ([\cosh(\frac{\theta}{2}) + \vec{\sigma} \cdot \vec{n} \sinh(\frac{\theta}{2}) ] \xi)^\dagger \sigma^\mu [\cosh(\frac{\theta}{2}) + \vec{\sigma} \cdot \vec{n} \sinh(\frac{\theta}{2}) ] \xi [/itex]

[itex]\psi_R^\dagger \sigma^\mu \psi_R = \cosh^2(\frac{\theta}{2}) \xi^\dagger \sigma^\mu \xi + \cosh(\frac{\theta}{2}) \sinh(\frac{\theta}{2}) \xi^\dagger( \sigma^\mu (\vec{\sigma} \cdot \vec{n}) + (\vec{\sigma} \cdot \vec{n}) \sigma^\mu ) \xi + \sinh^2(\frac{\theta}{2}) \xi^\dagger (\vec{\sigma} \cdot \vec{n}) \sigma^\mu (\vec{\sigma} \cdot \vec{n}) \xi [/itex]

Using the following hyperbolic identities,

[itex]\sinh^2(\frac{\theta}{2}) = \frac{1}{2} [\cosh(\theta) - 1] = \frac{1}{2} [\gamma - 1] [/itex]

[itex]\cosh^2(\frac{\theta}{2}) = \frac{1}{2} [\cosh(\theta) + 1] = \frac{1}{2} [\gamma + 1] [/itex]

[itex]\cosh(\frac{\theta}{2}) \sinh(\frac{\theta}{2}) = \frac{1}{2} \sinh(\theta) = \frac{1}{2} \beta \gamma[/itex]

And defining the spin vector

[itex]\xi^\dagger \sigma^\mu \xi = s^\mu = (1,\vec{s}) [/itex]

I get the following :

[itex] \psi_R^\dagger \sigma^\mu \psi_R = \frac{1}{2} [ \gamma \xi^\dagger (\sigma^\mu + \sigma^i \sigma^\mu \sigma_i ) \xi + \xi^\dagger (\sigma^\mu -\sigma^i \sigma^\mu \sigma_i ) \xi + \beta_i \gamma \xi^\dagger \{\sigma^\mu, \sigma^i\} \xi ] [/itex]


[itex]\beta_i [/itex] being [itex]\beta n_i [/itex], and [itex]\sigma^i \sigma^\mu \sigma_i [/itex] coming from [itex](\vec{\sigma} \cdot \vec{n}) \sigma^\mu (\vec{\sigma} \cdot \vec{n})[/itex] (I can't find the exact proof to show it to be so, but I think this is correct - though it may be the problem). So far this is pretty much the results indicated. But then, if I try finding out the results using this identity :

[itex]\sigma^i \sigma^\mu \sigma_i = \bar{\sigma}^\mu[/itex]

I finally get this formula :


[itex] \psi_R^\dagger \sigma^\mu \psi_R = \frac{1}{2} [ \gamma \xi^\dagger (\sigma^\mu + \bar{\sigma}^\mu) \xi + \xi^\dagger (\sigma^\mu -\bar{\sigma}^\mu ) \xi + \beta_i \gamma \xi^\dagger \{\sigma^\mu, \sigma^i\} \xi ] [/itex]

Which gives the following components for the current :

[itex] \psi_R^\dagger \sigma^0 \psi_R = \gamma \xi^\dagger \xi + \beta_i \gamma \xi^\dagger \sigma^i \xi = \gamma + \gamma \beta_i s^i[/itex]

Which is the correct result, but for the spatial components I find

[itex] \psi_R^\dagger \sigma^j \psi_R = \xi^\dagger \sigma^j \xi + \beta^j \gamma \xi^\dagger \xi = s^j + \beta^j \gamma[/itex]

Which according to the book lacks a term in [itex]+ (\gamma-1) \beta^i s_i \beta^j [/itex], unless I am misinterpreting the notation used (the results are page 16).
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