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Can we treat $\psi^{c}$ as a field independent from $\psi$?

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When we derive the Dirac equation from the Lagrangian, $$ \mathcal{L}=\overline{\psi}i\gamma^{\mu}\partial_{\mu}\psi-m\overline{\psi}\psi, $$ we assume $\psi$ and $\overline{\psi}=\psi^{*^{T}}\gamma^{0}$ are independent. So when we take the derivative of the Lagrangian with respect to $\overline{\psi}$, we get the Dirac equation $$ 0=\partial_{\mu}\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\overline{\psi}\right)}=\frac{\partial\mathcal{L}}{\partial\overline{\psi}}=\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi. $$

Now if we include a term with charge conjugation, $\psi^{c}=-i\gamma^{2}\psi^{*}$, into the Lagrangian (like $\Delta\mathcal{L}=\overline{\psi}\psi^{c}$), does this $\psi^c$ depend on $\overline{\psi}$ or $\psi$? Why or why not?

If $\psi^{c}$ depends on $\psi$, why wouldn't the reason that $\overline{\psi}$ and $\psi$ are independent apply for $\psi^{c}$ and $\psi$?

If $\psi^{c}$ depends on $\overline{\psi}$, how should we take derivative of $\Delta\mathcal{L}$ with respect to $\overline{\psi}$?

This post imported from StackExchange Physics at 2014-05-04 11:36 (UCT), posted by SE-user Louis Yang
asked Apr 26, 2014 in Theoretical Physics by Louis Yang (90 points) [ no revision ]
Possible related? physics.stackexchange.com/q/89002/29216

This post imported from StackExchange Physics at 2014-05-04 11:36 (UCT), posted by SE-user BMS

2 Answers

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The Dirac spinor $\psi$ and its complex conjugate $\psi^*$ are not independent variables, but in some calculations one can treat them as such.

For the similar question about a complex scalar field $\phi$ and its complex conjugate $\phi^*$, see e.g. this Phys.SE post.

This post imported from StackExchange Physics at 2014-05-04 11:36 (UCT), posted by SE-user Qmechanic
answered Apr 26, 2014 by Qmechanic (2,790 points) [ no revision ]
Thanks for your brief answer. I am confused. For a complex variable $z$ one can always write it as real and imaginary parts $z=x+iy$. If you compute $\frac{\partial z^{*}}{\partial z}$ or $\frac{\partial z}{\partial z^{*}}$, they are both zero. So this is the same reason why one should take $\frac{\partial\phi^{*}}{\partial\phi}=0$, right?

This post imported from StackExchange Physics at 2014-05-04 11:36 (UCT), posted by SE-user Louis Yang
Recalling the precise definition of $\frac{\partial z^{*}}{\partial z}=0=\frac{\partial z}{\partial z^{*}}$, it does not necessarily imply that $z$ and $z^{*}$ are independent variables. On one hand, if $z^{*}$ denotes the complex conjugate of $z$ (so that $z$ and $z^{*}$ are not independent variables), then $\frac{\partial z^{*}}{\partial z}=0=\frac{\partial z}{\partial z^{*}}$ are merely consequences of pertinent definitions. On the other hand, if $z$ and $z^{*}$ are truly indep. complex variables, then $\frac{\partial z^{*}}{\partial z}=0=\frac{\partial z}{\partial z^{*}}$ is automatic.

This post imported from StackExchange Physics at 2014-05-04 11:36 (UCT), posted by SE-user Qmechanic
One can always write $x=\frac{z+z^{*}}{2}$ and $y=\frac{z-z^{*}}{2i}$. Then one can express the derivative as $\partial_{z}=\frac{\partial x}{\partial z}\partial_{x}+\frac{\partial y}{\partial z}\partial_{y}=\frac{\partial_{x}-i\partial_{y}}{2} $ So one get $\frac{\partial z^{*}}{\partial z}=\frac{\partial z}{\partial z^{*}}=0$. Maybe "independent" is a not a good word to describe it, but at least it is the derivative that enter the derivation of Euler-Lagrange equation.

This post imported from StackExchange Physics at 2014-05-04 11:36 (UCT), posted by SE-user Louis Yang
+ 0 like - 0 dislike

Yes, when we want to obtain the equation of motion using Euler-Lagrange equation, we should treat $\psi$ and $\psi^c$ independent, but $\overline{\psi}$ and $\psi^c$ dependent. The reason for this is that we can simply expressed $\psi^c$ in terms of $\overline{\psi}$ by $$ \psi^{c}=C\overline{\psi}^{T}, $$ where $C=-i\gamma^{2}\gamma^{0}$ is the charge conjugation matrix. So $\overline{\psi}$ and $\psi^c$ are the same degree of freedom.

For the derivative of $\overline{\psi}\psi^{c}$ with respect to $\overline{\psi}$, one should be really careful because $\psi$ is anticommuting. Since the derivative in Euler-Lagrange equation actually comes from the variation of Lagrangian, We should start from the variation \begin{eqnarray} \delta\left(\overline{\psi}\psi^{c}\right) & = & \delta\left(\overline{\psi}C\overline{\psi}^{T}\right)=\delta\left(\overline{\psi_{i}}C_{ij}\overline{\psi_{j}}\right)=\delta\left(\overline{\psi_{i}}\right)C_{ij}\overline{\psi_{j}}+\overline{\psi_{i}}C_{ij}\delta\overline{\psi_{j}}\\ & = & \delta\left(\overline{\psi_{i}}\right)C_{ij}\overline{\psi_{j}}-\delta\left(\overline{\psi_{j}}\right)C_{ij}\overline{\psi_{i}}, \end{eqnarray} where I use the anticommutation of the fields to get the minus sign for the last step. Now notice that $C^{T}=C^{+}=-C$. So the last term is \begin{equation} -\delta\left(\overline{\psi_{j}}\right)C_{ij}\overline{\psi_{i}}=\delta\left(\overline{\psi_{j}}\right)C_{ji}\overline{\psi_{i}}=\delta\left(\overline{\psi_{i}}\right)C_{ij}\overline{\psi_{j}}. \end{equation} and we get $\delta\left(\overline{\psi}\psi^{c}\right)=2\delta\left(\overline{\psi}\right)C\overline{\psi}^{T}.$ Therefore, the equation of motion from this term is \begin{equation} \frac{\partial}{\partial\overline{\psi}}\left(\overline{\psi}\psi^{c}\right)=2C\overline{\psi}^{T}=2\psi^{c}. \end{equation}

This post imported from StackExchange Physics at 2014-05-04 11:36 (UCT), posted by SE-user Louis Yang
answered Apr 27, 2014 by Louis Yang (90 points) [ no revision ]

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