- #1

JD_PM

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- Homework Statement
- For a Dirac field, the transformations

$$\psi(x) \rightarrow \psi'(x) = e^{i \alpha \gamma_5} \psi(x), \ \ \ \ \ \ \ \ \psi^{\dagger} (x) \rightarrow \psi^{\dagger'}(x) = e^{-i \alpha \gamma_5} \psi(x)$$

where ##\alpha## is an arbitrary real parameter, are called chiral phase transformations.

Given the Dirac-Lagrangian density (and the definition of ##\bar{\psi}##):

$$\mathcal{L} = c \bar{\psi}\Big( i \hbar \gamma^{\mu} \partial_{\mu} - mc\Big) \psi, \ \ \ \ \ \ \ \ \ \bar{\psi} = \psi^{\dagger} \gamma^0$$

Where ##\gamma^0## is a type of Dirac matrix.

a) Show that the current associated to ##\mathcal{L}## in the limit ##m = 0## is ##j^{\mu} = \bar{\psi} \gamma^{\mu} \gamma_5 \psi##

b) Show that the equations of motion related to ##\mathcal{L}## (and ##m \neq 0##) are

$$\psi_L = \frac 1 2 (1 - \gamma_5) \psi, \ \ \ \ \ \ \psi_R = \frac 1 2 (1 + \gamma_5) \psi$$

And also show that they decouple in the limit ##m = 0##

c) After solving a) b) you have all you need to show and explain that the ##m=0## Dirac-Lagrangian density ##\mathcal{L} = i \hbar c \bar{\psi_L}\gamma^{\mu} \partial_{\mu} \psi_L## describes zero-mass fermions with negative helicity only, and zero-mass antifermions with positive helicity only. (This field is called the Weyl field and can be used to describe the neutrinos in weak interactions in the approximation of zero mass).

Source: Exercise 4.5, QFT by Mandl and Shaw.

- Relevant Equations
- $$j^{\mu}_{\nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi_a )} X_a (\phi) - F^{\mu} (\phi), \ \ \ \ \ \ \ \ \delta \mathcal{L} = \partial_{\mu} F^{\mu}; \ X_a (\phi) = \delta \phi_a$$

$$\partial_{\mu} \Big( \frac{\mathcal{L}}{\partial(\partial_{\mu} \psi)} \Big) - \frac{\partial \mathcal{L}}{\partial \psi} = 0 \ \ \ \ (1)$$

$$\partial_{\mu} \Big( \frac{\mathcal{L}}{\partial(\partial_{\mu} \psi^{\dagger})} \Big) - \frac{\partial \mathcal{L}}{\partial \psi^{\dagger}} = 0 \ \ \ \ (2)$$

a) Here I used the definition of current (reference: Tong notes, page 14)

$$j^{\mu}_{\nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi_a )} X_a (\phi) - F^{\mu} (\phi), \ \ \ \ \ \ \ \ \delta \mathcal{L} = \partial_{\mu} F^{\mu}; \ X_a (\phi) = \delta \phi_a$$

In our problem we get (assuming ##\delta \mathcal{L} = 0##):

$$j^{\mu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi^{\dagger})} \delta \psi^{\dagger} + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi)} \delta \psi$$

We know that ##\mathcal{L} = i \hbar c \bar{\psi}\gamma^{\mu} \partial_{\mu} \psi## , ##\delta \psi^{\dagger} = i \alpha \gamma_5 \psi## and ##\delta \psi^{\dagger} = -i \alpha \gamma_5 \psi^{\dagger}##

On the one hand, as ##\mathcal{L}## has no explicit dependence on ##\partial_{\mu} \psi^{\dagger}## we get

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi^{\dagger})} = 0$$

On the other hand we get

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi)} = i c \hbar \bar{\psi} \gamma^{\mu}$$

Thus the current I get is

$$j^{\mu} = i c \hbar \bar{\psi} \gamma^{\mu} \delta \psi = - c \alpha \hbar \bar{\psi} \gamma^{\mu} \gamma_5 \psi$$

The provided current is

$$j^{\mu} = \bar{\psi} \gamma^{\mu} \gamma_5 \psi$$

As you can see, my current and the provided one differ; I get a constant factor ##k := -c \alpha \hbar## Is that OK?b) Here it is all about using E-L equations

$$\partial_{\mu} \Big( \frac{\mathcal{L}}{\partial(\partial_{\mu} \psi)} \Big) - \frac{\partial \mathcal{L}}{\partial \psi} = 0 \ \ \ \ (1)$$

$$\partial_{\mu} \Big( \frac{\mathcal{L}}{\partial(\partial_{\mu} \psi^{\dagger})} \Big) - \frac{\partial \mathcal{L}}{\partial \psi^{\dagger}} = 0 \ \ \ \ (2)$$

We will get two equations of motion, as expected.

Out of ##(1)## we get:

$$i \hbar c \bar{\psi} \partial_{\mu} \gamma^{\mu} - \bar{\psi} mc^2 = 0$$

Out of ##(2)## we get (note I have used the definition ##\bar{\psi} = \psi^{\dagger} \gamma^0##):

$$-i \hbar c \gamma^0 \gamma^{\mu} \partial_{\mu} \psi + c^2 \gamma^0 m \psi = 0$$

But these are not the provided solutions.

Are my solutions OK? Is there a way of going from these solutions to the provided ones?

c) As suggested, let's first get a) and b) done.

$$j^{\mu}_{\nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi_a )} X_a (\phi) - F^{\mu} (\phi), \ \ \ \ \ \ \ \ \delta \mathcal{L} = \partial_{\mu} F^{\mu}; \ X_a (\phi) = \delta \phi_a$$

In our problem we get (assuming ##\delta \mathcal{L} = 0##):

$$j^{\mu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi^{\dagger})} \delta \psi^{\dagger} + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi)} \delta \psi$$

We know that ##\mathcal{L} = i \hbar c \bar{\psi}\gamma^{\mu} \partial_{\mu} \psi## , ##\delta \psi^{\dagger} = i \alpha \gamma_5 \psi## and ##\delta \psi^{\dagger} = -i \alpha \gamma_5 \psi^{\dagger}##

On the one hand, as ##\mathcal{L}## has no explicit dependence on ##\partial_{\mu} \psi^{\dagger}## we get

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi^{\dagger})} = 0$$

On the other hand we get

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi)} = i c \hbar \bar{\psi} \gamma^{\mu}$$

Thus the current I get is

$$j^{\mu} = i c \hbar \bar{\psi} \gamma^{\mu} \delta \psi = - c \alpha \hbar \bar{\psi} \gamma^{\mu} \gamma_5 \psi$$

The provided current is

$$j^{\mu} = \bar{\psi} \gamma^{\mu} \gamma_5 \psi$$

As you can see, my current and the provided one differ; I get a constant factor ##k := -c \alpha \hbar## Is that OK?b) Here it is all about using E-L equations

$$\partial_{\mu} \Big( \frac{\mathcal{L}}{\partial(\partial_{\mu} \psi)} \Big) - \frac{\partial \mathcal{L}}{\partial \psi} = 0 \ \ \ \ (1)$$

$$\partial_{\mu} \Big( \frac{\mathcal{L}}{\partial(\partial_{\mu} \psi^{\dagger})} \Big) - \frac{\partial \mathcal{L}}{\partial \psi^{\dagger}} = 0 \ \ \ \ (2)$$

We will get two equations of motion, as expected.

Out of ##(1)## we get:

$$i \hbar c \bar{\psi} \partial_{\mu} \gamma^{\mu} - \bar{\psi} mc^2 = 0$$

Out of ##(2)## we get (note I have used the definition ##\bar{\psi} = \psi^{\dagger} \gamma^0##):

$$-i \hbar c \gamma^0 \gamma^{\mu} \partial_{\mu} \psi + c^2 \gamma^0 m \psi = 0$$

But these are not the provided solutions.

Are my solutions OK? Is there a way of going from these solutions to the provided ones?

c) As suggested, let's first get a) and b) done.