Trick for conserved currents in classical field theory

[doggydan42]

Homework Statement

Use the trick for the lagrangian to calculate the conserved current associated with a rotation.

Relevant Equations

$$\mathcal{L} = \partial_\mu \psi^* \partial^\mu \psi - V(\psi^* \psi)$$
$$\delta \mathcal{L} = \partial_\mu(\alpha(x)) j^\mu$$

First I found the equations of motion for both fields:
$$\partial_\mu \partial^\mu \psi = -\frac{\partial V(\psi^* \psi)}{\psi^*}$$
The eq. of motion with the other field is simply found by $\psi \rightarrow \psi^*$ and $\psi^* \rightarrow \psi$ due to the symmetry between the two fields.

The transformation is with $\delta \psi = i\alpha \psi$ and $\delta \psi^* = -i\alpha \psi^*$. The trick has to do with letting alpha be a function and writing
$$\delta \mathcal{L} = \partial_\mu(\alpha(x)) h^\mu$$, where $h^\mu$ is the conserved current.

I got $$\delta \mathcal{L} = i(\partial^\mu \psi^*)\psi(\partial_\mu \alpha) + i(\partial^\mu \psi^*)(\partial_\mu\psi)\alpha - i(\partial^\mu \psi)\psi^*(\partial_\mu \alpha) - i(\partial^\mu \psi)(\partial_\mu\psi^*)\alpha + (\partial_\mu\partial^\mu\psi^*)i\alpha\psi - (\partial_\mu\partial^\mu\psi)i\alpha\psi^*$$

Using $(\partial_\mu\partial^\mu\psi^*)i\alpha\psi - (\partial_\mu\partial^\mu\psi)i\alpha\psi^* = i\alpha \partial_\mu(\psi\partial^\mu\psi^* - \psi^*\partial^\mu\psi) = \alpha \partial_\mu(h^\mu)$ defining $h^\mu$. This gives

$$\delta \mathcal{L} = (\partial_\mu \alpha) h^\mu - \alpha \partial_\mu h^\mu$$.

Since I already know that $h^\mu$ is the conserved quantity, I know this will give $\delta \mathcal{L} = (\partial_\mu \alpha) h^\mu$
So how would I show that $\partial_\mu h^\mu = 0$ without assuming it to be the conserved quantity? I also feel like I'm not using the trick properly, as it is supposed to be a trick to make it easier to solve.

 

[mark726]

To show that $\partial_\mu h^\mu = 0$ without assuming it to be the conserved quantity, we can use the fact that the equations of motion for both fields are given by
$$\partial_\mu \partial^\mu \psi = -\frac{\partial V(\psi^* \psi)}{\psi^*}$$
and
$$\partial_\mu \partial^\mu \psi^* = -\frac{\partial V(\psi^* \psi)}{\psi}.$$
Substitute these equations into the expression for $\delta \mathcal{L}$ and use the fact that $\delta \mathcal{L} = \partial_\mu(\alpha(x)) h^\mu$, we get
$$\delta \mathcal{L} = i(\partial^\mu \psi^*)\psi(\partial_\mu \alpha) + i(\partial^\mu \psi^*)(\partial_\mu\psi)\alpha - i(\partial^\mu \psi)\psi^*(\partial_\mu \alpha) - i(\partial^\mu \psi)(\partial_\mu\psi^*)\alpha + \partial_\mu(\alpha(x)) \partial^\mu(\psi\partial_\mu \psi^* - \psi^*\partial_\mu \psi).$$
Using the product rule for derivatives, we can write
$$\partial_\mu(\psi\partial^\mu \psi^* - \psi^*\partial^\mu \psi) = (\partial_\mu\psi)(\partial^\mu \psi^*) + \psi\partial_\mu(\partial^\mu \psi^*) - \psi^*\partial_\mu(\partial^\mu \psi).$$
Substituting this into the expression for $\delta \mathcal{L}$, we get
$$\delta \mathcal{L} = i(\partial^\mu \psi^*)\psi(\partial_\mu \alpha) + i(\partial^\mu \psi^*)(\partial_\mu\psi)\alpha - i(\partial^\mu \psi)\psi^*(\partial_\mu \alpha) - i(\partial^\mu \psi)(\partial_\mu\psi^*)\alpha + \partial_\mu(\alpha(x)) ((\partial_\mu\psi)(\partial^\mu \psi^*) + \psi\partial_\