- #1

doggydan42

- 170

- 18

- 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.

$$\partial_\mu \partial^\mu \psi = -\frac{\partial V(\psi^* \psi)}{\psi^*}$$

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.