How can one obtain conserved charges from a symmetry transformation in QFT?

[center o bass]

Homework Statement

From the Lagrangian density

$$L = \frac{1}2 \partial_\mu \phi_a \partial^\mu \phi_a - \frac{1}2 \phi_a \phi_a,$$

where a = 1,2,3 and the transformation

$$\phi_a \to \phi _a + \theta \epsilon_{abc} n_b \phi_c$$

show that one gets the conserved charges

$$Q_a = \int d^3x \epsilon_{abc}\dot{\phi}_b \phi_c.$$

Homework Equations

The transformation is a symmetry of the Lagrangian so by Noethers theorem
we got a conserved current which is given by

$$j^\mu = \frac{\partial L}{\partial(\partial_\mu \phi_a)} \delta \phi_a = \partial^\mu \phi_a \epsilon_{abc} n_b \phi_c$$

The Attempt at a Solution

The obvious conserved charge is just

$$Q = \int d^3x j^0 = \int d^3 x \dot \phi_a \epsilon_{abc} nb \phi_c$$
but this is not the 3 different charges in the expression for Q_a. There is no normal vector n in that expresion and the time differentiated field has got b-index instead of an a index.
How can one get from the conserved current to the expression for these charges?
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[vela]

Your expression for the conserved current doesn't look correct. In the middle term, you sum over a, but in the expression on the right, a is a free index.

 

[center o bass]

Your expression for the conserved current doesn't look correct. In the middle term, you sum over a, but in the expression on the right, a is a free index.

I agree and I have corrected it now. That was just a typo. Do you have any suggestion on how to go from the corrected expression to the conserved charge?

 

[TSny]

$$Q_a = \int d^3x \epsilon_{abc}\dot{\phi}_a \phi_c.$$

Is there a typo here for the subscript on the left side?

The obvious conserved charge is just

$$Q = \int d^3x j^0 = \int d^3 x \dot \phi_a \epsilon_{abc} nb \phi_c$$
but this is not the 3 different charges in the expression for Q_a.

Is the Lagrangian invariant under the transformation for any choice of the vector n_b? If so, you should be able to get three independent conserved charges by choosing three independent vectors for n. For example, what would the conserved charge be if you choose n₁ = 1 and n₂ = n₃ = 0?

 

[center o bass]

Is there a typo here for the subscript on the left side?Is the Lagrangian invariant under the transformation for any choice of the vector n_b? If so, you should be able to get three independent conserved charges by choosing three independent vectors for n. For example, what would the conserved charge be if you choose n₁ = 1 and n₂ = n₃ = 0?

Yes there was yet another typo. I have corrected that one too now. That's true.
Hmm.. The conserved current satisfy

$$\partial_\mu (\partial^\mu \phi_a \epsilon_{abc} n_b \phi_c) = 0$$

I agree with you that one should be able to get 3 independent conserved charges (and also currents), so what if i chose all components to be zero except the a'th component?
I.e. choose

$$n_b = \delta _{ba}.$$

Would it then be correct of me to write

$$\partial_\mu (\partial^\mu \phi_a \epsilon_{abc} \delta_{ab} \phi_c) =\partial_\mu ( \epsilon_{abc} \partial^\mu \phi_b \phi_c)= 0$$

so that i get

$$(j^\mu)_a = \epsilon_{abc} \partial^\mu \phi_b \phi_c?$$

 

[TSny]

Yes there was yet another typo. I have corrected that one too now. That's true.
Hmm.. The conserved current satisfy

$$\partial_\mu (\partial^\mu \phi_a \epsilon_{abc} n_b \phi_c) = 0$$

I agree with you that one should be able to get 3 independent conserved charges (and also currents), so what if i chose all components to be zero except the a'th component?
I.e. choose

$$n_b = \delta _{ba}.$$

You don't want to use a subscript "a" here, since you are already using that symbol as a dummy summation index in j^μ.

Would it then be correct of me to write

$$\partial_\mu (\partial^\mu \phi_a \epsilon_{abc} \delta_{ab} \phi_c) =\partial_\mu ( \epsilon_{abc} \partial^\mu \phi_b \phi_c)= 0$$

See, the subscript "a" is occurring three times here on the left. That's not good.

You might try again by letting $$n_b = \delta _{bd}.$$ so that you're letting n_d be the nonzero component of n.

You can actually go to your result for Q in your original post: $$Q = \int d^3x j^0 = \int d^3 x \dot \phi_a \epsilon_{abc} nb \phi_c$$ and make this substitution rather than starting way back at the expression for j^μ.