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Conserved current; Klein-Gordon

  • Thread starter Landau
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Landau
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Homework Statement



Given the Lagrangian density of a complex relativistic scalar field
[itex]\mathcal L=\frac{1}{2}\partial^\nu\phi^{*}\partial_\nu\phi-\frac{1}{2}m^2\phi^{*}\phi[/itex]
where * stands for complex conjugation, compute the conserved current (using Noether's theorem).

Homework Equations



I can (should) use [itex]J^k=-\frac{\partial \mathcal L}{\partial (\partial_k \phi_I)}\phi_I[/itex], where summation over I is implied (which just means we have two terms, one for [itex]\phi[/itex] and one for [itex]\phi^{*}[/itex]).

The Attempt at a Solution



Well, I computed [itex]\frac{\partial \mathcal L}{\partial (\partial_k \phi)}=\partial^k\phi^{*}[/itex], and similarly [itex]\frac{\partial \mathcal L}{\partial (\partial_k \phi^{*})}=\partial^k\phi[/itex].

Combining these, we simply get [itex]J^k=-\frac{\partial \mathcal L}{\partial (\partial_k \phi_I)}\phi_I=-\frac{\partial \mathcal L}{\partial (\partial_k \phi)}\phi-\frac{\partial \mathcal L}{\partial (\partial_k \phi^*)}\phi^*=-\frac{1}{2}\left(\partial^k\phi^{*}\phi+\partial^k\phi\phi^{*}\right)[/itex]

But the correct answer should be [itex]=-\frac{1}{2}\left(\partial^k\phi^{*}\phi-\partial^k\phi\phi^{*}\right)[/itex], differing in a minus sign.

(Just looking at the Lagrangian, [itex]\phi[/itex] and [itex]\phi^*[/itex] are symmetric, right? So the terms in the conserved current should also come in symmetric...but in the correct answer they aren't.)

Apparently [itex]\frac{\partial \mathcal L}{\partial (\partial_k \phi^{*})}=-\partial^k\phi[/itex]?
 
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Answers and Replies

  • #2
George Jones
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Apparently [itex]\frac{\partial \mathcal L}{\partial (\partial_k \phi^{*})}=-\partial^k\phi[/itex]?
No.

While [itex]\phi[/itex] and [itex]\phi^*[/itex] appear symmetrically in the Lagrangian, they don't come into play symmetrically in the application of Noether's theorem to a Klein-Gordon system.

The Lagrangian is invariant under the continuous transformation [itex]\phi \rightarrow e^{i\alpha} \phi[/itex], which means [itex]\phi^* \rightarrow e^{-i\alpha} \phi^*[/itex].
 

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