In the first SuSy lecture last week following theory of two real scalar fields has been considered as a first example:

$$\mathcal{L}=(\partial_\mu \phi_1)^2/2+(\partial_\mu \phi_2)^2/2-m^2(\phi_1^2+\phi_2^2)/2$$

which has the usual equation of motion $$(\Box+m^2)\phi_i=0;\quad i=1,2$$ Then the professor has written down following conserved currents for this theory:


Using the equation of motion, it is easy to show that these currents are indeed conserved for all $n$.

As far as I can see, these are no Noether currents, at least I cannot see how to get something with arbitrary many indices using the standard formula for Noether currents; furthermore I don't see which symmetry of the Lagrangian they could reflect.

My questions now are:

  1. Are there really no symmetries corresponding to these currents or do I overlook something? If there are no such symmetries and the formula for Noether's currents doesn't apply, is there any way to get such currents or is it rather an educated guess?

  2. Is there any way to intuitively visualize such currents with many indices? Can anything be said about their physical meaning (if there is some)?

  • $\begingroup$ Could you provide some additional clarification? By 'conserved currents that are not Noether currents,' do you mean conserved quantities which do not correspond to symmetries of the action? $\endgroup$
    – JamalS
    Commented Apr 13, 2014 at 18:17
  • $\begingroup$ OP wrote (v2): It is easy to show that these currents are indeed conserved for all $n$. Do you mean wrt. the first index only? $\endgroup$
    – Qmechanic
    Commented Apr 13, 2014 at 18:30
  • $\begingroup$ JamalS: That's basically the (first) question: The given currents are conserved and I don't see whether they correspond to any symmetry. $\endgroup$
    – Photon
    Commented Apr 13, 2014 at 18:43
  • $\begingroup$ Qmechanic: Exactly, "for all n" is meant as in "for all currents". The divergence is to be taken w.r.t the first index only in all currents. $\endgroup$
    – Photon
    Commented Apr 13, 2014 at 18:44

2 Answers 2


The Lagrangian density reads

$$ \tag{1} {\cal L}~=~\frac{1}{2}\sum_{i=1}^2\left( \partial_{\mu}\phi^i \partial^{\mu}\phi^i -m^2\phi^i\phi^i\right). $$

Consider an infinitesimal transformation

$$\tag{2} \delta \phi^i~=~\epsilon Y^i_{\nu_2\ldots \nu_n},$$

with generators

$$\tag{3} Y^1_{\nu_2\ldots \nu_n}~:=~ -(-\partial_{\nu_2})\cdots(-\partial_{\nu_n})\phi^2, $$


$$\tag{4} Y^2_{\nu_2\ldots \nu_n}~:=~ \partial_{\nu_2}\cdots\partial_{\nu_n}\phi^1. $$

Here $\nu_2,\ldots ,\nu_n\in\{0,1,2,3\}$ are fixed spacetime indices.

We claim that (i) the infinitesimal transformation (2) is a quasisymmetry of the Lagrangian density (1), and moreover (ii) the corresponding Noether current

$$\tag{5} J^{\mu}_{\nu_2\ldots \nu_n}~=~\partial^{\mu}\phi^2~ \partial_{\nu_2}\cdots\partial_{\nu_n}\phi^1 -\phi^2 ~\partial^{\mu}\partial_{\nu_2}\cdots\partial_{\nu_n}\phi^1$$

is conserved on-shell

$$\tag{6} d_{\mu}J^{\mu}_{\nu_2\ldots \nu_n}~\approx~0.$$

We leave the proof to the reader.


What you wrote is a lagrangian of a free theory. All free theories have infinite-dimensional symmetry, called higher-spin symmetry. The current you gave is a standard Noether current corresponding to one of the generators of the higher-spin symmetry. Usual space-time symmetry is a subalgebra of the higher-spin algebra.

As Maldacena a Zhiboedov showed in 3d, the presence of higher-spin currents implies the theory is free, http://arxiv.org/abs/1112.1016


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