# J=L+S derivation

1. Jan 13, 2016

### avikarto

I am trying to derive that $\vec{J}=\vec{L}+\vec{S}$ (in free space) starting from the general expression $\vec{J}=\int dV\epsilon_0\: \vec{r}\times\left(\vec{E}\times \vec{B}\right)$. I get to the point where I have that

$\vec{J}=\epsilon_0\int dV E_i\left(\vec{r}\times\nabla\right)A_i+\epsilon_0\int dV \vec{E}\times\vec{A}-\epsilon_0\int dV\left(\vec{E}\cdot\nabla\right)\left(\vec{r}\times\vec{A}\right)$

Equating the first two terms with $\vec{L}$ and $\vec{S}$ respectively, I can't figure out how to make the final term zero. The text I am following indicates that I should use the divergence theorem to make a surface integral vanish, but in order to do so I need to have a divergence, which I don't seem to have. Going through the tensor notation, all I can seem to rearrange this to is

$\left(\vec{E}\cdot\nabla\right)\left(\vec{r}\times\vec{A}\right)=\vec{r}\times\left(\vec{E}\cdot\nabla\right)\vec{A}+\vec{E}\times\vec{A}$

Last edited: Jan 13, 2016
2. Jan 13, 2016

### vanhees71

Where is this from? A split of the total angular momentum of the electromagnetic field into an orbital and a spin part doesn't make physical sense, because you cannot do this split in a unique gauge-invariant way. The total angular momentum of the electromagnetic field (in Heaviside-Lorentz units) reads
$$\vec{J}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \frac{1}{c} \vec{x} \times (\vec{E} \times \vec{B}),$$

3. Jan 13, 2016

### avikarto

The fact that J for light can be split cleanly into SAM-like and OAM-like terms is a consequence of Noether's theorem, and is found as such in many texts. See, for example, this page. The book that I am following at the moment is "The Angular Momentum of Light" by David Andrews and Mohamed Babiker.

4. Jan 14, 2016

### vanhees71

Well, the split is not gauge invariant. The single terms called OAM and SAM (standing for orbital and spin angular momentum) are meaningless from a physics point of view, because they are for sure unobservable, because they are not gauge invariant. Of course, $\vec{J}$ is gauge invariant and the total angular momentum of the electromagnetic field (with the origin of the spatial coordinate system as the reference point).

5. Jan 14, 2016

### samalkhaiat

Okay, before I show you how to do the split, I need to make few points here. The conserved canonical Noether current associated with Lorentz invariance is given by
$$M^{\mu \nu \rho} = x^{\mu}T^{\nu\rho} - x^{\nu}T^{\mu\rho} + S^{\mu\nu\rho} ,$$
where $T^{\mu\nu}$ is the canonical energy-momentum tensor which, for the electromagnetic field, is given by
$$T^{\mu\nu} = \partial^{\mu}A_{\sigma}F^{\nu\sigma} - \frac{1}{4} \eta^{\mu\nu} F^{2} .$$
Notice that this energy-momentum tensor is not symmetric and also not gauge invariant. And the tensor $S^{\mu\nu\sigma}$ determines the polarization properties of the field. For the electromagnetic field, it is given by
$$S^{\mu\nu\sigma} = A^{\mu}F^{\nu \sigma} - A^{\nu}F^{\mu\sigma} .$$
The angular momentum vector $J_{i}$ is obtained from the tensor $M^{\mu\nu\rho}$ according to
$$J_{i} = \frac{1}{2} \epsilon_{ijk} \int d^{3}x \ M^{jk0} .$$
For em-field, you can easily show that this becomes
$$J_{i} = \int d^{3}x \ ( \vec{E} \times \vec{A} )_{i} + \int d^{3}x \ E_{j} ( \vec{x} \times \vec{\nabla})_{i} A_{j} . \ \ \ (1)$$
So, the canonical moment tensor $M^{\mu\nu\rho}$ leads directly to
$$J_{i} = \frac{1}{2} \epsilon_{ijk} \int d^{3}x \ M^{jk0} = S_{i} + L_{i} . \ \ \ (2)$$
The expression you wrote for the angular momentum vector $J_{i}$ comes not from the canonical moment tenser $M^{\mu\nu\rho}$, but from the so-called Belinfante moment tensor:
$$J^{\mu\nu\rho} = x^{\mu} \theta^{\nu\rho} - x^{\nu} \theta^{\mu\rho} ,$$
where $\theta^{\mu\nu} = \theta^{\nu\mu}$ is the symmetric and gauge invariant (Belinfante) energy-momentum tensor. For em-field, it is given by
$$\theta^{\mu\nu} = F^{\mu}{}_{\rho}F^{\nu\rho} - \frac{1}{4} \eta^{\mu\nu}F^{2} .$$
Indeed, using the relation
$$\theta^{j0} = (\vec{E} \times \vec{B})_{j},$$
you can easily obtain the expression you wrote for the angular momentum 3-vector
$$\frac{1}{2}\epsilon_{ijk} \int d^{3}x \ J^{jk0} = \int d^{3}x \ ( \vec{x} \times \vec{E} \times \vec{B} )_{i} . \ \ \ \ (3)$$
You can also show (for any Poincare invariant field theory) that there exists a tensor $\chi^{\mu\nu\rho\tau} = - \chi^{\mu\nu\tau\rho}$ such that
$$J^{\mu\nu\rho} = M^{\mu\nu\rho} + \partial_{\tau} \chi^{\mu\nu\rho\tau} .$$
Integrating the $ij0$ component of this equation, using the anti-symmetry property of $\chi$ and the divergence theorem, we obtain
$$\int d^{3}x \ (J^{ij0} - M^{ij0}) = \int d^{3}x \ \partial_{k} \chi^{ij0k} = 0 .$$
Thus
$$\int d^{3}x \ \frac{1}{2}\epsilon_{kij} J^{ij0} = \int d^{3}x \ \frac{1}{2}\epsilon_{kij} M^{ij0} .$$
Now, in this equation, if you substitute (1), (2) and (3) you get
$$J_{k} = \int d^{3}x \ \frac{1}{2}\epsilon_{kij} J^{ij0} = S_{k} + L_{k} .$$
Of course, this method is valid for any Poincare invariant field theory including Maxwell’s theory.
Okay, if you are not happy with above method, let’s now do the split in the same way that you tried to do. Consider the integrand in equation (3)
$$\left( \vec{x} \times \vec{E} \times \vec{B} \right)_{i} = \epsilon_{ijk}\epsilon_{klm} x_{j}E_{l}B_{m} .$$
Using the relation $B_{m} = \epsilon_{mrs}\nabla_{r}A_{s}$, we get
$$\left( \vec{x} \times \vec{E} \times \vec{B} \right)_{i} = \epsilon_{ijk}\epsilon_{klm}\epsilon_{mrs}x_{j}E_{l}\nabla_{r}A_{s} . \ \ (4)$$
Now, write
$$x_{j}E_{l}\nabla_{r}A_{s} = \nabla_{r}( x_{j}E_{l}A_{s} ) - A_{s} \nabla_{r}( x_{j} E_{l}) .$$
Remember that we are integrating, so we can neglect the first term because it vanishes by the divergence theorem. So, we can just use
$$x_{j}E_{l}\nabla_{r}A_{s} = - A_{s} \delta_{rj} E_{l} - A_{s} x_{j} \nabla_{r} E_{l} . \ \ \ (5)$$
Let us also use the identity
$$\epsilon_{klm} \epsilon_{mrs} = \delta_{kr}\delta_{ls} - \delta_{ks}\delta_{lr} . \ \ \ \ (6)$$
Substituting (5) and (6) in (4) leads to
$$\left( \vec{x} \times \vec{E} \times \vec{B} \right)_{i} = - ( \epsilon_{ijr}\delta_{ls} - \epsilon_{ijs}\delta_{lr}) ( \delta_{rj} A_{s} E_{l} + x_{j} A_{s} \nabla_{r}E_{l}) .$$
Expanding and using $\epsilon_{ijr}\delta_{rj} = 0$ and the Maxwell’s equation $\nabla_{r}E_{r} = 0$, we obtain
$$\left( \vec{x} \times \vec{E} \times \vec{B} \right)_{i} = \epsilon_{ils} E_{l} A_{s} - \epsilon_{ijr} x_{j} A_{l} \nabla_{r}E_{l} . \ \ \ (7)$$
Let’s write
$$x_{j} A_{l} \nabla_{r}E_{l} = \nabla_{r}( x_{j} A_{l} E_{l} ) - \delta_{rj} A_{l} E_{l} - x_{j} E_{l} \nabla_{r}A_{l} .$$
Again, we neglect the divergence term:
$$x_{j} A_{l} \nabla_{r}E_{l} = - \delta_{rj} A_{l} E_{l} - x_{j} E_{l} \nabla_{r}A_{l} .$$
Inserting this in (7) and using $\epsilon_{ijr}\delta_{rj} = 0$, we get (now under the integration sign)
$$\int d^{3}x \ \left( \vec{x} \times \vec{E} \times \vec{B} \right)_{i} = \int d^{3}x \Big \{ \epsilon_{ils} E_{l} A_{s} + E_{l} ( \epsilon_{ijr} x_{j} \nabla_{r}) A_{l} \Big \}$$
or
$$\int d^{3}x \ \left( \vec{x} \times \vec{E} \times \vec{B} \right)_{i} = \int d^{3}x \ ( \vec{E} \times \vec{A} )_{i} + \int d^{3}x \ E_{j} ( \vec{x} \times \vec{\nabla})_{i} A_{j} .$$
Okay, let me finish by giving you a problem to think about.
Consider a circularly polarized plane wave travelling in the z-direction. In this case, the momentum density $\theta^{k0} = ( \vec{E} \times \vec{B})_{k}$ is a vector pointing exactly in the z-direction. Thus,
$$J^{120} = x^{1}\theta^{20} - x^{2}\theta^{10} = 0 .$$
For the z-component of the angular momentum vector, we therefore find $J_{3} = 0$? So what happened to the spin angular momentum of this beam of light?

Last edited: Jan 14, 2016
6. Jan 14, 2016

### vanhees71

This is all fine math, but what's the physics behind this split? In relativistic (quantum) field theory a split in orbital and spin angular momentum is at least tricky. In the case of gauge fields it's even unphysical, because gauge dependent quantities can never be observables.

The canonical energy-momentum tensor of gauge fields is not observable. If divergence free you can define an energy-momentum vector by integrating over a complete space-like hypersurface (e.g., $t=\text{const}$ in an arbitrary inertial (Minkowski) frame), which is observable (e.g., in the recoil of objects when hit by electromagnetic radiation; or heating up material by radiation).

The only place where the local energy-momentum-stress tensor becomes observable is in the context of general relativity as a source of gravity, and there it's clearly not the canonical energy-momentum tensor but the one you get from varying the Einstein-Hilbert action with the matter and radiation fields coupled to the metric in an approriate way, leading to the Belinfante energy-momentum tensor which is symmetric and gauge invariant, as it must be for an observable quantity.

The same holds true for the angular-momentum tensor density.

You can also derive gauge-independent energy-momentum and angular-momentum tensors via Noether's theorem applied to the space-time Poincare symmetries. However, one must take into account that a gauge field is only determined up to a gauge transformation, and thus you have to extend the transformation properties under Poincare transformations as those of a vector field (for the em. field) modulo a gauge transformation, which includes arbitrary gauge field components, which finally can be chosen such as to lead to gauge independent expressions for the Noether currents (i.e., in this case energy-momentum and angular-momentum tensor).

Further for plane waves $\propto \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x})$ the total energy and momentum and angular momentum diverge. They are not realizable in nature. To show that helicity eigenmodes have the correct meaning with regard to the projection of the total angular momentum (!!!) in direction to the momentum of the wave, you have to use appropriately defined wave packets with finite total energy, momentum, and angular momentum.

I've written this up, however in German. Perhaps you can get the idea from looking at the formulae:

For the derivation of gauge-invariant energy-momentum and angular-momentum tensors, see

http://theory.gsi.de/~vanhees/faq/edyn/node7.html
http://theory.gsi.de/~vanhees/faq/edyn/node8.html

and for the wave-packet approach to quasi-plane waves:

http://theory.gsi.de/~vanhees/faq/edyn/node22.html

7. Jan 14, 2016

### avikarto

The OAM and SAM themselves may very well be directly un-observable in the beam, however their effects can be clearly and distinctly visualized by their effects on a particle that gets hit by the beam. A beam carrying only SAM for instance would rotate the particle about the particle's own central axis, while a beam carrying only OAM would rotate the particle about the central axis of the beam. In this respect, the two separate components of $J$ are quite physical.

8. Jan 14, 2016

### avikarto

Thank you very much, this was extremely helpful in understanding the problem. I appreciate the effort!

9. Jan 15, 2016

### vanhees71

Ok, different fields lead to different motion of charged particles, but how can you say that one wave carries only SAM and another only OAM, when you cannot even physically define such waves. Again, the split is unphysical!

10. Jan 15, 2016

### avikarto

Another way to visualize the split is to think of a plane wave under two polarizations. A linearly polarized plane wave has no SAM, while a circularly polarized plane wave does have SAM. In the case of any polarization, however, no plane wave can ever have OAM. In order to have any OAM, you must have some component of $\vec{J}$ in the direction of propagation, and therefore some component of $\vec{E}$ or $\vec{B}$ in the direction of propagation. A plane wave has neither, and therefore no OAM. So, using a plane wave as the elementary example we can directly see the split, for these waves can have SAM without having OAM.

For the details, see this paper from Yao and Padgett.

Of course, you could always come back and say that a plane wave its self is unphysical since these will never truly arise in nature, but the idea should be clear from this simple example. Besides, many things can be very reasonably approximated as plane waves, in which case the arguments hold.

Last edited: Jan 15, 2016
11. Jan 15, 2016

### samalkhaiat

Yes, because the question in the OP was about the math. No body (apart from you!) asked about the physical meaning of the decomposition.
Google “Spin of the Proton” or “The Angular momentum Controversy”. You can also look at the papers bellow.
Yes I know that because I have been working on the angular momentum of the quarks and gluons constituent of the Nucleon for more than 15 years.
Non sense, consider any radiation field with $e^{-i\omega t}$ time dependence (for example Laguerre-Gaussian laser modes). Using the Maxwell equation
$$\mathbf{B} = - \frac{i}{\omega} \nabla \times \mathbf{E},$$
and repeating the math of my previous post, we obtain the following gauge invariant decomposition into SAM and OAM
\begin{align*} \mathbf{J} &= \frac{1}{2} \int d^3 x \, \mathbf{x} \times \left( \mathbf{E}^{*} \times \mathbf{B} + \mathbf{E} \times \mathbf{B}^{*} \right) \\ &= - \frac{i}{\omega} \int d^3 x \, \left( \mathbf{E}^{*} \times \mathbf{E} + E^{*}_{k} ( \mathbf{x} \times \nabla ) E_{k} \right) \end{align*}

Again non sense. The Hamiltonian $H$ and canonical momentum $\mathbf{P}$ of a charged particle moving in em-field are not gauge invariant but they are good observables
$$H = \frac{(\mathbf{P} - e \mathbf{A})^{2}}{2m} + e A^{0} .$$
You seem to “equate” gauge-invariance with measurability (observablity) which is not correct.
In QFT an operator $\mathcal{O}$ is an observable if, for all physical states $|\Phi \rangle \in \mathcal{V}_{phy}$, $\mathcal{O}|\Phi \rangle$ is itself a physical state, i.e., $\mathcal{O}|\Phi \rangle \in \mathcal{V}_{phy}$. In gauge field theories, the physical states of $\mathcal{V}_{phy}$ are defined by means of an appropriate subsidiary condition. In QED, it is given by
$$B^{(+)}(x) | \Phi \rangle = 0 ,$$
where $B^{(+)}(x)$ is the positive frequency part of the gauge fixing field $B(x) = B^{(+)}(x) + B^{(-)}(x)$. So, in gauge field theories, operator $\mathcal{O}$ is an observable if
$$B^{(+)}(x) \left(\mathcal{O} | \Phi \rangle \right) = 0 , \ \ \ \forall |\Phi \rangle \in \mathcal{V}_{phy} . \ \ \ \ \ \ \ (1)$$
This does not necessarily imply that $\mathcal{O}$ is gauge invariant, i.e., under gauge transformation $U$, Eq(1) does not imply
$$U^{\dagger}\mathcal{O} U = \mathcal{O} .$$
Equivalently, the condition (1) can be restated as follow:
An operator $\mathcal{O}$ is an observable if
$$[ B^{(+)}, \mathcal{O}] | \Phi \rangle = 0 , \ \ \forall | \Phi \rangle \in \mathcal{V}_{phy} . \ \ \ (2)$$
From translation invariance, we have
$$[ B^{(+)} (x) , P^{\mu} ] = i \partial^{\mu} B^{(+)}(x) .$$
This means that
$$[ B^{(+)} (x) , P^{\mu} ] | \Phi \rangle = i\partial^{\mu} ( B^{(+)}| \Phi \rangle ) = 0 .$$
Thus, the momentum operator $P^{\mu}$ is an observable in gauge field theories, i.e., it’s eigenvalues are physically measurable quantities. But, is $P^{\mu}$ gauge-invariant operator? If you think it is, then get ready for a surprise.
Theorem: In any theory, which is invariant under a local c-number gauge transformations, the total momentum $P^{\mu}$(defined as the generator of space-time translation) and angular momentum $M^{ij}$(defined as the generator of $SO(3)$) operators cannot be gauge invariant.
Proof: Let $Q$ be the generator of a c-number gauge transformation:
$$\delta_{\Lambda} A^{\mu} = [ i Q , A^{\mu} ] = \partial^{\mu} \Lambda , \ \ \ (3)$$
where, $\Lambda (x)$ is a (bounded) c-number field satisfying
$$\partial_{\mu}\partial^{\mu} \Lambda = 0 .$$
Now, consider the Jacobi identity
$$[P^{\mu} , [ Q , A^{\nu} ]] + [ A^{\nu} , [ P^{\mu} , Q ]] = [ Q , [ P^{\mu} , A^{\nu} ]] .$$
The first term on the left is zero because, by (3), $[ Q , A^{\nu}]$ is a c-number and therefore commute with $P^{\mu}$.
Form translation invariance, we have
$$[ i P^{\mu} , A^{\nu} ] = \partial^{\mu}A^{\nu} .$$
Thus
$$[ Q , [ i P^{\mu} , A^{\nu} ]] = \partial^{\mu} [ Q , A^{\nu} ] ,$$
or, by (3)
$$[ Q , [ P^{\mu} , A^{\nu} ]] = - \partial^{\mu} \partial^{\nu} \Lambda \neq 0 .$$
Therefore, from the Jacobi identity, it follows that
$$[ A^{\nu} , [ Q , P^{\mu} ] ] = \partial^{\mu} \partial^{\nu} \Lambda \neq 0 .$$
Thus
$$\delta_{\Lambda}P^{\mu} \equiv [i Q , P^{\mu}] \neq 0 .$$
Similarly, we can show that
$$\delta_{\Lambda}M^{ij} \equiv [i Q , M^{ij}] \neq 0 .$$
This shows that the total momentum and the angular momentum operators cannot be gauge invariant, even though they are pretty good observables. What we measure are not operators but matrix elements of operators between the physical states of the theory which are defined by appropriate subsidiary conditions. Indeed, we can show that $\langle \Psi | P^{\mu} | \Phi \rangle$ and $\langle \Psi | J_{k} | \Phi \rangle$ are gauge invariant for all $(| \Phi \rangle , | \Psi \rangle ) \in \mathcal{V}_{phy}$
The above theorem can also be proved in QCD.
You must be joking. I think I worth more than these Kindergarten stuff of yours.

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