# Angular Momentum of the Dirac field

I'm going through the Peskin & Shroeder's discussion on the Dirac field, and I am struggling with a couple of claims they make about angular momentum. First of all, the angular momentum operator is given by:

$$\mathbf{J} = \int d^3 \mathbf{x} \, \psi^{\dagger}(x) \left[ \mathbf{x} \times (-i \nabla) + \frac{1}{2}\mathbf{\Sigma} \right] \psi(x)$$

where $$\mathbf{\Sigma}$$ are, in the chiral representation, the duplicated Pauli matrices: $$\Sigma^i = \begin{pmatrix} \sigma^i & 0 \\ 0 & \sigma^i \end{pmatrix}$$. Now, the book claims:

• The division of angular momentum into spin and orbital parts is not easy for relativistic fermions. Since the linear momentum of the field is: $$\mathbf{P} = \int d^3 \mathbf{x} \, \psi^{\dagger}(x) (-i \nabla) \psi(x)$$ it seems for me quite natural to interpret the first term of $$\mathbf{J}$$ as the orbital angular momentum and the second one as the spin. Why is this not valid in general?

• The angular momentum operator (at least the z-component, although I think this should be true for any component) annihilates de vacuum state $$|0\rangle$$. However, this is not straightforward to me since there will be a term proportional to $$a^{\dagger}b^{\dagger}|0\rangle$$ when calculating $$J^z |0\rangle$$ which I cannot see how to cancel.

• For states $$b^{s\dagger}_\mathbf{0} |0\rangle$$, and assuming the previous result to be true, we can write: $$J^z b^{s\dagger}_\mathbf{0} |0\rangle = [J^z, b^{s\dagger}_\mathbf{0}] |0\rangle$$ It is then supposed to be straightforward to show that the first term of $$J^z$$ (i.e., the orbital part) does not contribute to this expression, but I do not know how to prove this.

I am using the following conventions for the field:

$$\psi(x) = \int \frac{d^3 \mathbf{p}}{(2\pi)^3} \frac{1}{\sqrt{2E_{\mathbf{p}}}} \sum_{s} \left[ a^s_{\mathbf{p}} u^s(p) e^{-ip\cdot x} + b^{s\dagger}_{\mathbf{p}} v^s(p) e^{ip\cdot x} \right]$$

I know I have not given many results I obtained on my own but that's because all I could do was basically dirty calculations in a piece of paper that don't seem useful to the discussion...

EDIT (to provide more information after the answer by Name YYY)

i) First of all, the z component of the angular momentum operator is given explicitly in terms of creation and annihilation operators by:

$$J^z = \int d^3 \mathbf{x} \frac{d^3 \mathbf{p} d^3 \mathbf{k} \, e^{-i(\mathbf{p}-\mathbf{k})\cdot\mathbf{x}} }{(2\pi)^6 \sqrt{4 E_{\mathbf{p}} E_{\mathbf{k}}}} \times \sum_{r,s} \left[ b^r_{-\mathbf{p}} v^{r \dagger} (-\mathbf{p}) + a^{r \dagger}_{\mathbf{p}} u^{r \dagger} (\mathbf{p}) \right] \left[ (\mathbf{x} \times \mathbf{k})^z + \frac{\Sigma^3}{2} \right] \left[ a^s_{\mathbf{k}} u^{s} (\mathbf{k}) + b^{s \dagger}_{-\mathbf{k}} v^{s} (-\mathbf{k}) \right]$$

where I am evaluating at $$t=0$$ the fields since, being $$\mathbf{J}$$ a conserved quantity, that does not affect the result. Now, the problem is that I cannot integrate over $$\mathbf{x}$$ (because I have an $$\mathbf{x}$$ in the vector product), which is the usual trick to get a $$\delta^{(3)}(\mathbf{p}-\mathbf{k})$$ and then use the orthogonality relation between the $$u$$'s and $$v$$'s. Without this, when applying $$J^z$$ to $$|0\rangle$$, you get terms of the form $$a^{r \dagger}_{\mathbf{p}} b^{s \dagger}_{-\mathbf{k}} |0\rangle$$ which does not necessarily vanish, since the factor is $$u^{r\dagger} (\mathbf{p})v^{s} (-\mathbf{k})$$ - with possibly the matrix $$\Sigma^3$$ in the middle in the spin term.

ii) Now, regarding the vanishing orbital part of the angular momentum when acting on $$b^{s \dagger}_{\mathbf{0}}|0\rangle$$, the heuristic argument seems a bit cheating to me... The Hamiltonian is also a scalar, and you can expand it using an integral over all possible values of the momentum, which is a vector. Similarly, it could be possible to expand the angular momentum of the zero momentum state with an integral over all possible momenta. Apart from that, I could prove that the orbital angular momentum does not contribute to $$[J^z, a^{t\dagger}_\mathbf{0}] |0\rangle$$ because the only non-vanishing commutator (acting on the vacuum) using the previous expression for $$J^z$$ is:

$$[a^{r \dagger}_\mathbf{p} a^{s}_\mathbf{k}, a^{t\dagger}_\mathbf{0}] |0\rangle = (2\pi)^3 \delta^{3}(\mathbf{k}) \delta^{st} a^{r \dagger}_\mathbf{p} |0\rangle$$

but this makes the angular part of the angular momentum zero since it contains the vector product $$\mathbf{x} \times \mathbf{k}$$. The problem with the $$b$$'s is that the analogous non-vanishing commutator is:

$$[b^{r }_{-\mathbf{p}} b^{s \dagger}_{-\mathbf{k}}, b^{t\dagger}_\mathbf{0}] |0\rangle = - (2\pi)^3 \delta^{3}(\mathbf{p}) \delta^{rt} b^{s \dagger}_{-\mathbf{k}} |0\rangle$$

and this contains $$\delta^{3}(\mathbf{p})$$ instead of $$\delta^{3}(\mathbf{k})$$, so it is not ovbious how to cancel the orbital part of the angular momentum...

I hope my problems with this calculation are more clear now... Thank you for your help!

• To handle the integration over $\bf{x}$, note that the differential operator only acts on the exponential. I can be switched by a differential operator on momentum – Iván Mauricio Burbano Jun 5 '19 at 21:40

## 1 Answer

it seems for me quite natural to interpret the first term of J as the orbital angular momentum and the second one as the spin. Why is this not valid in general?

The reason is that the more relativistic particle is, the less well defined is the definition of spin. Formally this is related to the fact that the square of the Pauli-Lubanski operator $\hat{W}_{\mu}$, which defines the spin $s$ of representation (one-particle state), becomes zero for massless particles. Instead of spin, massless particles (we can think about massive ultrarelativistic particle as about massless one) are characterized by helicity $\hat{h}$ - the projection of spin on momentum direction - which can take only two values: $$\hat{h} \equiv \hat{\mathbf W} \cdot \frac{\hat{\mathbf p}}{|\mathbf p|} = \pm s$$ You can heuristically think about that in a following way: when particle (boson or fermion of arbitrary spin, it doesn't matter) becomes ultrarelativistic, then due to relativistic aberration effect the distribution of projections of spin onto direction of motion becomes only parallel and antiparallel.

However, this is not straightforward to me since there will be a term proportional to

Don't forget about relations $$u^{\dagger}_{s}(\mathbf p)v_{s{'}}(-\mathbf p) = 0$$ This is important, since this quantity is in the front of $$a_{s{'}}^{\dagger}(\mathbf p)b_{s}^{\dagger}(-\mathbf p)|0\rangle$$

does not contribute to this expression, but I do not know how to prove this.

Since, as you've shown by answering on your second question, $$\hat{\mathbf J}_{z}|0\rangle = 0,$$ you may write: $$\hat{\mathbf J}_{z}\hat{a}^{\dagger}_{s}(\mathbf p)|0\rangle \equiv \hat{\mathbf J}_{z}\hat{a}^{\dagger}_{s}(\mathbf p)|0\rangle \mp \hat{a}^{\dagger}_{s}(\mathbf p)\hat{\mathbf J}_{z}|0\rangle \equiv [\hat{\mathbf J}_{z}, \hat{a}_{s}^{\dagger}(\mathbf p)]_{\mp}|0\rangle ,$$ which proves the relation.

Or you've asked about proof than the orbital part doesn't make contribution for zero momentum one-particle state? Then it is most easy to see by the following heuristic thinking: since the out state has zero momentum, then the quantity $$\mathbf L_{\mathbf 0} = \int d^{3}\mathbf x \hat{\Psi}^{\dagger}(\mathbf x)[\mathbf x \times (-i\mathbf{\nabla})]\hat{\Psi}(\mathbf x)\hat{a}_{s}^{\dagger}(0)|\rangle,$$ hasn't depend on any vector. Since it is, however, (pseudo)vector by the definition, it must depend on some vector. To avoid this contradiction, we come to the statement that this quantity is equal to zero.

• Ok with the first answer, but the other two arguments does not seem really convincing... I'll edit my post with some more calculations so that you know exactly where my problem is! – Alex V. Feb 22 '16 at 18:36