# Question about surface term in QFT problem

I am trying to follow the solution of the following problem (Srednicki 39.2):

To show that: $$J_z b_s^\dagger(p\hat z)|0\rangle=\frac{1}{2}\ s\ b_s^\dagger(p\hat z)\ |0\rangle,$$ where $J_z$ is the angular momentum in the $z$ direction, $b_s^\dagger$ is the creation operator of the Dirac field, and $s$ is the spin index (taking values +1 and -1). This problem shows a particular case: the spin eigenvalues of the state corresponding to a single particle travelling in the $z$ direction are +1/2 and -1/2.

Using the expression of the commutator:

$[\Psi (x),M^{\mu \nu}] = -i(x^\mu \partial^\nu - x^\nu \partial^\mu)\Psi (x) + S^{\mu\nu}\Psi(x)$

where $M^{\mu\nu}$ are the infinitesimal generators of the Lorentz group and $J_z=M^{12}$, using also the fact that:

$J_z |0\rangle =0 \rightarrow J_z b_s^\dagger(p\hat z)|0\rangle = [J_z, b_s^\dagger(p\hat z)]|0\rangle$

and using finally the expression of the creation operator in terms of the fields, then we arrive at the following expression:

$J_z b_s^\dagger(pz)|0\rangle = \int d^3x\ e^{ipx^3}i(x^1\partial^2 -x^2\partial^1)\overline\Psi(x) \gamma^0 u_s(\vec p)|0\rangle + \int d^3x\ e^{ipx^3}\overline\Psi(x)S^{12} \gamma^0 u_s(\vec p)|0\rangle$

Now the standard procedure (as far as I am aware) is to show that the first integral vanishes because it is equivalent to a "surface term" evaluated at the infinite, where the boundary conditions assumed for the fields are such that they vanish at infinite fast enough. I have no problem in following the procedure from this point on. My problem is precisely with the "surface term".

I can elaborate the first integral a little bit more. Noticing that the exponential is constant for the differential operator within brackets, and that also $\gamma^0$ and $u_s$ are constants for this operator, we can write:

$i\int d^3x\ (x^1\partial^2 -x^2\partial^1)(\overline\Psi(x) \gamma^0\ e^{ipx^3})\ u_s(\vec p)\ |0\rangle$

How to go from this to show that this integral is in fact a surface term and therefore it vanishes?. Do we have to use here some sort of the Stokes' theorem?.

It is probably a very simple question, but I am stuck at this point. I just do not see it. Hope somebody can help me.

## 1 Answer

The fact that your integral is a surface term is indeed due to Stokes theorem: $$\int d^3x\ (x^1\partial^2 -x^2\partial^1)(\overline\Psi(x) \gamma^0\ e^{ipx^3})\\ =\int d^3x\ \partial^2\left(x^1 \overline\Psi(x) \gamma^0\ e^{ipx^3}\right)-(1\leftrightarrow 2)\\ =\int dx_1 dx_3 \left(\int dx_2 \partial^2\left(x^1 \overline\Psi(x) \gamma^0\ e^{ipx^3}\right)\right)-(1\leftrightarrow 2)\\ =\int dx_1 dx_3 \left(\left.x^1 \overline\Psi(x) \gamma^0\ e^{ipx^3}\right|_{x_2=-\infty}^{x_2=\infty}\right)-(1\leftrightarrow 2)$$ The notation $(1\leftrightarrow 2)$ indicates the same term again with all indices 1 and 2 switched. You can see in this case that Stoke's theorem (or the divergence theorem) just reduce to the fundamental theorem of calculus which you use to compute the $x_2$-integral.

Now, you have to use the (usual) argument that the fields you consider (here $\overline\Psi(x)$) vanish at spatial infinity, in particular at $x_2=\pm\infty$. The reason for this assumption is for example discussed here.