# Angular Momentum Operator in Quantum Field Theory

I'm following along with David Tong's QFT course and am trying to derive the result shown in question 6 on his 2nd problem sheet, but am running into problems when applying it to the free real scalar field. This isn't a particularly insightful problem, but I am stuck and keep computing that it is 0. What I am trying to compute shows up in this question

This is that we can start with the total angular momentum as defined by: $$Q_i = \epsilon_{ijk}\int d^3 x (x^j T^{0k}-x^k T^{0j}) = 2\epsilon_{ijk}\int d^3 x x^j T^{0k}$$ with $T^{\mu \nu}$ the stress-energy tensor and arrive at the result: $$Q_i = -i\epsilon_{ijk}\int \frac{d^3 p}{(2\pi)^3} a^\dagger_\vec{p}\left( p_j \frac{\partial}{\partial p^k} - p_k\frac{\partial}{\partial p^j}\right) a_\vec{p}$$

It is not hard to start using $T^{0i} = \Pi \partial^i \phi$ from our Lagrangian $\mathcal{L}=\frac{1}{2}\partial_\mu \phi \partial^\mu \phi - \frac{1}{2}m^2\phi^2$, and then using our expansion of the field in terms of our creation and annihilation operators as $$\phi = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2E_\vec{p}}}\left(a_\vec{p} e^{i\vec{p}\cdot\vec{x}} + a^\dagger_\vec{p} e^{-i\vec{p}\cdot\vec{x}}\right)$$ and $$\Pi = -i\int \frac{d^3q}{(2\pi)^3} \sqrt{\frac{E_\vec{q}}{2}}\left(a_\vec{q} e^{i\vec{q}\cdot\vec{x}} - a^\dagger_\vec{q} e^{-i\vec{q}\cdot\vec{x}}\right)$$ Writing down the derivative: $$\partial^i\phi = -i\int \frac{d^3p}{(2\pi)^3} \frac{p^i}{\sqrt{2E_\vec{p}}}\left(a_\vec{p} e^{i\vec{p}\cdot\vec{x}} - a^\dagger_\vec{p} e^{-i\vec{p}\cdot\vec{x}}\right)$$

This gives me $$Q_i = \epsilon_{ijk}\int \frac{d^3p d^3q}{(2\pi)^6}\sqrt{\frac{E_\vec{q}}{E_\vec{p}}} p^k\int d^3x x^j(a_\vec{q}e^{i\vec{q}\cdot\vec{x}}-a^\dagger_\vec{q}e^{-i\vec{q}\cdot\vec{x}})(a_\vec{p}e^{i\vec{p}\cdot\vec{x}}-a^\dagger_\vec{p}e^{-i\vec{p}\cdot\vec{x}})$$

We can expand the product of annihilation and creation operators, and note that $x^j$ is really the derivative of the exponential terms with respect to $p^j$ which can be pulled out of the integral.

After doing this, integrating by parts to remove the derivative of the delta function left behind, we are left with $$Q_i = -i \epsilon_{ijk}\int\frac{d^3p d^3q}{(2\pi)^6}\left[\delta^{(3)}(\vec{p}+\vec{q})\frac{\partial}{\partial p^j} \left( \sqrt{\frac{E_\vec{q}}{E_\vec{p}}} p_k(a^\dagger_\vec{q} a^\dagger_\vec{p} - a_\vec{q} a_\vec{p})\right) + \delta^{(3)}(\vec{p}-\vec{q})\frac{\partial}{\partial p^j} \left( \sqrt{\frac{E_\vec{q}}{E_\vec{p}}} p_k(a^\dagger_\vec{q} a_\vec{p} - a_\vec{q} a^\dagger_\vec{p})\right)\right]$$

Integrating over $\vec{q}$ kills the energy ratios in both cases due to the dispersion relation, and the $p^k$ term can safely be taken outside the integral since otherwise we have a term like $\epsilon_{ijk}\frac{\partial p^k}{\partial p^j} = \epsilon_{ijk}\delta_j^k = 0$

Finally, this gives us $$Q_i = -i\epsilon_{ijk}\int\frac{d^3p}{(2\pi)^3}p^k\frac{\partial}{\partial p^j}(a^\dagger_{-\vec{p}} a^\dagger_\vec{p} - a_{-\vec{p}}a_\vec{p} + a^\dagger_\vec{p}a_\vec{p} - a_\vec{p}a^\dagger_\vec{p})$$

To keep going, we notice that since $\phi$ is a real scalar field, we can equate itself to its complex conjugate and we can switch $\vec{p} \rightarrow -\vec{p}$ to find that $a_{-\vec{p}} = a^\dagger_\vec{p}$. But plugging this in the above form of $Q_i$ gives 0. Any ideas what is going on here?

It looks like you've missed a sign: \begin{align} Q_i &= \epsilon_{ijk} \int_{p,q} \sqrt{\frac{E_q}{E_p}}p^k \int d^3x (a_qe^{iq\cdot x}-a_q^\dagger e^{-iq\cdot x}) \left(a_p (-i\frac{\partial}{\partial p^j})e^{ip\cdot x} -a_p^\dagger (i\frac{\partial}{\partial p^j})e^{-ip\cdot x}\right). \end{align} Then it looks like the term that goes like $a_q^\dagger a_p^\dagger$ should have the opposite sign.
• I've started from this line, and come to the same answer as before. After factoring out a $-i$, we still end up with terms like $(a_q a_p - a^\dagger_q a^\dagger_p)$ and $(a_q a^\dagger_p - a^\dagger_q a_p)$. Is there a problem moving the derivative through the terms depending on $\vec{q}$, since we end up taking $\vec{q} \rightarrow \pm \vec{p}$? – drglove Mar 27 '14 at 17:34