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How to compute the normal ordered angular momentum of a Klein-Gordon real scalar in terms of ladder operators?

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I'm trying to compute the angular momentum $$Q_i=-2\epsilon_{ijk}\int{d^3x}\,x^kT^{0j}\tag{1}$$ where ${T^\mu}_\nu=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\partial_\nu\phi-{\delta^\mu}_\nu\mathcal{L}$ with the Lagrangian $\mathcal{L}=\frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\frac{1}{2}m^2\phi^2$ on the real scalar field $\phi=\phi(x)$.

This is basically the same question and I'm following the same lectures by David Tong, but the answer doesn't seem successful to me and I'm not sure if I have some mistake or if my question is ultimately about normal ordering. As stated there, the final expression of $Q_i$ in terms of ladder operators should be $$Q_i=-i\epsilon_{ijk}\int\frac{d^3p}{(2\pi)^3} a^\dagger_\vec{p}\left(p_j\frac{\partial}{\partial{p}^k}-p_k\frac{\partial}{\partial{p}^j}\right) a_\vec{p}\tag{2}$$ How is this correctly calculated?

Here's what I did: As $T^{0j}=\dot\phi\partial^j\phi$ and \begin{align}\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)\tag{3}\\ \dot\phi=-i\int\frac{d^3p}{(2\pi)^3}\sqrt{\frac{E_\vec{p}}{2}}\left(a_\vec{p} e^{i\vec{p}\cdot\vec{x}}-a^\dagger_\vec{p} e^{-i\vec{p}\cdot\vec{x}}\right)\tag{4}\\ \partial^j\phi=-i\int\frac{d^3p}{(2\pi)^3}\frac{p^j}{\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)\tag{5}\end{align} then \begin{align}&Q_i=-2\epsilon_{ijk}\int{d^3x}\,x^k\dot\phi_p(x)\partial^j\phi_q(x)\\ &=\epsilon_{ijk}\sqrt{\frac{E_\vec{p}}{E_\vec{q}}}\int\frac{d^3pd^3qd^3x}{(2\pi)^6}x^kq^j\left(a_\vec{p}e^{i\vec{p}\cdot\vec{x}}-a^\dagger_\vec{p}e^{-i\vec{p}\cdot\vec{x}}\right)\left(a_\vec{q}e^{i\vec{q}\cdot\vec{x}}-a^\dagger_\vec{q}e^{-i\vec{q}\cdot\vec{x}}\right)\\ &=\cdots\left(a_\vec{p}a_\vec{q}e^{i(\vec{p}+\vec{q})\cdot\vec{x}}+a_\vec{p}^{\dagger}a_\vec{q}^{\dagger}e^{-i(\vec{p}+\vec{q})\cdot\vec{x}}-a_\vec{p}a^\dagger_\vec{q}e^{i(\vec{p}-\vec{q})\cdot\vec{x}}-a^\dagger_\vec{p}a_\vec{q}e^{-i(\vec{p}-\vec{q})\cdot\vec{x}}\right)\tag{6}\end{align} Now, \begin{align}\int{d^3x}\,x^ke^{i(\vec{p}\pm\vec{q})\cdot\vec{x}}=\mp{i}\frac{\partial}{\partial{p}^k}\int{d^3x}\,e^{i(\vec{p}\pm\vec{q})\cdot\vec{x}}=\mp{i}\,(2\pi)^3\frac{\partial}{\partial{p}^k}\delta(\vec{p}\pm\vec{q})\tag{7}\end{align} and so \begin{align}Q_i=-i\epsilon_{ijk}&\sqrt{\frac{E_\vec{p}}{E_\vec{q}}}\int\frac{d^3pd^3q}{(2\pi)^3}\\&q^j\left[(a_\vec{p}a_\vec{q}-a^\dagger_\vec{p}a^\dagger_\vec{q})\frac{\partial}{\partial{q}^k}\delta(\vec{p}+\vec{q})+(a_\vec{p}a^\dagger_\vec{q}-a^\dagger_\vec{p}a_\vec{q})\frac{\partial}{\partial{q}^k}\delta(\vec{p}-\vec{q})\right]\tag{8}\end{align} Now, integrating by parts on $q$, for example, for the first term \begin{align}\epsilon_{ijk}a_\vec{p}\int{d^3q}\,q^ja_\vec{q}\frac{\partial}{\partial{q}^k}\delta(\vec{p}+\vec{q})&=-\epsilon_{ijk}a_\vec{p}\int{d^3q}\,q^j\left[\frac{\partial}{\partial{q}^k}a_\vec{q}\right]\delta(\vec{p}+\vec{q})\\ &=\epsilon_{ijk}a_\vec{p}p^j\frac{\partial}{\partial(-p)^k}a_{-\vec{p}}\tag{9}\end{align} and this is where I differ from the question I mentioned at the beginning. Integrating all terms in $q$, \begin{align}Q_i=-i\epsilon_{ijk}&\int\frac{d^3p}{(2\pi)^3}\\&p^j\left[a_\vec{p}\frac{\partial}{\partial(-p)^k}a_{-\vec{p}}-a^\dagger_\vec{p}\frac{\partial}{\partial(-p)^k}a^\dagger_{-\vec{p}}-a_\vec{p}\frac{\partial}{\partial{p}^k}a^\dagger_\vec{p}+a^\dagger_\vec{p}\frac{\partial}{\partial{p}^k}a_\vec{p}\right]\tag{10}\end{align} Now, I'm guessing the first two terms vanish upon integration because they are odd in $p$, leaving \begin{align}Q_i=-i\epsilon_{ijk}\int\frac{d^3p}{(2\pi)^3}p^j\left[a^\dagger_\vec{p}\frac{\partial}{\partial{p}^k}a_\vec{p}-a_\vec{p}\frac{\partial}{\partial{p}^k}a^\dagger_\vec{p}\right]\tag{11}\end{align} So, if this expression is correct, how is it normal ordered?

What I tried was \begin{align}:a^\dagger_\vec{p}(\partial_ka_\vec{p})-a_\vec{p}(\partial_ka^\dagger_\vec{p}):&=a^\dagger_\vec{p}(\partial_ka_\vec{p})-:a_\vec{p}(\partial_ka^\dagger_\vec{p}):\\ &=a^\dagger_\vec{p}(\partial_ka_\vec{p})-(\partial_ka^\dagger_\vec{p})a_\vec{p}\\ &=2a^\dagger_\vec{p}(\partial_ka_\vec{p})-\partial_k(a^\dagger_\vec{p}a_\vec{p})\tag{12}\end{align} (where of course I'm using $\partial_k=\frac{\partial}{\partial{p}^k}$) which would yield the correct answer if $\partial_k(a^\dagger_\vec{p}a_\vec{p})=0$. I also noticed that $\partial_k(a^\dagger_\vec{p}a_\vec{p})=\partial_k(a_\vec{p}a^\dagger_\vec{p})$ since $[a_\vec{p},a^\dagger_\vec{p}]=(2\pi)^3\delta(0)$. However, I'm not really sure what's going on, whether if I made a mistake before or how to do the normal ordering if eq. (11) is right.

This post imported from StackExchange Physics at 2014-10-23 07:16 (UTC), posted by SE-user Pedro Figueroa
asked Oct 21, 2014 in Theoretical Physics by Pedro Figueroa (85 points) [ no revision ]
retagged Feb 1, 2015 by dimension10

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