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

Question

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.

Answer 1

Almost everything is correct except for the reasoning at step (12). It is not true that $\partial_{p_k} (a_\mathbf{p}^\dagger a_\mathbf{p}) = 0$. In case differentiating an operator with respect to momentum is unfamiliar, note that it is defined just like every other derivative, $$\frac{\partial}{\partial p_k} \mathcal{O}(\mathbf{p}) \equiv \lim_{\epsilon \to 0} \frac{\mathcal{O}(\mathbf{p} + \epsilon \hat{\mathbf{k}}) - \mathcal{O}(\mathbf{p})}{\epsilon}.$$ The operator $\mathcal{O}(\mathbf{p}) = a_{\mathbf{p}}^\dagger a_{\mathbf{p}}$ clearly depends on what $\mathbf{p}$ is, because it represents the number of particles with momentum $\mathbf{p}$. So its derivative doesn't vanish.

Instead, note that the error term is $$\int d\mathbf{p} \, p^j \frac{\partial}{\partial p_k} (a_{\mathbf{p}}^\dagger a_{\mathbf{p}}) = \int d\mathbf{p} \, \frac{\partial}{\partial p_k} (p^j a_{\mathbf{p}}^\dagger a_{\mathbf{p}}) - \delta^{jk} a_{\mathbf{p}}^\dagger a_{\mathbf{p}}.$$ The second term vanishes by the antisymmetry of $\epsilon^{ijk}$, while the first term vanishes because it's just the integral of a total derivative. This is not any different than saying that $$\int_{-\infty}^\infty dx \, \frac{df}{dx} = 0$$ except that the letter $x$ has been replaced with $p$. You might protest that $p^j a_{\mathbf{p}}^\dagger a_{\mathbf{p}}$ does not vanish as $p \to \infty$, but as the integration bounds go to infinity, the boundary terms essentially count the number of increasingly high-energy particles, which goes to zero for any finite-energy configuration. (This is a bit mathematically cavalier, and this kind of reasoning could fail in other contexts, but it's good enough for the free scalar field.)

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To complete the derivation very explicitly, after normal ordering we're left with $$Q_i=-2 i\epsilon_{ijk}\int\frac{d^3p}{(2\pi)^3} p^j a^\dagger_\vec{p} \partial_{p_k} a_\vec{p} = -2 i\epsilon_{ijk}\int\frac{d^3p}{(2\pi)^3} a^\dagger_\vec{p} p^j \partial_{p_k} a_\vec{p}.$$ Since $\epsilon_{ijk}$ is antisymmetric, only the part of the integrand antisymmetric in $j$ and $k$ contributes, giving $$Q_i= -i\epsilon_{ijk}\int\frac{d^3p}{(2\pi)^3} a^\dagger_\vec{p} p^j \partial_{p_k} - p^k \partial_{p_j} a_\vec{p}.$$ This is exactly what we wanted to show.

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Edit: to provide even further detail we'll show how the unwanted first two terms in (10) vanish. The stated reasoning in the OP is incorrect; the terms do not just vanish by symmetry. The first term is proportional to $$I = \int d\mathbf{p} \, p^j a_p \frac{\partial}{\partial p_k} a_{-p}.$$ By changing variables from $p \to -p$, we find that $$I = \int d\mathbf{p} \, (-p^j) a_{-p} \frac{\partial}{\partial (-p)_k} a_p = \int d\mathbf{p} \, p^j a_{-p} \frac{\partial}{\partial p_k} a_p = \int d\mathbf{p} \, p^j \left(\frac{\partial}{\partial p_k} a_p\right) a_{-p}$$ since lowering operators commute. Therefore, adding these two expressions, $$2I = \int d\mathbf{p} \, p^j \frac{\partial}{\partial p_k} (a_p a_{-p}).$$ As before, we integrate by parts. The term with $\partial p^j / \partial p_k = \delta^{jk}$ again vanishes by the antisymmetry of $\epsilon^{ijk}$. The remaining boundary term can be argued to have no contribution for finite-energy contributions, i.e. its matrix elements between any two such configurations vanishes. The same argument can be applied to the second term in (10).

Answer 2

What you did seems fine to me, and indeed you can show that $\frac{\partial}{\partial{p}^k}(a_\vec{p}^\dagger{a}_\vec{p})=0$, just take the definitions $$a_\vec{p}=\int{d^3x}\,e^{-i\vec{p}\cdot\vec{x}}\left[\frac{E_\vec{p}}{2}\phi(\vec{x})+\frac{i}{E_\vec{p}}\dot\phi(\vec{x})\right]\\ a_\vec{p}^\dagger=\int{d^3x}\,e^{i\vec{p}\cdot\vec{x}}\left[\frac{E_\vec{p}}{2}\phi(\vec{x})-\frac{i}{E_\vec{p}}\dot\phi(\vec{x})\right]$$ which you should be able to get from your (3) and (4) eqs, and use $\partial_x(AB)=A(\partial_x{B})+(\partial_x{A}){B}$; as the ladder operators won't get commuted, the two terms in the R.H.S. will differ by just a sign, the negative one that comes out with the annihilation operator.