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.


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.)

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.

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).

  • $\begingroup$ Ok, that clarifies a lot. The two first terms of (10) however do not at all contain odd integrands. How to prove that they have a zero contribution? $\endgroup$ – jac Mar 2 at 15:24
  • $\begingroup$ In the reasoning used to show that what you call the 'error term' vanishes, you did not consider the factor $p^j$. It does not change the result however. One can use partial integration and the fact that $\epsilon_{ijk}\frac{\partial p^j}{\partial p_k}=0$ to get to the result that this term does not contribute. $\endgroup$ – jac Mar 2 at 15:33
  • $\begingroup$ @jac Thanks for the catch! I edited to address those two extra terms. $\endgroup$ – knzhou Mar 2 at 17:01
  • $\begingroup$ "since lowering operators commute" should be "since lowering operators and $\frac {\partial}{\partial p_k} commute. All clear to me now. $\endgroup$ – jac Mar 2 at 17:59
  • $\begingroup$ @jac No, I said that because I commuted $a_{-p}$ to the right in the last step. $\endgroup$ – knzhou Mar 3 at 0:43

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.

  • $\begingroup$ You say the two will differ by just a sign, but this is not entirely true, since you get for the derivative of the planar wave something like (without the factors ~ $p_k$: $$i(x-y)exp(ip(x-y))*(\phi(x)\phi(y)+\dot{\phi}(x)\dot{\phi}(y)+i/2(\dot{\phi}(x)\phi(y)-\phi(x)\dot{\phi}(y)))$$ and for the other : $$exp(ip(x-y))*((p_k/4)\phi(x)\phi(y)-(p_k/(E_k^2)\dot{\phi}(x)\dot{\phi}(y))$$ and it is not obvious for me that these cancel integrating wrt $x$ and $y$. $\endgroup$ – faero Mar 29 '15 at 12:04

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