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?

  • $\begingroup$ The idea that $a_{-\mathbf{p}} = a^\dagger_{\mathbf{p}}$ cannot be true. Otherwise $\Pi = 0$ always $\endgroup$ Mar 5, 2020 at 16:44

2 Answers 2


The problem here was when doing the integration by parts. One has to be very careful when remembering which operators act on which expressions. In my case, it turned out that the derivative was acting on the wrong terms when doing the integration by parts, which indeed did add an extra minus sign.


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.

  • $\begingroup$ 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 $\vec{q} \rightarrow \pm \vec{p}$? $\endgroup$
    – drglove
    Mar 27, 2014 at 17:34

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