- #1

Gleeson

- 30

- 4

- Homework Statement
- I've been studying Tong's QFT notes and am trying to do problem sheet 2, question 6 here: https://www.damtp.cam.ac.uk/user/tong/qft.html

We are asked to take the classical angular momentum of the field,

##Q_i = \epsilon_{ijk}\int d^3x \left(x^j T^{0k} - x^k T^{0j}\right)##,

and show that after normal ordering, the quantum operator ##Q_i## can be written as

##Q_i = -i \epsilon_{ijk}\int \frac{d^3p}{(2 \pi)^3}a_{\vec{p}}^{\dagger}\left(p^j \frac{\partial}{\partial p_k} - p^k \frac{\partial}{\partial p_j} \right) a_{\vec{p}} ##,

and hence to show that a stationary one-particle state has zero angular momentum.

- Relevant Equations
- None.

My attempt/questions:

I use ##T^{0i} = \dot{\phi}\partial^i \phi##, ##\dot{\phi} = \pi##, and antisymmetry of ##Q_i## to get:

##Q_i = 2\epsilon_{ijk}\int d^3x [x^j \partial^k \phi(\vec{x})] \pi(\vec{x})##.

I then plug in the expansions for ##\phi(\vec{x})## and ##\pi(\vec{x})## and multiply it out to get that:

##Q_i = -2i \epsilon_{ijk} \int \frac{d^3xd^3p d^3q}{(2\pi)^6}\sqrt{\frac{E_{\vec{q}}}{E_{\vec{p}}}} x^j i p^k\left(a_{\vec{p}}a_{\vec{q}} e^{i(p + q)x} - a_{\vec{p}}a_{\vec{q}}^{\dagger} e^{i(p - q)x} - a_{\vec{p}}^{\dagger}a_{\vec{q}} e^{i(q - p)x} + a_{\vec{p}}^{\dagger}a_{\vec{q}}^{\dagger} e^{-i(p + q)x} \right)##,

where ##x##, ##p##, and ##q## are all 3-vectors, but I'm dropping the arrow.Therefore

##Q_i = -2i \epsilon_{ijk} \int \frac{d^3p d^3q}{(2\pi)^3} \sqrt{\frac{E_{\vec{q}}}{E_{\vec{p}}}} p^k \left( a_pa_q \int \frac{d^3x}{(2\pi)^3} \frac{\partial}{\partial p_j}e^{i(p + q)x} - a_pa_q^{\dagger} \int \frac{d^3x}{(2\pi)^3} \frac{\partial}{\partial p_j}e^{i(p - q)x} + a_p^{\dagger}a_q \int \frac{d^3x}{(2\pi)^3} \frac{\partial}{\partial p_j}e^{i(q - p)x} - a_p^{\dagger}a_q \int \frac{d^3x}{(2\pi)^3} \frac{\partial}{\partial p_j}e^{-i(p + q)x}\right)##,

so

##Q_i = - 2 i \epsilon_{ijk}\int \frac{d^3p d^3q}{(2 \pi)^3} \sqrt{\frac{E_{\vec{q}}}{E_{\vec{p}}}}p^k \left((a_pa_q - a_p^{\dagger}a_q)\frac{\partial}{\partial p_j}\delta(p + q) + (a_p^{\dagger}a_q - a_p a_q^{\dagger})\frac{\partial}{\partial p_j}\delta(p - q) \right)##.

- I can add more intermediate lines of explanation to the above if needed

- Is this correct so far? If not, what is wrong?

- I believe I need to integrate by parts now, to get the derivatives off the delta functions. It looks like the ratio of Energies should eventually cancel due to a delta function, but I can't see how the next couple of lines should go. Won't the derivative end up on the delta Energy, and the Energy term will need to be derived before the delta function can act on it?

- If someone could walk me through the next couple of lines I would very much appreciate it.

Thanks

I use ##T^{0i} = \dot{\phi}\partial^i \phi##, ##\dot{\phi} = \pi##, and antisymmetry of ##Q_i## to get:

##Q_i = 2\epsilon_{ijk}\int d^3x [x^j \partial^k \phi(\vec{x})] \pi(\vec{x})##.

I then plug in the expansions for ##\phi(\vec{x})## and ##\pi(\vec{x})## and multiply it out to get that:

##Q_i = -2i \epsilon_{ijk} \int \frac{d^3xd^3p d^3q}{(2\pi)^6}\sqrt{\frac{E_{\vec{q}}}{E_{\vec{p}}}} x^j i p^k\left(a_{\vec{p}}a_{\vec{q}} e^{i(p + q)x} - a_{\vec{p}}a_{\vec{q}}^{\dagger} e^{i(p - q)x} - a_{\vec{p}}^{\dagger}a_{\vec{q}} e^{i(q - p)x} + a_{\vec{p}}^{\dagger}a_{\vec{q}}^{\dagger} e^{-i(p + q)x} \right)##,

where ##x##, ##p##, and ##q## are all 3-vectors, but I'm dropping the arrow.Therefore

##Q_i = -2i \epsilon_{ijk} \int \frac{d^3p d^3q}{(2\pi)^3} \sqrt{\frac{E_{\vec{q}}}{E_{\vec{p}}}} p^k \left( a_pa_q \int \frac{d^3x}{(2\pi)^3} \frac{\partial}{\partial p_j}e^{i(p + q)x} - a_pa_q^{\dagger} \int \frac{d^3x}{(2\pi)^3} \frac{\partial}{\partial p_j}e^{i(p - q)x} + a_p^{\dagger}a_q \int \frac{d^3x}{(2\pi)^3} \frac{\partial}{\partial p_j}e^{i(q - p)x} - a_p^{\dagger}a_q \int \frac{d^3x}{(2\pi)^3} \frac{\partial}{\partial p_j}e^{-i(p + q)x}\right)##,

so

##Q_i = - 2 i \epsilon_{ijk}\int \frac{d^3p d^3q}{(2 \pi)^3} \sqrt{\frac{E_{\vec{q}}}{E_{\vec{p}}}}p^k \left((a_pa_q - a_p^{\dagger}a_q)\frac{\partial}{\partial p_j}\delta(p + q) + (a_p^{\dagger}a_q - a_p a_q^{\dagger})\frac{\partial}{\partial p_j}\delta(p - q) \right)##.

- I can add more intermediate lines of explanation to the above if needed

- Is this correct so far? If not, what is wrong?

- I believe I need to integrate by parts now, to get the derivatives off the delta functions. It looks like the ratio of Energies should eventually cancel due to a delta function, but I can't see how the next couple of lines should go. Won't the derivative end up on the delta Energy, and the Energy term will need to be derived before the delta function can act on it?

- If someone could walk me through the next couple of lines I would very much appreciate it.

Thanks

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