Noether charge for Lorentz Transformation in terms of creation / annihilation operators

Question

I am trying to compute the conserved charge for a continuous Lorentz symmetry for a real scalar field in terms of creation / annihilation operators. So I have,

$$\mathcal{L} = \frac{1}{2} \partial_{\mu} \phi \partial^{\mu}\phi - \frac{1}{2}m^2 \phi^2$$

Following a similar argument as What conservation law corresponds to Lorentz boosts?, I am able to show that, the conserved charge is,

$$M^{\mu \nu} = \int \left( x^{\mu} T^{0 \nu} - x^{\nu} T^{0 \mu} \right) \mathrm{d}^3x$$

However, I need to express this in terms of creation and anhilation operators. So I started by writing down the terms for my stress-energy tensor.

For this problem, we are writing the Lorentz Transformation as :

$$\Lambda^{\mu}_{\nu} = \delta^{\mu}_{\nu} + \omega^{\mu}_{\nu}$$

where $\omega$ is anti-symmetric. Because of this, our stress-energy tensor is also going to be anti-symmetric and so $T^{00} = 0$. This allows me to write:

$$ T^{0\mu} = \frac{\partial \mathcal{L}}{\partial(\partial_0 \phi)} \partial^{\mu} \phi\; - \; \delta^{0 \mu} = \frac{1}{2} \partial^{0} \phi \; \partial^{\mu} \phi$$

Using this, I get,

$$M^{\mu \nu} = \frac{1}{2} \int_x x^{\nu} \partial^{0}\phi\; \partial^{\mu} \phi - x^{\mu} \partial^{0}\phi \; \partial^{\nu} \phi$$

If we're using the (1, -1, -1, -1) signature for our metric, we can write,

$$M^{\mu \nu} = \frac{1}{2} \int_x x^{\nu} \dot{\phi}\; \partial^{\mu} \phi - x^{\mu} \dot{\phi} \; \partial^{\nu} \phi = \frac{1}{2} \int_x x^{\nu}\; \Pi(x, t)\; \partial^{\mu} \phi - x^{\mu} \; \Pi(x, t) \; \partial^{\nu} \phi$$

where $\Pi(x, t)$ is our canonical conjugate momentum (density).

Now, I'm a little iffy about the following part but what I did next was, argue that because $x^{\mu}$ is going to be just a component of the position 4-vector (and so a number), I can move it outside the integral to write

$$ M^{\mu \nu} = \frac{1}{2} \left[ x^{\mu} \int \mathrm{d}^3 x \; \Pi(x, t) \partial_{\nu} \phi - x^{\nu} \int \mathrm{d}^3 x \; \Pi(x, t) \partial_{\mu} \phi \right]$$

The signs have flipped as I lowered the index. This gives,

$$ M^{\mu \nu} = \frac{1}{2} \left[ x^{\mu} P_{\nu} - x^{\nu} P_{\mu} \right] = \frac{1}{2} \left[ x^{\nu} P^{\mu} - x^{\mu} P^{\nu} \right] $$

This makes sense because this takes the form of angular-momenta but I don't know how to simplify this to get it in terms of a sum of a product of creation/anhilation operators, one of which is differentiated. I am stuck here.

Answer 1

You did not write down the creation and annihilation operators you title your question with. I'll let you write the full $T^{0j}=:\Pi \partial_j\Phi:~$ in terms of canonical 3D oscillator modes, $$ [a(k),a(k')]=[a^\dagger (k),a^\dagger (k')]=0, \\ [a(k),a^\dagger (k')=(2\pi)^3~2\omega_k ~\delta^3(k-k'),\\ \Phi({\mathbf x})=\int\frac{d^3k}{(2\pi)^3 2\omega_k} (a(k)e^{i{\mathbf k\cdot x}}+a^\dagger (k) e^{-i{\mathbf k\cdot x}}),\\ \Pi({\mathbf x})=-i\int\frac{d^3k}{(2\pi)^3 2 } (a(k)e^{i{\mathbf k\cdot x}}-a^\dagger (k) e^{-i{\mathbf k\cdot x}}), $$ for $\omega_k=\sqrt{m^2+{\mathbf k}^2}$. Scalar field theory is a simple repackaging of an infinity of oscillators: go to your favorite text. In any case, slipping the x s out of the x-integrals is plain unsound!

It is then evident then that $$ M_{0j}= i\int \frac{d^3k}{(2\pi)^3 2\omega_k} a^\dagger(k) \left(\omega_k \frac{\partial}{\partial k^j}\right)a(k)~, \\ M_{jl}= i\int \frac{d^3k}{(2\pi)^3 2\omega_k} a^\dagger(k) \left( k_j\frac{\partial}{\partial k^l} - k_l\frac{\partial}{\partial k^j} \right )a(k)~. $$ Can you now verify they lead to the requisite Lie algebraic commutators for the group?

Can you check the "classical" field increment of the action of the rotation piece on the field, $$ [M_{jl},\Phi(x)]\propto i \left( x_j\frac{\partial}{\partial x^l} - x_l\frac{\partial}{\partial x^j} \right )\Phi(x)~~? $$