# Temporal component of the normal ordered momentum operator

1. Dec 8, 2014

### mjordan2nd

1. The problem statement, all variables and given/known data

Consider the real scalar field with the Lagrangian $\mathcal{L}=\frac{1}{2}\partial_\mu \phi \partial^\mu \phi - \frac{1}{2} m^2 \phi^2$. Show that after normal ordering the conserved four-momentum $P^\mu = \int d^3x T^{0 \mu}$ takes the operator form

$$P^\mu = \int \frac{d^3p}{(2 \pi)^3} p^\mu a_p^\dagger a_p.$$

I have already showed that the three spatial components of the momentum operator satisfy the above. I'm left with showing that the temporal component of the normal-ordered momentum operator also satisfies the above.

2. Relevant equations

The classical temporal component, $T^{00},$ of the energy-mometum tensor is

$$T^{00}=\frac{1}{2} \dot{\phi}^2+\frac{1}{2} \left( \nabla \phi \right)^2 + \frac{1}{2}m^2 \phi^2.$$

To quantize this we use the following expansion for the fields

$$\phi(x) = \int \frac{d^3p}{(2 \pi)^3 \sqrt{2 E_p}} \left[ a_p e^{i \vec{p} \cdot \vec{x}} + a_p^\dagger e^{-i\vec{p} \cdot \vec{x}} \right].$$

Note that $a_p$ and $a_p^\dagger$ satisfy the typical commutation relations for creation and annihilation operators.

3. The attempt at a solution

After taking the appropriate derivatives, expansion, simplifications using delta functions, commutation relations, and imposing that everything lies on the mass-shell I have been able to show that

$$P^0 = \frac{1}{4} \int d^3p \left[ \left( a_p a_{-p} + a_p^\dagger a_{-p}^\dagger \right) \left( \frac{-2 \vec{p}^2}{E_p} \right) + \left( a_p a_p^\dagger + a_p^\dagger a_p \right) 2E_p \right],$$

where $E_p = p^0.$ I've been over this calculation twice, and am fairly confident that it is correct thus far, though I may still be wrong on that fact. If I only had the last two terms then this would be exactly what I was looking for. However, I can't see how to make the first two terms disappear in this case. For the spatial part I also had four terms, but instead of the first two terms being multiplied by $p^2$ they were only multiplied by $p$, making the first two terms odd, and thus disappear when integrated over the reals. In this case my first two terms are even, and so I'm a bit lost as to how to make them go away.

Any help would be appreciated.

Thanks.

2. Dec 9, 2014

### king vitamin

Unfortunately, it seems that your mistake was made in deriving the equation you've given, because the first two terms should cancel out.

I think your issue might be in neglecting that the exponents in the field expansion should contain time-dependence. For example, I would expect that your term $a_pa_{-p}$ should be multiplying a factor $e^{-2i\omega t}$ (if I got the sign in the exponent right).

3. Dec 9, 2014

### mjordan2nd

I wanted to explicitly do this in the Schrodinger picture. A friend and I were working on this together, and he decided to do this in the Heisenberg picture while I did it in the Schrodinger picture. I don't see where the exponentials would come from in the Schrodinger picture. I can post the gory details if you think that would be helpful.