Temporal component of the normal ordered momentum operator

[mjordan2nd]

Homework Statement

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.

Homework 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.

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.

 

[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).

 

[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.