Finding the normal ordered momentum operator for free theory

Question

I am asked to show that the normal ordered momentum operator for free theory is

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

The free theory Lagrangian is given by

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

I have shown that for this Lagrangian the energy-momentum tensor is

$$T^{\mu \nu} = \partial^\mu \phi \partial^\nu \phi - \eta^{\mu \nu}\mathcal{L}.$$

From here we can show that the energy is

$$p_0 = \int d^3x T^{00} = \int d^3x \left( \frac{1}{2} \phi^2 + \frac{1}{2} (\nabla \phi) ^2 + \frac{1}{2}m^2 \phi^2\right),$$

and the three-momentum is

$$p_i = \int d^3x T^{0i} = \int d^3x \:\dot{\phi} \partial^i \phi.$$

When quantizing these expressions, I am trying to work in the Schrodinger picture. In order to show the first expression, I will work with the spatial and temporal parts of the four-momentum separately due to the difference in their expressions in the classical field theory. At this point I'm only trying to show that the spatial part of the normal ordered momentum is consistent with the first equation of this post. Quantizing gives us

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

I am told that this expression actually contains the time part in the exponentials, but since this is the Schrodinger picture we choose to define this operator at some fixed time which we conveniently choose to be $t=0.$ Given that, the time derivative of the above expression is

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

and

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

Per my rough calculation this means that

$$\dot{\phi} \partial^i \phi=\int \frac{d^3p d^3k}{(2 \pi)^6} \sqrt{\frac{E_p}{4 E_k}} k_i \left[ a_p a_k e^{i(p+k)x} - a_p a_k^\dagger e^{i(p-k)x} - a_p^\dagger a_k e^{i(k-p)x} + a_p^\dagger a_k^\dagger e^{-i(p+k)x} \right].$$

Then, using the exponential definition of the delta function and integrating through x and p I found

$$p^i = \int d^3x \dot{\phi} \partial^i \phi= \int \frac{d^3k}{(2 \pi)^3} \frac{1}{2} k_i \left[a_k a_{-k} - a_k a_k^\dagger - a_k^\dagger a_k + a_k^\dagger a_{-k}^\dagger \right].$$

I appear to have made a mistake in sign somewhere but I'm not too worried about that -- I can go back and find it on my own time. That said,, this is something like what I want except for the first and last terms in square brackets. If I do this in the Heisenberg picture I will get an exponentially decaying factor attached to the first and last terms which I think will make those two terms disappear but I feel like I should be able to get the same result in the Schrodinger picture, and I can't quite understand why I'm not.

Answer 1

The first and last terms strictly and completely disappear. They are equal to zero because $k_i a_k a_{-k} $ (just like its hermitian conjugate) is an odd function of $\vec k$ – a function obeying $f(-\vec k) = -f(\vec k)$ – and the integral of an odd function vanishes because the contribution from the region around $\vec k$ is exactly cancelled by the contribution from the vicinity of $-\vec k$.

Without words, you may divide the $k_i a_k a_{-k}$ term as $$\frac 12 (k_i a_k a_{-k} + k_i a_k a_{-k} ) $$ and write the result as a sum of two integrals. In one of them, you may use the substitution $k\to (-k)$ which changes $k_i a_k a_{-k}$ to $-k_i a_{-k} a_k =-k_i a_k a_{-k}$ which is the same thing as the term you didn't change so they exactly cancel.