It is claimed in [1] (3.2.1) that the momentum components of the energy-momentum tensor was found to be

\label{eqn:noetherCurrentScalarField:20}
\Be_n \int d^3 x T^{0 n} = \int d^3 k \Bk a_k^\dagger a_k.

I don’t see this result anywhere, so let’s calculate it.

First, from the Noether current for the scalar field Lagrangian in question, what is the energy-momentum tensor explicitly?

\label{eqn:noetherCurrentScalarField:40}
\begin{aligned}
T^{\mu \nu}
&= \Pi^\mu \partial^\nu \phi – g^{\mu \nu} \LL \\
&= \Pi^\mu \partial^\nu \phi – g^{\mu \nu} \inv{2} \lr{ \partial_\alpha \phi \partial^\alpha \phi – \mu^2 \phi^2 } \\
&= \Pi^\mu \Pi^\nu – g^{\mu \nu} \inv{2} \lr{ \Pi_\alpha \Pi^\alpha – \mu^2 \phi^2 } \\
&= \Pi^\mu \Pi^\nu – \inv{2} g^{\mu \nu} g_{\alpha\beta} \Pi^\beta \Pi^\alpha + \inv{2} g^{\mu \nu} \mu^2 \phi^2.
\end{aligned}

Consider some special cases for the indexes. For $$\mu = \nu = 0$$, the result is the Hamiltonian density

\label{eqn:noetherCurrentScalarField:200}
\begin{aligned}
T^{00}
&= \Pi^0 \Pi^0 – \inv{2} g^{0 0} \Pi_\alpha \Pi^\alpha + \inv{2} g^{0 0} \mu^2 \phi^2 \\
&= \Pi^0 \Pi^0 – \inv{2} \Pi_\alpha \Pi^\alpha + \inv{2} \mu^2 \phi^2 \\
&= \inv{2} \Pi^0 \Pi^0 – \inv{2} \Pi_n \Pi^n + \inv{2} \mu^2 \phi^2 \\
&= \inv{2} \Pi^2 + \inv{2} (\spacegrad \phi)^2 + \inv{2} \mu^2 \phi^2,
\end{aligned}

where $$\Pi^2 = (\partial_0 \phi)^2 \ne \partial^2 \phi$$. For any $$\mu \ne \nu$$ the off diagonal metric elements are zero, leaving just
\label{eqn:noetherCurrentScalarField:220}
T^{\mu\nu} = \Pi^\mu \Pi^\nu.

Finally, when $$n \ne 0$$, the remaining diagonal terms are
\label{eqn:noetherCurrentScalarField:240}
\begin{aligned}
T^{nn}
&= \Pi^n \Pi^n – \inv{2} g^{n n} \Pi_\alpha \Pi^\alpha + \inv{2} g^{n n} n^2 \phi^2 \\
&= \Pi^n \Pi^n + \inv{2} \Pi_\alpha \Pi^\alpha – \inv{2} \mu^2 \phi^2 \\
&= \inv{2} \Pi^2 + \Pi^n \Pi^n – \inv{2} \Pi^m \Pi^m – \inv{2} \mu^2 \phi^2 \\
&= \inv{2} \Pi^2 + \inv{2} \Pi^n \Pi^n – \inv{2} \sum_{m\ne n,0} \Pi^m \Pi^m – \inv{2} \mu^2 \phi^2 \\
&= \inv{2} \sum_{m = n,0} \Pi^m \Pi^m – \inv{2} \sum_{m\ne n,0} \Pi^m \Pi^m – \inv{2} \mu^2 \phi^2.
\end{aligned}

The canonical momenta are

\label{eqn:noetherCurrentScalarField:60}
\Pi^\mu
=
\partial^\mu
\int \frac{d^3 k}{(2\pi)^{3/2} \sqrt{ 2 \omega_k }} \lr{ a_k e^{-i k \cdot x} + a_k^\dagger e^{i k \cdot x} },

but
\label{eqn:noetherCurrentScalarField:80}
\begin{aligned}
\partial^\mu e^{i k \cdot x}
&=
\partial^\mu \exp\lr{ i k^\alpha x_\alpha } \\
&=
i k^\mu \exp\lr{ i k \cdot x },
\end{aligned}

so
\label{eqn:noetherCurrentScalarField:100}
\begin{aligned}
\Pi^\mu
&=
i
\int \frac{d^3 k k^\mu}{(2\pi)^{3/2} \sqrt{ 2 \omega_k }} \lr{ – a_k e^{-i k \cdot x} + a_k^\dagger e^{i k \cdot x} } \\
&=
i
\int \frac{d^3 k k^\mu}{(2\pi)^{3/2} \sqrt{ 2 \omega_k }} \lr{ – a_k e^{-i \omega_k t + \Bk \cdot \Bx} + a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx} }.
\end{aligned}

This gives
\label{eqn:noetherCurrentScalarField:120}
\begin{aligned}
\int d^3 x \Pi^\mu \Pi^\nu
&=
-\inv{2} \int d^3 x \inv{(2\pi)^3}
\int d^3 k d^3 j \frac{k^\mu j^\nu}{\sqrt{\omega_k \omega_j}}
\lr{ – a_k e^{-i \omega_k t + \Bk \cdot \Bx} + a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx} }
\lr{ – a_j e^{-i \omega_j t + \Bj \cdot \Bx} + a_j^\dagger e^{i \omega_j t – i \Bj \cdot \Bx} } \\
&=
-\inv{2} \int d^3 x \inv{(2\pi)^3}
\int d^3 k d^3 j \frac{k^\mu j^\nu}{\sqrt{\omega_k \omega_j}}
\lr{
a_k a_j e^{-i (\omega_j + \omega_k) t + (\Bj + \Bk) \cdot \Bx}
– a_k a_j^\dagger e^{i (\omega_j – \omega_k) t – i (\Bj -\Bk) \cdot \Bx}
– a_k^\dagger a_j e^{-i (\omega_j -\omega_k) t – (\Bk – \Bj) \cdot \Bx}
+ a_k^\dagger a_j^\dagger e^{i (\omega_j + \omega_k) t – i (\Bj + \Bk) \cdot \Bx}
} \\
&=
-\inv{2}
\int d^3 k d^3 j \frac{k^\mu j^\nu}{\sqrt{\omega_k \omega_j}}
\lr{
a_k a_j e^{-i (\omega_j + \omega_k) t } \delta^3(\Bj + \Bk)
– a_k a_j^\dagger e^{i (\omega_j – \omega_k) t } \delta^3(\Bj -\Bk)
– a_k^\dagger a_j e^{-i (\omega_j -\omega_k) t } \delta^3 (\Bk – \Bj)
+ a_k^\dagger a_j^\dagger e^{i (\omega_j + \omega_k) t } \delta^3 (\Bj + \Bk)
}.
\end{aligned}

There are two cases here to consider. The first is $$\nu = 0$$, for which

\label{eqn:noetherCurrentScalarField:140}
\int d^3 x \Pi^\mu \Pi^0
=
-\inv{2}
\int d^3 k k^\mu
\lr{
a_k a_{-k} e^{-2 i \omega_k t }
– a_k a_k^\dagger
– a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{2 i \omega_k t }
}.

For $$\nu \ne 0$$

\label{eqn:noetherCurrentScalarField:160}
\begin{aligned}
\int d^3 x \Pi^\mu \Pi^\nu
&=
-\inv{2}
\int d^3 k \frac{k^\mu k^\nu}{\omega_k}
\lr{
– a_k a_{-k} e^{- 2 i \omega_k t }
– a_k a_k^\dagger
– a_k^\dagger a_k
– a_k^\dagger a_{-k}^\dagger e^{ 2 i \omega_k t }
} \\
&=
\inv{2}
\int d^3 k \frac{k^\mu k^\nu}{\omega_k}
\lr{
a_k a_{-k} e^{- 2 i \omega_k t }
+ a_k a_k^\dagger
+ a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{ 2 i \omega_k t }
}.
\end{aligned}

Here’s a summary of these products

\label{eqn:noetherCurrentScalarField:300}
\int d^3 x \Pi^0 \Pi^0
=
-\inv{2}
\int d^3 k \omega_k
\lr{
a_k a_{-k} e^{-2 i \omega_k t }
– a_k a_k^\dagger
– a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{2 i \omega_k t }
},

\label{eqn:noetherCurrentScalarField:280}
\int d^3 x \Pi^n \Pi^0
= \int d^3 x \Pi^0 \Pi^n
=
-\inv{2}
\int d^3 k k^n
\lr{
a_k a_{-k} e^{-2 i \omega_k t }
– a_k a_k^\dagger
– a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{2 i \omega_k t }
},

\label{eqn:noetherCurrentScalarField:340}
\int d^3 x \Pi^m \Pi^n
=
\inv{2}
\int d^3 k \frac{k^m k^n}{\omega_k}
\lr{
a_k a_{-k} e^{- 2 i \omega_k t }
+ a_k a_k^\dagger
+ a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{ 2 i \omega_k t }
}.

For the mass term it was previously found that

\label{eqn:noetherCurrentScalarField:180}
\inv{2} \int d^3 x \mu^2 \phi^2
=
\frac{\mu^2}{4}
\int
d^3 k
\inv{ \omega_k }
\lr{
a_{-k} a_k e^{- 2 i \omega_k t }
+a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
+a_k a_k^\dagger
+a_k^\dagger a_k
}.

The Hamiltonian component has been previously calculated, and resolves to

\label{eqn:noetherCurrentScalarField:360}
\int d^3 x T^{00}
=
\inv{2}
\int d^3 k
\omega_k
\lr{
a_k a_k^\dagger
+ a_k^\dagger a_k
}.

The other diagonal components, for $$r \ne s \ne t$$ are
\label{eqn:noetherCurrentScalarField:380}
\begin{aligned}
\int d^3 x T^{rr}
&=
\int d^3 x
\lr{
\inv{2} \sum_{m = r,0} \Pi^m \Pi^m – \inv{2} \sum_{m = s,t} \Pi^m \Pi^m – \inv{2} \mu^2 \phi^2
} \\
&=
\inv{4}
\int d^3 k \frac{(k^r)^2 – (k^s)^2 – (k^t)^2 – \mu^2}{\omega_k}
\lr{
a_k a_{-k} e^{- 2 i \omega_k t }
+ a_k a_k^\dagger
+ a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{ 2 i \omega_k t }
}
-\inv{4}
\int d^3 k \omega_k
\lr{
a_k a_{-k} e^{-2 i \omega_k t }
– a_k a_k^\dagger
– a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{2 i \omega_k t }
} \\
&=
\inv{4}
\int d^3 k \frac{(k^r)^2 – (k^s)^2 – (k^t)^2 – \mu^2 – \omega_k^2}{\omega_k}
\lr{
a_k a_{-k} e^{- 2 i \omega_k t }
+ a_k^\dagger a_{-k}^\dagger e^{ 2 i \omega_k t }
}
+
\inv{4}
\int d^3 k \frac{(k^r)^2 – (k^s)^2 – (k^t)^2 – \mu^2 + \omega_k^2}{\omega_k}
\lr{
a_k a_k^\dagger
+ a_k^\dagger a_k
} \\
&=
\inv{2}
\int d^3 k \frac{ (k^r)^2 – \omega_k^2}{\omega_k}
\lr{
a_k a_{-k} e^{- 2 i \omega_k t }
+ a_k^\dagger a_{-k}^\dagger e^{ 2 i \omega_k t }
}
+
\inv{2}
\int d^3 k \frac{ (k^r)^2}{\omega_k}
\lr{
a_k a_k^\dagger
+ a_k^\dagger a_k
}.
\end{aligned}

This doesn’t have the nice cancelation that killed the time dependent terms in the Hamiltonian. Such cancellation also doesn’t appear in the off diagonal energy-momentum tensor components, which are

\label{eqn:noetherCurrentScalarField:400}
\begin{aligned}
\int d^3 x T^{n 0}
&=
\int d^3 x T^{n 0} \\
&=
-\inv{2}
\int d^3 k k^n
\lr{
a_k a_{-k} e^{-2 i \omega_k t }
– a_k a_k^\dagger
– a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{2 i \omega_k t }
},
\end{aligned}

and for $$m \ne n \ne 0$$
\label{eqn:noetherCurrentScalarField:420}
\int d^3 x T^{m n}
=
\inv{2}
\int d^3 k \frac{k^m k^n}{\omega_k}
\lr{
a_k a_{-k} e^{- 2 i \omega_k t }
+ a_k a_k^\dagger
+ a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{ 2 i \omega_k t }
}.

The \ref{eqn:noetherCurrentScalarField:400} result has time dependence that the stated result does not (but is linear in $$\Bk$$ as desired)? Did I miss something?

# References

[1] Michael Luke. PHY2403F Lecture Notes: Quantum Field Theory, 2015. URL https://piazza.com/utoronto.ca/fall2015/phy2403f/resources. [Online; accessed 02-Jan-2016].