Derivation of momentum in QFT - from Energy-Momentum Tensor The conserved 4-momentum operator for the complex scalar field $\psi = \frac{1}{\sqrt{2}}(\psi_1 + i\psi_2)$ is given in terms of the mode operators in $\psi$ and $\psi^{\dagger}$ as $$P^{\nu} = \int \frac{d^3 p}{(2\pi)^3 }\frac{1}{2 \omega(p)} p^{\nu} (a^{\dagger}(p)  a(p) + b^{\dagger}(p) b(p))$$
This is just stated in my notes but I would like to see how to get to it using the mode operators. The lagrangian for the complex scalar field is $$ \mathcal L = \partial_{\mu} \psi^{\dagger} \partial^{\mu} \psi - m^2 \psi^{\dagger} \psi.$$ The the stress energy tensor associated with this theory is $$T^{\mu \nu} = \frac{\partial \mathcal L}{\partial (\partial_{\mu}\psi)} \partial^{\nu} \psi + \partial^{\nu} \psi^{\dagger} \frac{\partial \mathcal L}{\partial (\partial_{\mu} \psi^{\dagger})}  - \mathcal L\delta^{\mu \nu},$$ which using the lagrangian gives $$T^{\mu \nu} = \partial^{\mu} \psi^{\dagger} \partial^{\nu} \psi + \partial^{\nu} \psi^{\dagger}\partial^{\mu} \psi  - \mathcal L\delta^{\mu \nu}$$
Then $$P^{\nu} = \int T^{0 \nu} d^3 x = \int (\partial^{0} \psi^{\dagger} \partial^{\nu} \psi + \partial^{\nu} \psi^{\dagger}\partial^{0}\psi  - \mathcal L\delta^{0\nu}) d^3 x $$so$$P^0= \int (\partial^{0} \psi^{\dagger} \partial^{0} \psi + \partial^{0} \psi^{\dagger}\partial^{0}\psi - \partial_0 \psi^{\dagger} \partial^0 \psi - \partial_i \psi^{\dagger} \partial^i \psi + m^2 \psi^{\dagger}\psi) d^3 x $$
Similarly, I obtain $$P^i = \int d^3 x (\partial_0 \psi^{\dagger} \partial^i \psi + \partial^i \psi^{\dagger} \partial_0 \psi)$$
I understand how the expression for $P^0$ is derived using the integral I  have wrote above but the expression for $P^i$ is incorrect by a sign. I see in my notes they have indeed the integral expression for $P^i$ that I got but a minus in front. But I am not sure about the source of this minus.  Perhaps I am missing something conceptually in the derivation of $P^i$ then. Thanks for any comments.
 A: Inserting the expansion
$$
\psi=\int\frac{d^3p}{(2\pi)^32\omega_p}(a_pe^{-ipx}+b_p^\dagger e^{ipx})
$$
into the expression for the Hamiltonian
$$
H=\int d^3x(\dot{\psi}^\dagger\dot{\psi}+\nabla\psi^\dagger\cdot\nabla\psi+m^2\psi^\dagger\psi)
$$
we get
$$
H=\int d^3x\int\int\frac{d^3p}{(2\pi)^32\omega_p}\frac{d^3p^{\prime}}{(2\pi)^32\omega_p^{\prime}}(A+B+C)
$$
where
$$
\begin{array}{l}
A=\omega_p\omega_{p^\prime}(a_p^\dagger e^{ipx}-b_p e^{-ipx})(a_{p^\prime}e^{-i{p^\prime}x}-b^\dagger_{p^\prime}e^{i{p^\prime}x})\\
B=\vec{p}\cdot \vec{p}^\prime(a_p^\dagger e^{ipx}-b_p e^{-ipx})(a_{p^\prime}e^{-i{p^\prime}x}-b^\dagger_{p^\prime}e^{i{p^\prime}x})\\
C=m^2(a_p^\dagger e^{ipx}+b_p e^{-ipx})(a_{p^\prime}e^{-i{p^\prime}x}+b^\dagger_{p^\prime}e^{i{p^\prime}x})
\end{array}
$$
The integration over $x$ yields two kinds of combinations: $a_p^\dagger a_{p^{\prime}}(2\pi)^3\delta(\vec{p}-\vec{p}^{\prime}),b_p b_{p^{\prime}}^\dagger(2\pi)^3\delta(\vec{p}-\vec{p}^{\prime})$ and $a_p^{\dagger} b_{p^{\prime}}^\dagger(2\pi)^3\delta(\vec{p}+\vec{p}^{\prime}),b_p a_{p^{\prime}}(2\pi)^3\delta(\vec{p}+\vec{p}^{\prime})$. The expectation value of the latter ones in any momentum eigenstate is obviously zero, so they have no contribution to the Hamiltonian. Integrating over $p^{\prime}$ and using the relation  $\omega_p^2=|\vec{p}|^2+m^2$  we'll get
$$
H=\int \frac{d^3p}{(2\pi)^32\omega_p}\omega_p(a_p^\dagger a_p+b_p b_p^\dagger)
$$
A: For the complex momentum,
$$
T^{\mu\nu}= \partial^{\mu}\phi^{\dagger}(x)\partial^{\nu}\phi(x) +  
    \partial^{\nu}\phi^{\dagger}(x)\partial^{\mu}\phi(x) - g^{\mu}_{\nu}\mathcal{L}
$$
Now you can consider two separate cases:


*

*$T^{0i}$ which gives the 3-momentum, $P^{i}$ i.e. $g^{0}_{i} = (0,0,0)$

*$T^{00}$ which gives the hamiltonian, $H = P^{0}$ i.e. $g^{0}_{i} = 1$ (dependent on metric may, be -1)


working from here is straightforward and this method is quoted in:


*

*Greiner, pg. 82

*Weinberg, pg. 310


If would also strongly suggest looking at page 286 of Schwabl as he has an interesting result involving,
$$
\mathbf{P} = \sum_{p} \mathbf{\mathbf{p}} 
\Big( \hat{n}_{a(\ mathbf{p})} + \hat{n}_{b(\ mathbf{p})} \Big)
$$
