# Derivation of momentum in QFT - from Energy-Momentum Tensor [closed]

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.

## closed as off-topic by ACuriousMind♦, Gert, JamalS, Kyle Kanos, John DuffieldNov 21 '15 at 9:19

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• I'm doing the same problem: The $m^2$ term just won't go anywhere for me. I first thought perhaps it was because the metric, $g_{0}^{\nu}=\left(1,0,0,0\right)$ and this part has no time dependence. However, that would not work for the Hamiltonian which contains the $m$ term when using $T^{00}$... – Alexander McFarlane Nov 20 '15 at 12:01
• It's a bit of a tedious derivation which gets a bit easier when you realize $P^0$ is just the Hamiltonian, but still, it's just being careful, especially with how the derivatives act on the mods expansion and remembering what $\omega(p)$ is. It's done pretty explicitly in chapter 1.5 of these notes. – ACuriousMind Nov 20 '15 at 13:08
• In your calculation, what happens to the terms without $m^2$? If you include those, you should be able to factorize out an $\omega(p)^2 = (m^2 + \vec p^2)$ from the integrand, which cancels the $\omega(p)^2$ in the denominator. (Note that $p^0 / \omega(p) = 1$.) – Noiralef Nov 20 '15 at 13:17
• @Noiralef: Yes exactly, I use the energy momentum relationship to write $p_0^2 - p_i^2 = m^2$ which allows me to combine the terms that did not depend on m. Basically I have $$\frac{p_0^2 - p_i^2}{w(p)^2} + \frac{m^2}{w(p)^2} = 2\frac{m^2}{w(p)^2}$$ as written in the last equation in my post. This is where I am stuck. – CAF Nov 20 '15 at 13:22
• To those who decided to close my thread, please consider the edit I have now made to reopen it. – CAF Nov 21 '15 at 9:46

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 $$H=\int \frac{d^3p}{(2\pi)^32\omega_p}\omega_p(a_p^\dagger a_p+b_p b_p^\dagger)$$

• Thanks for your answer. Could you please explain though why 'the expectation values of the latter ones in any momentum eigenstate is obviously zero'? is it because those terms come with an oscillating exponential and therefore vanish over all p? – CAF Nov 21 '15 at 8:59
• @CAF For example, $a_p^\dagger b_{p^{\prime}}^\dagger$ creates a particle with momentum $p$ and an antiparticle with momentum $p^{\prime}$, thus applying it to a momentum eigenstate produces another eigenstate that differs from the original one, so their inner product is zero. – Xavier Nov 21 '15 at 10:16
• So do you mean that $a_p^{\dagger}b_{p'}^{\dagger}|0\rangle = |u_p, \bar v_{p'}\rangle$? But in what way will the eigenstate differ if we apply this operator onto the state again? Won't it just produce a state with two particles of momentum $p$ and two antiparticles of momentum $p'$? – CAF Nov 21 '15 at 10:31
• @CAF Not exactly. Let $|\alpha\rangle$ be any eigenstate of momentum and particle number, then $\langle \alpha|a_p^\dagger a_{p^\prime}|\alpha \rangle$ vanishes unless $p=p^{\prime}$. Here we have $\delta(\vec{p}-\vec{p}^{\prime})$, so it remains; but $\langle \alpha|a_p^\dagger b_{p^\prime}^\dagger|\alpha \rangle$ will always vanish. Because the matrix of the sum of those terms without a creation and an annihilation operator being in pairs is diagonal, it vanishes as well. – Xavier Nov 21 '15 at 15:26
• @CAF $\partial^\mu$ is defined as $\partial^\mu=\frac{\partial}{\partial x_{\mu}}$, and for $i=1,2,3$, $\partial^i=\frac{\partial}{\partial x_i}=-\frac{\partial}{\partial x^i}$, thus $\partial^i(p^j x^j)=-p^i$. You make a mistake about the sign here. – Xavier Nov 21 '15 at 16:51

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:

1. $T^{0i}$ which gives the 3-momentum, $P^{i}$ i.e. $g^{0}_{i} = (0,0,0)$
2. $T^{00}$ 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,

$$\mathbf{P} = \sum_{p} \mathbf{\mathbf{p}} \Big( \hat{n}_{a(\ mathbf{p})} + \hat{n}_{b(\ mathbf{p})} \Big)$$

• Thanks for your response. Two things I don't understand: It is obvious (by inspection) that the lagrangian density vanishes for field configurations that solve the Dirac equation, but this is not so clear for the lagrangian for a complex scalar field. Also, as far as I can see, $\mathcal L$ for a complex scalar field does contain a term of the form $\partial_{\mu}\psi^{\dagger}$ (As it must otherwise $\mathcal L$ is not real). Please tell me if you agree with this. – CAF Nov 20 '15 at 16:44
• I actually got told by my professor that it was something to do with the $g^{i0}$ actually being a summation over the 3-field rather than the 4-field, hence it only contains $\mathbf{0}$. It seems to be one of those derivations where every book and resource uses the annoying phrase "it is clear to see" when it really isn't that clear to a student learning the topic! Fundamentally though, you can safely know that it does vanish to finish your proof. I will try and get more details on exactly why over the next day or so – Alexander McFarlane Nov 20 '15 at 17:25
• Yes I agree with you for the evaluation of $P^i$. In that case there is component $g^{0i}$ multiplying $\mathcal L$ so indeed it vanishes. However, in the case of $P^0$ this is not true and is the case I am having difficulty with. Although, I did derive it using the observation by ACuriousMind but I still would like to see why my method is not giving me what I want. – CAF Nov 20 '15 at 17:31
• @CAF I have the formal answer: $g^{\nu}_0$ can be thought of in two separate cases 1) $g^{i}_0 = (0,0,0)$ and 2) $g^{0}_0 = 1$ (or -1 depending on your metric definition). 1) will give you the definition of $P^{\mu}$ and 2) will give you the definition of $H=P^{0}$. This is precisely the reason why the two definition are always explicitly separated out in all literature, else you would just be able to combine them as one. – Alexander McFarlane Nov 20 '15 at 18:00
• Just wondering, what starting expression did you use for computing $P^i$? I had $$P^i = \int (\partial_o \psi^{\dagger} \partial^i \psi + \partial^i \psi^{\dagger} \partial_o \psi) d^3 x$$ which follows exactly from the definition of the stress energy tensor yet does not give the right answer. I thought I had $P^i$ but I noticed a sign error in what I had previously. – CAF Nov 20 '15 at 20:23