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Assume we have a real Klein Gordon field $\phi(x,y,z,t)$, and we do the non-relativistic expansion of it in terms of a complex field $\psi(x,y,z,t)$ $$\phi=\frac{1}{\sqrt{2m}}(\psi e^{-imt}+\psi^* e^{imt}).\tag{1}$$ Notice that in the non-relativistic limit, $\psi$ obeys the Schrodinger equation $$i\dot{\psi}=-\frac{\nabla^2\psi}{2m}.\tag{2}$$ We have the energy momentum tensor of the Klein Gordon field $$T^{00}=\frac{1}{2}\dot{\phi}^2+\frac{1}{2}(\nabla \phi)^2+\frac{1}{2}m^2\phi^2.\tag{3}$$ If we substitute $\phi$ in terms of $\psi$ and neglect all terms that are highly oscillating (i.e. containing factor of $e^{\pm 2imt}$ after the substitution ), we have the following expression of the $T^{00}$.

$$T^{00}=\frac{1}{4m}(2\dot{\psi}\dot{\psi^*}+2im\psi^*\dot{\psi}-2im\psi\dot{\psi^*}+4m^2|\psi|^2+2(\nabla \psi)\cdot(\nabla \psi^*)).\tag{4}$$

Then the energy of the Klein Gordon field should be given by $\int d^3x T^{00}$. Then the $4m^2|\psi|^2 $ term will integrate to give m, which is the rest energy of the particle

The $2\dot{\psi}\dot{\psi^*}$ term can be neglected, as it is suppressed by the non-relativistic factor compared with other term, for example

$$(\nabla \psi)\cdot(\nabla \psi^*)\sim p^2\psi^2,\tag{5}$$ $$\dot{\psi}\dot{\psi^*}\sim \frac{p^4}{m^2}\psi^2,\tag{6}$$ Where $p$ is approximately the momentum of the $\psi$ field, and $\frac{p}{m}\sim \frac{v}{c}\ll 1$ according to non-relativistic assumption.

However, it seems to me that the kinetic energy term is with the wrong coefficient, because $$\frac{i\psi^*\dot{\psi}-i\psi\dot{\psi^*}}{2}+\frac{1}{2m}(\nabla \psi)\cdot (\nabla \psi^*)=\psi^*(-\frac{\nabla^2}{m}\psi).\tag{7}$$ The equation makes use of the Schrodinger equation and is up to total spatial derivative, which plays no role in integration

Then, it seems to me that $$\int d^3x T^{00}=m+\langle\psi|-\frac{\nabla^2}{m}|\psi\rangle=rest \quad energy+ 2\times kinetic\quad energy.\tag{8}$$ That is, the coefficient of the kinetic energy is different from the naive expectation $E_{rel}=E_{rest}+E_{kin}$ by 2.

Can anyone tell me what is happening here? Why is it off by a factor of 2? I also have the question: When we talk about energy, which energy are we talking about? Which energy is observable or physical? For example, the field may represents some axion cloud that evolves around a gravitational potential. And the gravitational potential provider (some star, for example) may exchange the energy with the axion cloud. Which energy should I use when we discuss this "backreaction" problem? The energy of the $\phi$ field or the energy of the $\psi$ field?

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2 Answers 2

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  1. We can ignore the cross-terms in the Klein-Gordon (KG) action from OP's expansion $$\Phi(\vec{x},t)~=~\frac{\hbar}{\sqrt{2m}}\left(\exp\left(-\frac{imc^2t}{\hbar}\right)\psi(\vec{x},t) +\exp\left(\frac{imc^2t}{\hbar}\right)\psi(\vec{x},t)^{\ast}\right)\tag{1}$$ in the limit $c\to\infty$ due to Riemann-Lebesgue lemma. In order words, the particle-antiparticle interaction terms disappear, so we might as well focus on the particle sector alone $$\Phi(\vec{x},t)~=~\frac{\hbar}{\sqrt{2m}}\exp\left(-\frac{imc^2t}{\hbar}\right)\psi(\vec{x},t).\tag{1'}$$ (The anti-particle sector is similar.)

  2. We will follow standard conventions: For a real (complex) KG field $\Phi$, the action terms have (have not) a $1/2$ symmetry factor, respectively. Eq. $(1)$ is a real KG field, while eq. $(1')$ is a complex KG field.

  3. It is straightforward to derive the corresponding Schrödinger Lagrangian density $$\begin{align} {\cal L}_{KG}~=~&\left|\frac{1}{c}\partial_t \Phi\right|^2 - |\nabla \Phi|^2 -|\frac{mc}{\hbar}\Phi|^2\cr \stackrel{(1')}{\longrightarrow}&\underbrace{\frac{i\hbar}{2}\left(\psi^{\ast}\partial_t\psi-\partial_t \psi^{\ast}\psi \right)}_{\text{symplectic terms}} -\underbrace{\frac{\hbar^2}{2m}|\nabla \psi|^2}_{\text{non-rel. Hamiltonian density}}\cr &\quad {\rm for }\quad c~\to~\infty,\end{align}\tag{A}$$ cf. e.g. my Phys.SE answer here.

  4. The relativistic KG energy density is dominated by the rest energy $$\begin{align} {\cal T}^{00}_{KG}~=~&\left|\frac{1}{c}\partial_t \Phi\right|^2 + |\nabla \Phi|^2 +|\frac{mc}{\hbar}\Phi|^2\cr \stackrel{(1')}{\longrightarrow}&\frac{i\hbar}{2}\left(\psi^{\ast}\partial_t\psi-\partial_t \psi^{\ast}\psi \right) +\frac{\hbar^2}{2m}|\nabla \psi|^2 +\underbrace{mc^2|\psi|^2}_{\text{rest energy density}}\cr \stackrel{\begin{array}{c}\text{TDSE}\cr\text{IBP}\end{array}}{\sim}& \frac{\hbar^2}{m}|\nabla \psi|^2 +\underbrace{mc^2|\psi|^2}_{\text{rest energy density}}\cr &\quad {\rm for }\quad c~\to~\infty,\end{align}\tag{B}$$ and hence less useful in the non-relativistic limit.

  5. If we include the anti-particle sector and recall the normalization convention from section 2, then the RHS of eqs. (A) & (B) remain the same. Therefore eq. (B) agrees with OP's result (8).

  6. OP's issue is related to that the Legendre transformation from the Lagrangian to the Hamiltonian formulation grows singular in the limit $c\to\infty$, see e.g. this Phys.SE post. It is important to recognize that the symplectic terms are not part of the non-relativistic Hamiltonian in eq. (A); compare with OP's eq. (7).

  7. Whereas there is a smooth transition between the relativistic and non-relativistic actions, note that the non-relativistic notion of energy (which excludes rest energy) is a departure from the relativistic notion of energy (which includes rest energy). This feature is similar to what happens for the point particle, cf. e.g. my Phys.SE answer here.

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  • $\begingroup$ Thanks for the answer. I also would like to know, which energy should we actually use when we talks about energy conservation? The one with the factor 2 or the one without? $\endgroup$
    – Tan Tixuan
    May 13, 2022 at 6:06
  • $\begingroup$ That follows from section 2. $\endgroup$
    – Qmechanic
    May 13, 2022 at 6:30
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You are mixing the real Klein-Gordon field, which describes a neutral particle with the complex Schrödinger field. That is not consistent.

The correct approach starts with the complex Klein-Gordon field. Then write $$\phi = e^{imt} \psi \,.$$ This gives $$\dot \phi = im e^{imt} \psi + e^{imt} \dot \psi \,.$$ The non-relativistic result is obtained by neglecting terms that are nonlinear in $\dot \phi$, which happens in the energy-momentum tensor, or contain the second time derivative, which happens in the equation of motion.

An approach that does not work is to take the non-relativistic limit of the Lagrangian and then derive the equation of motion in terms of $\psi$.

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  • $\begingroup$ Hi, thanks for the answer. However, I think the first equation in my post is kind of quite standard techniques, for example, in arxiv.org/abs/1704.05081, equation (4). The same techniques is used to a real porca field $\endgroup$
    – Tan Tixuan
    May 9, 2022 at 9:13
  • $\begingroup$ Besides, I also checked the complex case. It seems the same problem still exists, i.e. the coefficient of the kinetic energy is off by 2 $\endgroup$
    – Tan Tixuan
    May 9, 2022 at 9:28

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