# Klein-Gordon equation in the non-relativistic and semiclassical limit in a Wigner approach

I would like to analyse the semiclassical and non-relativistic limit of the Klein-Gordon equation, $$$$\frac{1}{c^2} \partial_t^2 \phi - \Delta \phi + \frac{M^2 c^2}{\hbar^2} \phi =0. \label{KGEscalar}$$$$

Therefore I used a Wigner transformation and passed to the limit $$\epsilon = \frac{1}{c^2}=\hbar \to 0$$. The evolution equation for the Wigner density $$\omega^{\epsilon}$$ in the limit which I obtained in both cases is $$$$\frac{\partial}{\partial t} \omega^{0} =0.$$$$ This means that the "quasi probability" to detect a particle at a specific location in the phase space is constant, i.e. it is zero when we demand any normalization conditions.

I mean this is not surprisingly when we consider the Klein-Gordon equation above, if we let $$\epsilon \to 0$$ this of course would lead to problems. But this seems really strange to me from a physical point of view. In the nonrelativistic limit we have that the rest energy become unbounded and therefore can cause problems. But what happens in the semiclassical limit, i.e. $$\hbar \to 0$$? At the moment I'm not sure how to intepret this result.

To be clear, I defined the Wigner transformation for $$L^2$$ functions $$f, g$$ as

$$$$\omega^{\epsilon} (f, g) (x, \xi) := \int _{\mathbb{R}} f(x-\frac{v} {2} \epsilon) g(x-\frac{v} {2} \epsilon) e^{i v \xi } d v.$$$$

Then choose $$f=g= \phi$$. The evolution equation for $$\omega^{\epsilon}_q (\phi, \phi)$$ can easily be obtained by a Fourier transformation $$\mathscr{F} (x \mapsto \zeta)$$and reads $$$$\partial_t \omega^{\epsilon}_q - i \zeta \int _{-\frac{1}{2} }^{\frac{1}{2} } \nabla \lambda_q (\xi - s \epsilon \zeta) ds=0.$$$$ Where $$\lambda_q$$ are the eigenvalues of the associated symbol of the differential operator of the Klein Gordon equation. The constants $$c$$ and $$\hbar$$ enter in the symbol.

• Frankly, I have no idea where the last equation emerged from... the single time derivative? the absence of a star product? As I indicate in my answer, it is so much easier to look at all the plane-wave solutions of the K-G equations than pretend to solve it from first principles in phase space... Commented Nov 19, 2018 at 23:22
• The single time derivative simply comes from expressing the KGE as a first order system of equations. The last equation come from manipulating the expressions defined above and computing the eigenvalues of the associated symbol of the differential operator. I would like to understand the limits of the KGE in the framework of pseudo differential operators, therefore I chose the approach above Commented Nov 19, 2018 at 23:31
• Should be linear in $\omega$... actually the star convolution should vanish before any limit.... Commented Nov 20, 2018 at 4:01
• I'm not sure which convolution operation you mean, but when one is passing to the limit $\epsilon \to 0$, the evolution equation is in general non linear, e.g. when one is considering the Schrödinger equation with a harmonic oszillator potential instead Commented Nov 20, 2018 at 8:30
• Moyal’s equation is linear. Be more explicit in your ill defined evolution equation. Commented Nov 20, 2018 at 12:13

I would go to the nonrelativistic limit in a different way. Substitute $$\Psi = e^{-mc^2/\hbar} \phi$$ and neglect the second time derivative of $$\phi$$. The result is the Schrödinger equation, so it is sufficient to study the Schrödinger equation in the semiclassical limit.

Assuming you are considering the plane-wave solution of the K-G equation, take it to be in 1+1 for computational simplicity, $$\phi \propto \exp \left (-it\sqrt{k^2 + M^2c^2/\hbar^2} +ikx\right ),$$ the (real) Wigner transform of its density matrix would be proportional to the unavoidable $$\int dy e^{-iyp} e^{-ik(x-y\hbar/2)+ik(x+y\hbar/2 ) } \propto \delta (p-\hbar k).$$

The $$k\mapsto -k$$ case is, equivalently, also a solution. c does not enter because the time dependence washes out (energy-momentum conservation), while $$\hbar$$ may be absorbed into the units of the initial (conserved) wavenumber k. As un-normalizable plane waves, issues of normalization are moot. Perhaps you wish to provide more clearly and explicitly your normalization considerations.

In consequence, as you might expect for a free wave, the Wigner function is nothing but conservation of momentum, in any limit. You never had to take such.

It is independent of time, of course, and a plain flat line parallel to the x-axis, as in the non-relativistic case.

• The nonrelativistic limit of the Klein-Gordon equation is the Schrodinger equation. You can find it by substituting $\psi =e^{imc^2t/\hbar}\phi$ and neglecting the second order time derivative of $\phi$. Commented May 17, 2020 at 9:52
• Obviously. What is your point? What is your Wigner transform? You are looking elsewhere! Commented May 17, 2020 at 11:33
• I removed the Wigner transformation by Occam's razor. Commented Jan 15, 2021 at 17:49
• But that's all the OP is asking about, no? Commented Jan 15, 2021 at 18:44