Klein-Gordon equation in the nonrelativistic limit

[George Keeling]

TL;DR

Don't understand reasoning in book.

I am reading Relativistic Quantum Mechanics Wave Equations, 3rd ed. - W. Greiner and I'm on section 1.3 looking at a solution $\psi$ to the Klein-Gordon equation in the nonrelativistic limit.

The solution is split up:
$$\psi=\phi\exp{\left[-\frac{i}{\hbar}m_0c^2t\right]}$$and we then show that in the nonrelativistic limit $\phi$ obeys the Schrödinger equation (or "the free Schrödinger equation for spinless particles"). So far so good. But then Grainer writes "As the type of particle which is described by a wave equation does not depend upon whether the particle is relativistic or nonrelativistic, we infer that the Klein-Gordon equation describes spin-zero particles".

If $\psi$ had obeyed the Schrödinger equation in the nonrelativistic limit, I would understand the inference but not $\phi$. Can anybody enlighten me?

 

[JimWhoKnew]

If $\psi$ had obeyed the Schrödinger equation in the nonrelativistic limit, I would understand the inference but not $\phi$. Can anybody enlighten me?

Try to back-substitute$$\phi=\psi\exp{\left[\frac{i}{\hbar}m_0c^2t\right]}$$in equation 1.32 .
What equation does $\psi$ obey in this non-relativistic limit?
What is its spin?
Express the solutions for this NR $\psi$ in terms of $~\vec{r},t,\vec{k}~$, and compare with the relativistic Klein-Gordon solutions.

Additional ways to look at it:
- At any given time $\psi$ differs from $\phi$ by a global phase factor.
- In the formalism of NRQM $\psi$ is gauge equivalent to $\phi$.
- If the relativistic $\psi$ had non-zero spin, it would have possessed more than one (field) degrees of freedom. These would have been carried over to the NR $\phi$ .

 

[George Keeling]

Oh yes. I quickly get to
$$i\hbar\frac{\partial\psi}{\partial t}=-\frac{\hbar^2}{2m_0}\nabla^2\psi+m_0c^2\psi$$Which is the Schrödinger equation for a free particle in a (rather large) constant potential of $m_0c^2$. I guess that will do. And adding a constant potential is the gauge equivalent thing, I think. What a shame Greiner didn't add that in.

Thank you!

 

[JimWhoKnew]

Oh yes. I quickly get to
$$i\hbar\frac{\partial\psi}{\partial t}=-\frac{\hbar^2}{2m_0}\nabla^2\psi+m_0c^2\psi$$Which is the Schrödinger equation for a free particle in a (rather large) constant potential of $m_0c^2$.

From relativity, having the rest mass as the limit of the energy when the kinetic contribution goes to zero, is a welcome feature ($~E=\sqrt{m_0^2+p^2}~$).

And adding a constant potential is the gauge equivalent thing, I think.

Classically, the equations of motion are invariant under addition of a constant to the Hamiltonian (energy spectrum shift). The more general symmetry is the invariance of the Euler-Lagrange equations under addition of a total derivative $~\frac{d}{dt}F(q,t)~$ to the Lagrangian. This symmetry is the origin of the gauge in the present context. A discussion of its consequences in NR quantization can be found in "Photons and Atoms" by Cohen-Tannoudji et al.

What a shame Greiner didn't add that in.

Since Greiner promises to discuss spin thoroughly later on, and does so, I think he can be excused.