# On Schrödinger equation and $U= 2mV$ potential

If we consider $\hbar=c=1$ we can write stationnary Schrödinger equation for a free particle as $$\left(\frac{k^2}{2m} + V - \frac{k_0^2}{2m}\right)\Psi = 0$$

I see often the use of a reduced potential $U=2mV$ for convenience. We then have $$(k^2 - k_0^2 + U)\Psi = 0$$

I guess that from the fact that we assume $H_0 = k^2/2m$ in the first place, we take the non relativistic limit that the kinetic energy is much smaller that the mass energy.

When working at higher kinetic energies, relativistic effects cannot be ignored. Kinetic energy is no longer $k^2/2m$. Though, in a lot of papers, I see that they define $U = 2\omega V$ where $\omega$ is the total energy $\omega=\sqrt{k^2+m^2}$ and still solve the same Schrödinger equation.

How can we justify an equation of the type $$(k^2 - k_0^2 + 2\omega V)\Psi = 0$$

Where does this $2\omega$ appear ? Does this come from a Klein Gordon equation correspondance? In here they find at the very end (3.137) some sort of equivalence between Schrödinger and Klein Gordon equation. The potential $V$ in the Klein Gordon equation (I'm no expert in K-G nor relativistic physics but does the potential concept exist for K-G eq? Is it the same object as in Schrödinger equation?) has a sort of equivalent $(2EV-V^2)/2m$ potential in the Schrödinger equation... But if this $\omega$ came from here, where does the $-V^2/2m$ disappear ? Or maybe it can be neglected?

• I'm a bit confused: Are you in the non-relativistic setting where the Schrödinger equation is valid, or in the relativistic setting where the Klein-Gordon or Dirac equation govern the dynamics? It seems you're asking why we don't take a relativistic expression into account for the non-relativistic case. Jan 11, 2017 at 16:35
• @ACuriousMind That's where lies my problem... All this seems pretty weird to me. Take Phys. Rev. C. 19 - 929, equation (4) for example (I maybe can add the paper if you can't access it, but I don't know if I'm allowed to...). They start from pionic atoms, so $k$ is the momentum of the pion and it's basically close to zero -> we solve a Schrödinger equation with $(\nabla−k_0^2+2\mu V)\Psi=0$, the aim being to find a potential $V$ at this energy. Up to there, ok. But then, they want to derive this potential for 50 MeV incident pions... They then use $(\nabla−k_0^2+2\omega V)\Psi=0$... ?? Jan 11, 2017 at 17:02
• They seem to consider the fact the pions now have a non negligible kinetic energy (50 MeV compared to 135 MeV of their rest mass is quite something) by having this $\omega$ replace the $\mu$. But where does this come from? That is my question... Jan 11, 2017 at 17:03
• In here they find at the very end (3.137) some sort of equivalent between Schrödinger and Klein Gordon equation, where the potential $V$ in the Klein Gordon equation (I'm no expert in K-G nor relativistic physics but does the potential concept exist for K-G eq?) finds an equivalent $(2EV-V^2)/2m$ Schrödinger potential... But if this $\omega$ came from here, where does the $-V^2/2m$ disappear ? Or maybe it can be neglected? Jan 11, 2017 at 17:27

In the end I found how to get to this form.

Schrödinger equation has the form $$(T+V-E_0)\psi = 0$$

Relativistically, the kinetic energy $T$ is given by $$T=\sqrt{k^2+m^2}-m$$ where again, $\hbar=c=1$ and all factors of these constants have been omitted.

Just as a quick notice, we get back to the low energy approximation considering $k<<m$. Then \begin{align} \sqrt{k^2+m^2} -m &= m\sqrt{1+k^2/m^2} -m \\ &\simeq m(1+k^2/2m^2)-m \quad\text{Taylor expansion}\\ \Rightarrow T \simeq k^2/2m& \end{align}

which is the non relativistic value for the kinetic energy.

Here we do not make this assumption since we are a bit higher in energy. The incident energy $E_0$ is given by the initial kinetic energy so it reads similarly $$E_0 = \sqrt{k_0^2+m^2}-m$$

If we replace this into the Schrödinger equation we have \begin{align} \left(\sqrt{k^2+m^2}-\sqrt{k_0^2+m^2}+V\right)\psi &= 0 \\ \left(\left[\sqrt{k^2+m^2}-\sqrt{k_0^2+m^2}\right]\cdot \frac{\sqrt{k^2+m^2}+\sqrt{k_0^2+m^2}}{\sqrt{k^2+m^2}+\sqrt{k_0^2+m^2}}+V\right)\psi &= 0 \\ \left(\frac{(k^2+m^2)-(k_0^2+m^2)}{\sqrt{k^2+m^2}+\sqrt{k_0^2+m^2}}+V\right)\psi &= 0 \end{align}

If we now define $\omega = \sqrt{k_0^2+m^2}$, the incident energy, we have $$\left(k^2 - k_0^2 + (\omega+\sqrt{k^2+m^2}) V\right)\psi=0$$

If we now make the approximation $\sqrt{k^2+m^2}\simeq \sqrt{k_0^2+m^2} =\omega$ we can rewrite this $$\left(k^2 - k_0^2 + 2\omega V\right)\psi=0$$

which is what we were looking for.