# Schrödinger equation with gravitational potential

In "On the Gravitization of Quantum Mechanics 1: Quantum State Reduction" by Roger Penrose, illustrates two different approaches to incorporating gravitational field into a quantum system.

Penrose states that these two wavefunctions should be related by a phase factor, maintaining consistency with Einstein’s equivalence principle.

$$i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \nabla^2 \psi + mgz \psi$$ $$i\hbar \frac{\partial \phi}{\partial t} = -\frac{\hbar^2}{2m} \nabla^2 \phi$$ $$\psi = \phi \exp \left( \frac{i}{\hbar} \left( -mgtz + \frac{1}{6}m g^2 t^3 \right) \right)$$ Could someone provide a detailed derivation of the phase factor that relates the Newtonian and Einsteinian wavefunctions?

• Nice question (+1) According to Wikipedia, "The key idea is that since space–time metric should be well defined, nature “dislikes” these space–time superpositions and suppresses them by collapsing the wave function to one of the two localized states." Could you elaborate in what way nature "dislikes" non well-defined metrics? How exactly will a non well-defined metric cause further problems downstream? Commented Aug 7 at 6:50
• I was able to find a derivation here: pubs.aip.org/aapt/ajp/article/84/11/879/1043874/…
– d_b
Commented Aug 7 at 7:27
• Most of the question looks AI-generated. If you didn't write it, you need to cite a source for it. Commented Aug 8 at 17:28

Einstein's equivalence principle can be stated in several ways. In its strong form, it implies for a body subjected to a gravitational field that, in every point of space, there exists a local frame of reference where the said body is free-falling $$-$$ which is equivalent to linearize the gravitational field by Taylor expansion $$-$$ and that this frame of reference can't be distinguished experimentally from a referential frame without gravitation, that is an inertial frame where the body's motion is completely free.
For the sake of simplicitly, let's consider a one-dimensional space with a frame of reference $$(z,t)$$ where a body of mass $$m$$ follows a free motion. In consequence, its wavefunction $$\phi(z,t)$$ satisfies the following Schrödinger equation : $$i\hbar\frac{\partial}{\partial t}\phi(z,t) = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial z^2}\phi(z,t) \tag{1}$$ Now, let's consider another body of mass $$m$$, which is subjected to a gravitational field in the referential frame $$(z',t')$$. This gravitational field may be approximated locally by the linear potential $$V(z',t') = mgz'$$, which leads to a local free fall motion by Newton's law, i.e. $$\frac{\mathrm{d}^2z'}{\mathrm{d}t'^2} = -g$$. Let's name $$\psi(z',t')$$ the wavefunction associated to the said (locally) free-falling body. Its Schrödinger equation is given by $$i\hbar\frac{\partial}{\partial t'}\psi(z',t') = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial z'^2}\psi(z',t') + V(z',t')\psi(z',t') \tag{2}$$
How are the two frames of reference related to each other ? Assuming a non-relativistic context, one can take $$t' = t$$. Then, the transformation $$\begin{cases} z = z' + \frac{1}{2}gt'^2 \\ t \,= t' \\ \end{cases}$$ produces an inertial frame actually, because the free fall in the frame $$(z',t')$$ is mapped to a free motion with respect to $$(z,t)$$, given that
$$\frac{\mathrm{d}^2z}{\mathrm{d}t^2} = \frac{\mathrm{d}^2}{\mathrm{d}t'^2}\left(z' + \frac{1}{2}gt'^2\right) = \frac{\mathrm{d}^2z'}{\mathrm{d}t'^2} + g = -g+g = 0.$$
As said in the statement of Einstein's principle above, the free body and the free-falling body are experimentally indistinguishable inside the inertial frame $$(z,t)$$. In a quantum perspective, it means that their wavefunctions, expressed with respect to this same frame, can differ only by a global phase $$-$$ because a global phase cannot be measured experimentally. As a consequence, one should be able to find a solution of the form $$\psi(z,t) = \phi(z,t)e^{iS(z,t)/\hbar}$$. This form implies : \begin{align} \frac{\partial\psi}{\partial t'} &= \left(\frac{\partial t}{\partial t'}\frac{\partial}{\partial t} + \frac{\partial z}{\partial t'}\frac{\partial}{\partial z}\right) \phi\,e^{iS/\hbar} \\ &= \left(\frac{\partial}{\partial t} + gt\frac{\partial}{\partial z}\right) \phi\,e^{iS/\hbar} \\ &= \left(\phi_t + gt\phi_z + \frac{i}{\hbar}(S_t + gtS_z)\phi\right)e^{iS/\hbar} \end{align} where the subscripts denote partial derivatives and the arguments have been omitted in order to save up space. Similarly, one has : $$\frac{\partial^2\psi}{\partial z'^2} = \frac{\partial^2}{\partial z^2} \phi\,e^{iS/\hbar} = \left(\phi_{zz} + \frac{2i}{\hbar}S_z\phi_z - \frac{1}{\hbar^2}(S_{zz}+S_z^2)\phi\right)e^{iS/\hbar}$$
Finally, taking Eq. $$(1)$$ into account, one may rewrite Eq. $$(2)$$ as follows : $$i\hbar gt\phi_z - (S_t + gtS_z)\phi = -\frac{i\hbar}{m}S_z\phi_z + \frac{1}{2m}(S_{zz}+S_z^2)\phi + mg\left(z-\frac{1}{2}gt^2\right)\phi$$ This differential equation can be split into the system $$\begin{cases} 0 = gt + \frac{S_z}{m} \\ 0 = S_t + gtS_z + \frac{S_{zz}+S_z^2}{2m} + mg\left(z-\frac{1}{2}gt^2\right) \end{cases}$$ by gathering together the terms with respect to $$\phi_z$$ and $$\phi$$ respectively. The first equation leads to $$S(z,t) = -mgzt + f(t)$$, where $$f$$ is an arbitrary function of time. Thus, one has $$S_z = -mgt$$ and $$S_{zz} = 0$$, as well as $$S_t = -mgz + f'(t)$$, so that the second equation is reduced to $$0 = f'(t) - mg^2t^2$$, hence $$f(t) = \frac{1}{3}mg^2t^3$$ and thus $$S(z,t) = -mgzt + \frac{1}{3}mg^2t^3$$. Coming back to the initial frame of reference in the end, one ends up with the wanted solution, i.e. \begin{align} \psi(z',z') &= \phi(z',t') e^{iS(z',t')/\hbar} \\ &= \phi(z',t') \exp\left(\frac{i}{\hbar}\left(-mg\left(z'+\frac{1}{2}gt'^2\right)t' + \frac{1}{3}mg^2t'^3\right)\right) \\ &= \phi(z',t') \exp\left(-\frac{i}{\hbar}mgt'\left(z' + \frac{1}{6}gt'^2\right)\right) \end{align}