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Posts Tagged ‘Schrödinger’s equation’

## Gravity and Quantum Mechanics — The Simple Version

In general relativity, gravity is no longer considered a real force like the Lorentz force, but a fictitious one like the Coreolis force or centrifugal force in a rotating reference frame. To get rid of the force of gravity locally, all one has to do is switch to a reference frame that is accelerating at the same rate that objects in free fall are accelerating. The force of gravity term will then disappear from Newton’s second law.

Some claim that the same trick doesn’t work in quantum mechanics. This is not true. They just weren’t trying hard enough. Start with the wave equation for a particle in a constant gravitational field (Planck’s constant has been set to 1 and the units of mass are measured in seconds per metre squared as a result. See here.):

$-\frac{1}{2m} \psi_{zz} + mgz\psi = i\psi_t$ (1)

z is height and $V(z) = mgz$ is the gravitational potential energy. Thus in this coordinate system, objects fall in the -z direction. Now we’ll define new coordinates, u, such that when an approximately classical particle is in free fall in the old coordinate system, it is stationary in the new one. This implies $z = u - \frac{g}{2}t^2$. We’ll call our new wavefunction $\phi$. Now, we need to establish a relationship between the old wavefunction and the new one. A good first guess would be

$\psi(z, t) = \phi \left( z + \frac{g}{2} t^2, t \right) .$

There is a problem with this, though. Recall that given a pre-existing wavefunction $\psi$, to change its momentum by p one multiplies $\psi$ by $e^{i p z}$. There is no term like this in our first guess, which would mean that the particle has the same momentum in both reference frames. This is definitely not true. Thus an extra as-yet-unknown factor of $e^{i \gamma(z, t)}$ is needed, where $\gamma(z, t)$ must be determined. Our new coordinate transformation for the wavefunction is

$\psi(z, t) = e^{i \gamma(z, t)} \phi \left( z + \frac{1}{2} g t^2, t \right) .$ (2)

We would like to find a $\gamma$ such that $\psi$ obeys the free fall wave equation (1) and $\phi$ obeys the free particle wave equation,

$-\frac{1}{2m} \phi_{uu} = i \phi_t .$ (3)

To do this, first we replace $\psi$ in (1) with the right side of (2), resulting in the following:

$-\frac{1}{2m} \phi_{uu} - \frac{i}{m} \gamma_z \phi_u + \left( \frac{1}{2m} \left( \gamma_z^2 - i \gamma_{zz} \right) + mgz \right) \phi = i \phi_t + igt\phi_u - \gamma_t \phi$

The left and right sides of (3) are present in the above expression, but there are a bunch of extra terms. These terms will all cancel if

$-\frac{i}{m} \gamma_z = igt$ (4) and $\frac{1}{2m} \left( \gamma_z^2 - i \gamma_{zz} \right) + mgz = -\gamma_t$ (5).

Thus (3) is true if $\gamma$ is subject to conditions (4) and (5). Condition (4) implies $\gamma = -mgtz + c(t)$. Replacing $\gamma$ with this in condition (5) results in

$c = -\frac{mg^2}{6} t^3 .$

This means

$\gamma = -mgtz - \frac{mg^2}{6} t^3 .$

Including the $\gamma$, final formula for converting from $\phi$ to $\psi$ is

$\psi(z, t) = e^{-imgtz - \frac{img^2}{6}t^3} \phi \left( z + \frac{g}{2} t^2, t \right)$ (6).

Thus we have found a transformation of the original free fall wave equation to the free particle wave equation in complete analogy to the classical case. Now that we have $\gamma$, what is its significance? As was mentioned before, a factor of $e^{ipz}$ out in front of a wavefunction shifts the particle’s momentum by p. Such a factor is indeed present in (6). In our case, the difference in momentum from $\phi$ to $\psi$ is -mgt. This makes perfect sense.

But what about the other factor, $e^{-\frac{img^2}{6} t^3}$? It was necessary for the math to work out, but does it mean anything? It is just a constant of unit modulus out in front of the wavefunction. Thus it has no meaning at all and the state mentioned in (6) is completely equivalent to

$\psi(z, t) = e^{-imgtz} \phi \left( z + \frac{g}{2} t^2, t \right)$.

I was first shown this neat trick by Andy Randono in his 4th year undergraduate general relativity at the University of Waterloo. If you’re reading, thanks for teaching it to us!