# Ground state energy in a simple quantum gravity situation

I found this problem in an MIT undergrad QM problem set; it is problem number 2, part a, number iv. I'll summarize everything below, but here's the link: http://ocw.mit.edu/courses/physics/8-04-quantum-physics-i-spring-2013/assignments/MIT8_04S13_ps1.pdf

And here are the solutions: http://ocw.mit.edu/courses/physics/8-04-quantum-physics-i-spring-2013/assignments/MIT8_04S13_ps1_sol.pdf

Summary: basically, we have the energy of a particle given by $E=\frac{1}{2}mv^2+mgx$, in a system characterized by three physical parameters: the mass $m$ of the particle, the Earth's gravitational acceleration $g$, and Planck's constant $\hbar$. In this system, the characteristic length, time, speed, and energy can be found through dimensional analysis, which is what the problem set asks you to do.

I found (in agreement with the solutions) that

$$v=(\hbar g/m)^{1/3}\\ l=(\hbar^2/m^2g)^{1/3}\\ E=(\hbar^2mg^2)^{1/3}$$

However, later, part iv. of the problem asks for the following: use your dimensional analysis results to give a simple estimate for the ground state energy of this system.

What I did is the following: the lowest energy for the system will be given for the lowest values of $x$ and $p$, but we can't say that they're both zero; that would violate the uncertainty principle. So, we say that $$x=0\pm \sqrt{\frac{\hbar}{2}}\\ p=0\pm \sqrt{\frac{\hbar}{2}}$$

This way the uncertainty principle is satisfied (but right at the limit) and the energy is minimized. Plugging these values into the energy gives

$$E_{min}=\frac{\hbar}{4m}\pm mg\sqrt{\frac{\hbar}{2}}$$

And waving off the uncertainty in the energy, we'd have

$$E_{min}\approx \frac{\hbar}{4m}$$

I thought this whole procedure was fairly reasonable, and it checks out in that $E_{min}\rightarrow 0$ as $\hbar\rightarrow0$. The only thing that bothered me slightly was that it implied a possibility of the energy being negative because of the uncertainty, but maybe that could be possible with quantum gravity in this case. However, the MIT solution was completely different. Basically what they did was plug in the characteristic speed and length in to the energy formula, and out popped the following:

$$E_{min}=\frac{3}{2}(mg^2\hbar^2)^{1/3}$$

Is my method of doing it also acceptable, despite the fact that it gives a totally different result? After having looked at the MIT solution I think I still prefer my solution, mostly because it relies on only the uncertainty principle and the total energy of the system, while the MIT one relies on the characteristic length and speed in the system, which as far as I'm aware aren't as fundamental, and don't indicate anything having to do with minimum energies (or do they?).

Any help/input on this would be highly appreciated.

• Your estimates don't have the right dimensions, so they can't be right. The point is that there is only one energy scale in the whole problem, so that has to be the ground state energy, up to some factor of order one. – Holographer Jan 31 '15 at 22:27
• Ah very cool, thanks so much! And how about if I were to change the position and momentum so they'd have the right units (by multiplying by certain factors of g, h, and m), but still satisfy the uncertainty relation? – Physics Llama Feb 1 '15 at 2:41