Before I continue, I will pose two questions to the reader:

1. You have a quantum particle in a harmonic potential. What is the probability of finding the particle at some distance from the centre of the trap?

2. You have a particle undergoing Brownian diffusion in a harmonic potential. What is the probability of finding the particle at some distance from the centre of the trap?

If you answered "Gaussian!" to both questions, you are correct. For a harmonic (quadratic) potential, the Boltzmann distribution, which is e-to-the-minus-the-potential has the same functional form as the solution to the Schroedinger equation with that potential. Is this special? Is this interesting? Is this the only example of a potential where that is true?

I thought this picture was a good idea when I started making it and I can't be dissuaded now. |

There are quite a few connections between quantum and statistical mechanics. The diffusion equation and the Schroedinger equation are basically the same, if you replace the thermal energy with Planck's constant and make time imaginary. A lot of the theory for my Ph.D. thesis applied insights from established Schroedinger solutions to study diffusion solutions. There is also something called the Wick Rotation that I don't really understand that maps thermal systems onto quantum systems, and John Baez and Blake Pollard have an article called "Quantropy" discussing a lot of these connections.

Plugging the Boltzmann distribution into the solution of a physical differential equation in order to solve it isn't outlandish, the Poisson-Boltzmann equation to find the distribution of ions in a fluid is an example.

In 2013 my Ph.D. research took me in the direction of analysing experimental histograms to try to measure a thermal potential, by using the Boltzmann distribution. We were trying to show that these histograms are Gaussian*, implying that the potential is quadratic, and the standard deviation of the histograms would give us the spring constant. This got me thinking about the connection between the thermal particle on a spring, and the quantum particle on a spring.

Not really finding any physical significance to this connection, I looked for a mathematical reason, which starts by substituting the Boltzmann distribution into the Schroedinger equation. Because the units are different between thermal and quantum physics, we will write down the (one -dimensional time-independent) Schroedinger equation and Boltzmann distribution in a general form:

$-\alpha \frac{d^{2}\Psi}{d\,x^{2}}+V(x)\Psi=E\Psi$

and

$\Psi=\kappa e^{-\beta\,V(x)/2}$

I've added a factor of 1/2 in the exponential, because the probability amplitude is the square of the wavefunction, so the wavefunction in this case would be the square-root of the Boltzmann distribution.

Substituting the second equation into the first, we have a modified version of the Schroedinger equation that satisfies this constraint for the potential:

$-\alpha \frac{d^{2}}{d\,x^{2}}\left[\kappa e^{-\beta\,V(x)/2}\right]+V(x)\left(\kappa e^{-\beta\,V(x)/2}\right)=E\left(\kappa e^{-\beta\,V(x)/2}\right)$

I still have a free parameter to play with, which is the value of the potential at x=0, which would have to be balanced by a normalization constant. There is a trivial solution where V(0)=E=V(x), or I can solve for V(0) to kill the higher order terms. If I try to kill the quartic term, I get:

$V(0)=\frac{E\beta-1}{\beta}$

and if I plug this in, something cool happens, which is that every higher term dies. I'm just left with a quadratic potential!

$V(x)=\frac{E\beta-1}{\beta}+\frac{x^2}{\alpha\beta^2}$

We have shown that if we demand that the Schroedinger equation be solved by the Boltzmann distribution, one of the possible potentials is quadratic. Just to complete the quantum harmonic oscillator picture, we can kill y-intercept with $\beta=1/E$, and return to the classic Schroedinger equation with $\alpha=\hbar^{2}/2m$. Demanding that the spring constant is $m\omega^{2}$ tells us that the energy eigenvalue is $\hbar\omega/2$, the ground state energy of the quantum harmonic oscillator, so everything is tied up in a neat little package.

So, that particular value of V(0) gets me a quadratic potential, which is one of the few potentials for which there is a solution to the Schroedinger equation. Can this method be used to find others? What if I try to kill an arbitrary higher order term? It turns out that this does not work as well. You can choose V(0) to eliminate any term, but only that term and not the higher ones. Only the quartic term will cause the Taylor dominoes to collapse if you knock it over.

This argument can also be used for the excited states of the harmonic oscillator, e.g. multiplying the wavefunction by x before substituting it. This doesn't add much of interest. I also wanted to see if this would work as a general method of finding solvable potentials for the Schroedinger equation, trying various functions mapping between the wavefunction and the potential. It generally doesn't work, but I did find one solvable potential with this method:

The constraint $\Psi(x)=\frac{\beta}{V(x)}$ gives $V(x)=\frac{-E}{\cos{\sqrt{\frac{2mE}{\hbar^2}}x}-1}$ as an easily solvable Schroedinger potential. I guess that's kind of cool.

Through all this we investigated a mathematical connection between the Boltzmann distribution and the Schroedinger equation, but didn't get much physical insight. I asked John Baez, author of Quantropy, whether the connection was physically significant, and he gave me a "Yes," so I hope to learn more. Overall, I explored an idea that I had, and found that it lead somewhere kind of, but not overly, interesting. I think that's worthwhile.

*Of course, basically any large sample you measure will be Gaussian...stupid central limit theorem. This project unfortunately was a dead end, and took a lot of time and effort.

**This is one of my favourite math tricks. I highly recommend.

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ReplyDeleteHaving done the calculation on more than a napkin this time, I'm giving it another shot. In retrospect I probably shouldn't have deleted my previous comment but it doesn't matter, the results are the same.

DeleteAs I was saying, I think the equation (that you tried to solve using mapple) is solvable by hand.

First of, we can rearrange the equation to get

φ''(x) + (φ'(x))² + λφ(x) = 0

where I've chosen the notation λ = 4m/βħ² and φ = β/2 (E - V) for simplicity. This can be separated into two first order equations by using the substitution ω(φ) = φ'², which gives for the second equation

ω' + 2ω + 2λφ = 0.

This can be solved and we get the equation in φ

(φ'(x))² = λ(1/2 - φ)

which has for solutions

φ(x) = 1/2 - λ/4 (x - C)²

for all C. Substituting that result into the potential and the wavefunction gives

V(x) = E - 1/β + 2m/β²ħ² (x - C)²

ψ(x) = (2m/πβħ²)^(1/4) exp[ - m/βħ² (x - C)² ]

We do get the quadratic potential, but somehow it is the only solution here (which seems to differ from your study somehow). The integration parameter C corresponds to the freedom of spatial translation, which was expected. The choice of E also doesn't impact the wavefunction, which is reassuring. Finally, it is not particularly surprising that the normal distribution would show up (since we are dealing with the classical thermodynamic equilibrium) as well as the De Broglie thermal wavelength (2πβħ²/m)^(1/2), but it is interesting to see that the normal distribution is the only solution, and that it happens in the case of a quadratic potential.

Thanks for the solution! Perhaps I rely too much on computers to do my algebra.

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