Is massively collaborative mathematical physics possible?

In the spirit of the Polymath Project, I’d like to present some thoughts on a mathematical physics problem that could be interesting.  The problem is simple enough: compute the eigenvalues of the Hamiltonian for a perturbed quantum harmonic oscillator.  But before we get started, I’ll just state that we should try to follow Timothy Gowers’s 12 ground rules at the end of his Polymath kick-off post.  Also, I’ll credit Clark Alexander for walking me through this entire method, and encouraging this post.

The Unperturbed Quantum Harmonic Oscillator — A Survey of Notation

So, we have the quantum harmonic oscillator, easily represented and solved by using raising and lowering operators:

H_{0} = \frac{(p^2+x^2)}{2} \rightarrow H_{0} = BA+\frac{1}{2}

where the notation B=a^{\dagger} and A=a is employed, such that [A,B] = 1.

The spectrum of solutions to the Schrödinger Equation are familiar — the first three are shown in the following plot:

Unperturbed Eigenstates of the Harmonic Oscillator

The Quartic Perturbation to the Quantum Harmonic Oscillator — Rewriting the Hamiltonian

With the above notation, we can readily incorporate a perturbation of \frac{\lambda}{4}{\cdot}(x^{4}) into our Hamiltonian:

H_{4} = H_{0} + \frac{\lambda}{4}{\cdot}(x^{4}) \rightarrow H_{4} = H_{0} + \frac{\lambda}{4}{\cdot}((B+A)^{4})

H_{4} can be normal ordered to the following result:

(1)   \begin{eqnarray*} H_{4} & = & H_{0} + \frac{\lambda}{4}{\cdot}((B+A)^{4}) \\ & = & H_{0} + {\lambda}{\cdot}(0.25){\cdot}(B^{4}+A^{4}) + {\lambda}{\cdo\ t}(B^{3}A+BA^{3}) \nonumber \\ &   &  + {\lambda}{\cdot}(1.5){\cdot}(B^{2}+A^{2}) + {\lambda}{\cdot}(1.5\ ){\cdot}B^{2}A^{2} \nonumber \\ &   &  + {\lambda}{\cdot}(3){\cdot}BA + {\lambda}{\cdot}(0.75) \end{eqnarray*}

Intuitively, we know that the result of this perturbation will be a slight change in the resulting eigenstates making up the solution.  A plot of the resulting change for an arbitrarily chosen lambda is shown below:

Perturbed Eigenstates of the Harmonic Oscillator

Computing the Eigenvalues to First Order in \lambda in 3\frac{1}{2} Easy Steps

Step 1: Identify All Elements of the Lie Algebra

Elements of the Lie Algebra at first order (L_{m}^{(k)} where k=1) are determined by performing commutations with H_{0} and H_{4}, as identified below.  At first order, terms of order O(\lambda^{2}) are ignored, so only one commutation is required.

The first commutator:

(2)   \begin{eqnarray*} [H_{0},H_{4}] & = & [B{\cdot}A,{\lambda}{\cdot}(\frac{A+B}{\sqrt{2}})^{4}] \\ &   &  \\ & = & {\lambda}{\cdot}(B^{4}-A^{4}) + {\lambda}{\cdot}(2){\cdot}(B^{3}A-BA^{3}) + {\lambda}{\cdot}(3){\cdot}(B^{2}-A^{2}) \\ \end{eqnarray*}

 At this point, we can identify all the terms of the Lie Algebra to first order.

Nonperturbative Terms (\lambda^{k} where k=0)

(3)   \begin{eqnarray*} L_{0}^{(1)}  & = & I = 1\\ L_{1}^{(1)}  & = & BA \\ \end{eqnarray*}

First Order Terms (\lambda^{k} where k=1)

(4)   \begin{eqnarray*} L_{2}^{(1)}  & = & {\lambda}{\cdot}I = {\lambda} \\ L_{3}^{(1)}  & = & {\lambda}{\cdot}BA \\ L_{4}^{(1)}  & = & {\lambda}{\cdot}B^{2}A^{2} \\ &   & \\ L_{5}^{(1)}  & = & {\lambda}{\cdot}(B^{4}+A^{4}) \\ L_{6}^{(1)}  & = & {\lambda}{\cdot}(B^{3}A+BA^{3}) \\ L_{7}^{(1)}  & = & {\lambda}{\cdot}(B^{2}+A^{2}) \\ L_{8}^{(1)}  & = & {\lambda}{\cdot}(B^{4}-A^{4}) \\ L_{9}^{(1)}  & = & {\lambda}{\cdot}(B^{3}A-BA^{3}) \\ L_{10}^{(1)} & = & {\lambda}{\cdot}(B^{2}-A^{2}) \end{eqnarray*}

In this representation, we see the following:

(5)   \begin{eqnarray*} H_{0} & = & L_{1}^{(1)}+\frac{1}{2}L_{0}^{(1)} \\ H_{4} & = & H_{0} + (0.25)L_{5}^{(1)} + L_{6}^{(1)} \\ &   & + (1.5)L_{7}^{(1)} + (1.5)L_{4}^{(1)} \\ &   & + (3)L_{3}^{(1)} + (0.75)L_{2}^{(1)} \end{eqnarray*}

This representation is complete for our purposes because it satisfies two conditions:

  1. H_{4} can be completely represented by terms in the algebra.
  2. No two terms can be commuted to create a third non-trivial term not shown in the group.  (Remember, \lambda^{2} = 0).

Step 2: Construct a General Lie Group Element

In principle, the Lie group element could be constructed from all terms in the Lie algebra, like so:

U = \textrm{exp}(\sum\limits_{k=0}^{10} \alpha_{k}{\cdot}L_{k})

But, by nature of the Hammard lemma, we can choose to exclude all terms that commute with H_{0}.  So we construct U as follows:

U = \textrm{exp}(\alpha_{5}L_{5}+\alpha_{6}L_{6}+\alpha_{7}L_{7}+\alpha_{8}L_{8}+\alpha_{9}L_{9}+\alpha_{10}L_{10})

This gives us 6 constants we tune in order make this Lie group element a transformation of basis between perturbed and unperturbed eigenstates.

Step 3: Use the Hammard Lemma to Compute our Lie Group Element

It is our goal to choose a U such that the following is true:

(6)   \begin{eqnarray*} H_{4} - U^{\dagger}H_{0}U = \Lambda_{4} \end{eqnarray*}

where [U,\Lambda_{4}] = 0 + O(\lambda^{2})

U^{\dagger}H_{0}U = H_{0} + [-X,H_{0}] + \frac{1}{2!}([-X,[-X,H_{0}]]) + {\cdots}

where X = \alpha_{5}L_{5}+\alpha_{6}L_{6}+\alpha_{7}L_{7}+\alpha_{8}L_{8}+\alpha_{9}L_{9}+\alpha_{10}L_{10}.

To first order in \lambda this simplifies to:

(7)   \begin{eqnarray*} U^{\dagger}H_{0}U \approx H_{0} + [-X,H_{0}] \end{eqnarray*}

Performing the commutator of Equation 7 and normal ordering, we get the following:

(8)   \begin{eqnarray*} [-X,H_{0}] & = & [H_{0},X] \\ & = & [BA,X] \\ & = & {\lambda}{\cdot}[BA,\alpha_{5}(B^{4}+A^{4})+\alpha_{6}(B^{3}A+BA^{3}) \\ &   &                     +\alpha_{7}(B^{2}+A^{2})+\alpha_{8}(B^{4}-A^{4}) \\ &   &                     +\alpha_{9}(B^{3}A-BA^{3})+\alpha_{10}(B^{2}-A^{2})] \\ &   & \\ & = & {\lambda}{\cdot}( (4\alpha_8){\cdot}(B^{4}+A^{4}) + (4\alpha_5){\cdot}(B^{4}-A^{4}) \\ &   &  + (2\alpha_9){\cdot}(B^{3}A+BA^{3}) + (2\alpha_6){\cdot}(B^{3}A-BA^{3}) \\ &   &  + (2\alpha_{10}){\cdot}(B^{2}+A^{2}) + (2\alpha_7){\cdot}(B^{2}-A^{2})) \\ &   & \\ & = & (4\alpha_8){\cdot}L_{5} + (4\alpha_5){\cdot}L_{8} \\ &   &  + (2\alpha_9){\cdot}L_{6} + (2\alpha_6){\cdot}L_{9} \\ &   &  + (2\alpha_{10}){\cdot}L_{7} + (2\alpha_7){\cdot}L_{10} \\ &   & \end{eqnarray*}

Step 3\frac{1}{2}: Tune \alpha_{k} so \Lambda_{4} is a Number Operator

From Equations 5 and 7,

(9)   \begin{eqnarray*} \Lambda_{4} & = & H_{4} - U^{\dagger}H_{0}U \nonumber \\ &   & \nonumber \\ & = & (0.25-4\alpha_8)L_{5} + (1-2\alpha_9)L_{6} \\ &   & + (1.5-2\alpha_{10})L_{7} + (1.5)L_{4} \nonumber \\ &   & + (3)L_{3} + (0.75)L_{2} \nonumber \\ &   & + (-4\alpha_5){\cdot}L_{8} + (-2\alpha_6){\cdot}L_{9} \nonumber \\ &   & + (-2\alpha_7){\cdot}L_{10} \nonumber \end{eqnarray*}

Now, using our knowledge that \Lambda_{4} must commute with U, we know that \Lambda_{4} cannot have terms involving L_{5}, L_{6}, L_{7}, L_{8}, L_{9} or L_{10}.  Thus, the alphas must be tuned such that:

(10)   \begin{eqnarray*} (-4\alpha_5)        & = & 0 \\ \alpha_5    & = & 0            \\ &   &  \\ (-2\alpha_6)        & = & 0 \\ \alpha_6    & = & 0             \\ &   &\\ (-2\alpha_7)        & = & 0 \\ \alpha_7    & = & 0             \\ &   &   \\ (0.25-4\alpha_8)    & = & 0 \\ \alpha_8    & = & \frac{1}{16}  \\ &   &\\ (1-2\alpha_9)       & = & 0 \\ \alpha_9    & = & \frac{1}{2}   \\ &   &\\ (1.5-2\alpha_{10})  & = & 0 \\ \alpha_{10} & = & \frac{3}{4} \end{eqnarray*}

Which leaves:

(11)   \begin{eqnarray*} \Lambda_{4} & = & \frac{3}{2}L_{4} + 3L_{3} + \frac{3}{4}L_{2} \\ & = & \frac{3}{2}\lambda(B^{2}A^{2}) + 3\lambda(BA) + \frac{3}{4}\lambda \end{eqnarray*}

Huzzah!  We’ve done it.  This is the perturbation to the eigenvalue introduced by the quartic term.

What’s So Special About This Approach to Perturbation Theory?

While we were methodically plodding through, you might not have noticed, but there’s a few special aspects to this approach:

  1. The solution came down to performing a few commutators (Step 1), expanding the Hammard Lemma (Step 3), and solving a simple system of equations, (Step 3\frac{1}{2}).
  2. It stays just that simple for all orders in \lambda.
  3. It stays just that simple for all perturbations of the form x^{n}

Perhaps there’s more here.  Perhaps we could find a general form for the perturbation to n^{th} order (in \lambda) to the harmonic oscillator?  Or more?

Feel free to comment with questions and thoughts.  Where would you go from here?

Sunshine, Taxes, and Maps

I’ve recently been looking at economic incentives for solar power in the U.S.  I thought I’d consider what incentives might encourage the installation of solar panels for different locations across the country.

First, let’s see how much sunshine different parts of the country get annually:

Click for better resolution
This plot is generated in R using the shapefile GIS data provided by the National Renewable Energy Laboratory (NREL) via the Open Energy Information (OpenEI) platform.  It is based off a model that takes input from 14,000+ solar radiation stations across the country.

Naturally, increased sunlight can provide more incentive to install solar panels.  However, that is only a small part of the story.

What happens when we recreate the above map, but measure the economic value of installing the solar panels due to:

Clearly, there’s quite a bit of information to coalesce.  To simplify things, I consider two scenarios:

The Residential Scenario:

Source: Used with permission
  • 20 sq. meters of solar panelling installed on a single family residence, unobstructed
  • Solar panels show a 10% energy conversion efficiency (reasonable, if low)
  • Installation costs $50,000
  • Home owners bear a tax rate of approximately 15%

The resulting gross payout of tax incentives and energy production (not subtracting the cost of installation on the residence) extrapolated over time, looks as follows:  (Be sure to click for the high resolution version)

Click for better resolution

In addition to the residential scenario above, I also considered a more commercial scenario, which involved a different set of state level tax incentives, and a few other assumptions.

The Commercial Scenario:

Source: Used with permission
  • 200 sq. meters of solar panelling installed on a business, unobstructed
  • Solar panels show a 10% energy conversion efficiency
  • Installation costs $500,000
  • Corporation bears a tax rate of approximately 30%

The resulting gross payout of tax incentives and energy production (not subtracting the cost of installation on the residence) extrapolated over time, looks as follows:  (Be sure to click for the high resolution version)

Click for better resolution


Naturally, there are a huge number of assumptions baked into these maps.  (Stable tax rates, typical weather, steady energy prices … the list goes on.)  Nevertheless, I think it’s interesting to see, at least according to the data I used, how tax write-offs start out dominating the state-to-state variation in payout.  But, after a few decades, solar potential begins playing a more critical role.


A few things I learned along the way:

  • I taught myself quite a bit about R in this little study.
  • Using industry standard tools for GIS for the first time (namely becoming familiar with the structure of a “shapefile”).  This was probably the most challenging part.
  • In dealing with tax incentives, the number of stipulations and caveats can be absolutely crushing.  I found the best way to deal with these were to simply hypothesize a couple of scenarios, and run with the results.

Questions and comments are welcome.

If Madison Crime were Elevation

What if Madison, WI were mapped so that elevation represented crime?  This idea is directly inspired by Doug McCune’s post about mapping San Francisco crime as elevation from last year.

Accessing the data was a little tricky.  In Madison, the police put incident reports online, but some scrubbing of the data is required, and the reports needed to be geocoded.  After a little scripting to overcome these obstacles, I was ready to start mapping.

I divided crimes into 3 broad categories:

  • Robbery/Burglary/Theft
  • Drugs and Alcohol
  • Violent Crime

Here are the results:

Noticeable features:

  • The State Street area (at map-center), linking the capitol and the UW-Madison campus, has the largest spike of robbery activity.  (This will be a recurring theme…)
  • The shopping centers, in particular the malls, have significant numbers of reports.



Noticeable features:

  • State Street is again an unpleasant place to be.
  • The south side (Park St / Fish Hatchery Rd) has its fair share of violent crime, as does the shopping center where Verona Rd meets the Beltline Hwy.
  • Madison’s east side is pretty chilled out.  Perhaps this correlates with the better pizza options?



Noticeable features:

  • Not nearly so much trouble with State St
  • The biggest trouble in town seems to be on the east side.


A few things I learned along the way:

  • How to build a crude web-spider using cURL.
  • How to geocode addresses using the Google Geocoding API — remarkably user-friendly!
  • A few basics about MySQL to store the scraped and geocoded data.
  • Quite a bit about a ray-tracing program called POV-Ray.
  • How to animate graphics using imagemagick — though, the resulting animations (showing the distribution of crimes changing along with the time of day) are far too big to put on this site.

Please leave any questions or comments.