Is massively collaborative mathematical physics possible?

Posted on April 7, 2012 by Ed

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:

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:

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:

$H_{4}$ can be completely represented by terms in the algebra.
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:

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 $3\frac{1}{2}$ ).
It stays just that simple for all orders in $\lambda$ .
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?

13 thoughts on “Is massively collaborative mathematical physics possible?”

Clark Alexander on April 7, 2012 at 11:04 pm said:

There are already many reductions that can be made to simplify using commutators. Perhaps we can use d/dx and x as our other coordinate system. There is probably someway to move (large) commutators back and forth between fourier space(s).

Reply ↓
Clark Alexander on April 9, 2012 at 12:01 pm said:

So what we need is the following… If we want $Lambda_n$ to commute we need to find some operator $U$ so that

$U^* H_0 U = H_n + U^* f(N) U $

In this case f(N) will be a function of a number operator and we’ll have eigenvectors given by $U^* |j>$ where $N|j> = j |j>$ so that $|j>$ are the eigenvectors of the unperturbed oscillator.

In this case by the functional calculus (at least formally) $f(N)|j> = f(j) |j>$.

So we should be able to read off perturbations directly.

Reply ↓
Clark Alexander on April 10, 2012 at 12:52 am said:

Now H_0 = N + 1/2 which means it is already a function of N.

So we can move everything as a function of N to one side.

U^* (N+1/2 + f(N)) U = H_n.

So we need to figure out a few things…

1) How U^* N U works in general:
2) How to “build” an appropriate U

So our reduction now becomes:
g(N) = f(N) + N
then
U^* g(N) U = N + lambda x^n

Reply ↓
Clark Alexander on April 10, 2012 at 12:47 pm said:

Calling all the topologists out there:

Is there a way to setup a spectral sequence structure on the perturbed eigenvalues?

Consider that we need to compute these values…
1) lambda^k (in the power series)
2) x^n (the perturbation in the potential)
3) the m^th energy level for the x^n potential.

Therefore we’re looking for the terms E^k_{n,m}.

What’s known so far are
i) E^0_{n,m} all n,m
ii) E^1_{n,m} all n,m (solved by A.)
iii) E^k_{0,m} all k,m
iv) E^k_{1,m} all k,m
v) E^2_{4,m} all m (solved by A.)

What’s also known is that

E^k_{n,m} implies E^{k+1}_{n,0}
we get the next ground state if we know the entire previous solution.

Can we set up some sort of differential structure here?
Perhaps also we won’t have a d^2=0 but perhaps d^j = 0 for some other j…

Maybe
d^j : E^k_{n,m} rightarrow E^{k+1}_{n,m-j} or something like this…

Reply ↓
Clark Alexander on April 11, 2012 at 10:19 pm said:

If we move the U operators over we have

$g(N) = UNU^* + \lambda U(x^n)U^*$

In this case we may be able to convert $U$ from being in terms of $A$ and $B$
to being in terms of $\frac{d}{dx}$ and $x$ . Luckily (let’s call $p=\frac{d}{dx}$ for the moment)

We can mover the piece furthest to the right into
$\lambda (UxU^*)^n$
But $UxU^* = e^X x e^{-X} = [x,X] + 1/2 [[x,X],X] + \dots$

and if $X = f(p,x)$ then
$[x,f(p,x)] = \frac{\partial f}{\partial p} (p,x)$ which makes our commutators a little easier.

Reply ↓
Clark Alexander on April 14, 2012 at 11:29 am said:

In the same way that we can view $U$ and $U^*$ as differential operators in “standard coordinates” we can realize $A$ and $B$ as “differential operators” of $B$ and $A$ respectively.

$[A,B]=1$ means $A = \frac{d}{dB}$ and $B = - \frac{d}{dA}$ .

Reply ↓
Clark Alexander on April 14, 2012 at 6:04 pm said:

Now that I can write these things a little more nicely:

Let me write these three term energy values as
$E^k_n(m)$ where $k$ corresponds to the $\lambda^k$ correction, $n$ corresponds to $x^n$ the power of the overall perturbation, and $m$ corresponds to the energy level of the eigenstate.

So here we have:
$E^k_0(m) = m+1/2$ when $k=0$ and $0$ otherwise.
$E^k_1(m) = m+1/2$ when $k=0$ and $-1/2(\lambda^2)$ when $k=2$ $0$ otherwise.
$E^1_n(m) = 0$ if $n$ is odd.
$=\frac{1}{2^k}\left( \sum_{j=0}^k j! \left{\begin{array}{c} 2k\k \end{array}_{k-j}\right} \binom{m}{j} \right)$ when $n=2k$ .

Furthermore, we know
$E^2_4(m) = -51\binom{m}{3} - \frac{117}{2} \binom{m}{2} -36m - \frac{21}{8}$ .

The question becomes whether we can define some function (or sequence of functions) which may function as “differentials” so that
$d: E^{k_0}_{n_0}(m_0) \rightarrow E^{k_1}_{n_1}(m_1)$ .

Reply ↓
Clark Alexander on April 14, 2012 at 6:10 pm said:

Here’s a guess at the energy levels for $H_4$ .

$E^k_4(m) = \sum_{j=0}^{k+1} a^{(k)}_j\binom{m}{j}$

In this case $a^{(k)}_0$ will be our ground state correction for $\lambda^k$ .
One thing we can see easily is that $sgn(a^{(k)}_0) = -sgn(a^{(k+1)}_0$ .
In fact we know from Bender and Wu (ca.1963) $sgn(a^{(k)}_j) = -sgn(a^{(k+1)}_j$ .

Reply ↓
Clark Alexander on April 15, 2012 at 11:55 am said:

For the ground state of the the quartic oscillator, it appears we can make some rapid progress…

When calculating commutators that give ground state energies, the only surviving terms are of the form
$[B^n-A^n,B^n+A^n]$ These, of course, yield higher order terms, but the constant term given is simply
$-2(n!)$ .

When looking at the second order, we also know that
$[BA, B^mA^k \pm B^kA^m] = (m-k) B^mA^k \mp B^kA^m$

So our commutators will carry the extra terms…
$[\frac{c}{n}(B^n-A^n),c(B^n+A^n)]$ times an additional $\frac{\lambda^k}{k!}$ .

So it looks like the quartic ground state will be the summation of

$E_4(0) = \sum_{j=0}^{\infty} \frac{c_j^2}{(2j)^3}\cdot\frac{1}{j!}$ or something like this.

The problem is how fast the $c_j$ will diverge. It appears that this is in good standing for resummability if the $c_j$ grow larger than this.
Again, according to Bender and Wu, the overall terms grows asymptotically like
$~\left(\frac{-3}{2}\right)^k \Gamma(k+\frac{1}{2})$ which means that $c_j$ might diverge incredibly fast.

Reply ↓
Clark Alexander on April 15, 2012 at 11:56 am said:

For the ground state of the the quartic oscillator, it appears we can make some rapid progress…

When calculating commutators that give ground state energies, the only surviving terms are of the form
$[B^n-A^n,B^n+A^n]$ These, of course, yield higher order terms, but the constant term given is simply
$-2(n!)$ .

When looking at the second order, we also know that
$[BA, B^mA^k \pm B^kA^m] = (m-k) B^mA^k \mp B^kA^m$

So our commutators will carry the extra terms…
$[\frac{c}{n}(B^n-A^n),c(B^n+A^n)]$ times an additional $\frac{\lambda^k}{k!}$ .

So it looks like the quartic ground state will be the summation of

$E_4(0) = \sum_{j=0}^{\infty} \frac{c_j^2}{(2j)^3}\cdot\frac{1}{j!}$ or something like this.

The problem is how fast the $c_j$ will diverge. It appears that this is in good standing for resummability if the $c_j$ grow larger than this.
Again, according to Bender and Wu, the overall terms grows asymptotically like
$~\left(\frac{-3}{2}\right)^k \Gamma(k+\frac{1}{2})$ which means that $c_j$ might diverge incredibly fast.

Reply ↓
Clark Alexander on April 15, 2012 at 11:59 am said:

For the ground state of the the quartic oscillator, it appears we can make some rapid progress…

When calculating commutators that give ground state energies, the only surviving terms are of the form
$[B^n-A^n,B^n+A^n]$ These, of course, yield higher order terms, but the constant term given is simply
$-2(n!)$ .

When looking at the second order, we also know that
$[BA, B^mA^k \pm B^kA^m] = (m-k) B^mA^k \mp B^kA^m$

So our commutators will carry the extra terms…
$[\frac{c}{n}(B^n-A^n),c(B^n+A^n)]$ times an additional $\frac{\lambda^k}{k!}$ .

So it looks like the quartic ground state will be the summation of

$E_4(0) = \sum_{j=0}^{\infty} \frac{c_j^2}{(2j)^3}\cdot\frac{(2j)!}{j!}$ or something like this.

The problem is how fast the $c_j$ will diverge. It appears that this will be difficult to resume if the $c_j$ diverge at all.
Again, according to Bender and Wu, the overall terms grows asymptotically like
$~\left(\frac{-3}{2}\right)^k \Gamma(k+\frac{1}{2})$ which means that $c_j$ appear to diverge but at some algebraic rate, perhaps $c_j = (2j+1)$ .
Not sure…

Reply ↓
Clark Alexander on April 15, 2012 at 12:01 pm said:

Sorry for the three posts, I obviously made a few typing mistakes the first two times.

Reply ↓
Clark Alexander on May 5, 2012 at 9:45 am said:

If we simply set up the differential equation as a power series we arrive at the recurrence relation

$-\frac{(n+2)(n+1)}{2} c_{n+2} + \frac{1}{2}c_{n-2} + \lambda c_{n-4} - Ec_n = 0$

Sadly this is a 6th order equation rather than a second order equation.
So it seems we need to know $c_0$ through $c_5$ …

Reply ↓

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Is massively collaborative mathematical physics possible?

The Unperturbed Quantum Harmonic Oscillator — A Survey of Notation

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

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

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

13 thoughts on “Is massively collaborative mathematical physics possible?”

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The Unperturbed Quantum Harmonic Oscillator — A Survey of Notation

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

Computing the Eigenvalues to First Order in in Easy Steps

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

13 thoughts on “Is massively collaborative mathematical physics possible?”

Leave a Reply Cancel reply

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