Given a real-valued $K$-periodic potential $(v(0),\dots,v(K-1))$, the corresponding monodromy matrix reads
\begin{equation} M = \begin{pmatrix} -v(K-1)&-1\\1&0\end{pmatrix} \begin{pmatrix} -v(K-2)&-1\\1&0\end{pmatrix} \cdots \begin{pmatrix} -v(0)&-1\\1&0\end{pmatrix} \end{equation}If $v$ is a potential over the alphabet $\Sigma_\lambda = \{0,\lambda\}$ with $\lambda \in \mathbb{R}$, then the corresponding matrix entries of $M$ are polynomials in $\lambda$.
The location of the zeros of the entry $M_{2,1}$ is linked to the invertibility of the corresponding one-sided periodic Schrödinger operator $H_+$. More precisely, if $|\mathrm{tr}(M) | > 2$ and $M_{2,1} \neq 0$, then $H_+$ is invertible.
This notepad will systematically determine all potentials up to length $9$ over the alphabet $\Sigma_0$ and calculate the corresponding monodromy matrices, the zeros of the entry $M_{2,1}$ and the trace of $M$. See the section "Systematic Studies of $\{0, \lambda\}$-Valued Potentials" for details. In particular, all polynomial roots are given by closed-form algebraic expressions.
Run the cell below with Kmin = 1 and Kmax = 9 in order to generate all data for the proofs in the section "Systematic Studies of $\{0, \lambda\}$-Valued Potentials".
The cell will produce a dictionary pots, where the keys are given by stringified potentials and the corresponding values consist of the symbolic monodromy matrix (key='mon'), the zeros of the matrix entry (key='sol') and the value of the trace (key='trace').
After all results have been calculated. Run the cell that renders "Human Readable Output".
In order to access and further analyse the values in the dictionary for the potential v = [0,1,1] use the syntax
v = [0,1,1]
pots[str(v)]
More examples follow below in the analysis section.
It is possible to export the files in JSON, see the corresponding section at the end of this notebook.
# Run this cell once at startup
###############################
l = var('l')
def mon(v):
##########################
# Calculate the general monodromy matrix with scaling l
#
# Example:
# v = [1,1,0]; mon(v).expand()
# > [ l 1]
# > [l^2 - 1 l]
#
##########################
M = identity_matrix(2)
for vv in v:
M = matrix(2,2,[ - l*vv, -1, 1, 0])*M
return M
def analyse_v(v):
##########################
# calculate monodromy, zeros and check trace.
# return a dictionary containing symbolic elements
#
# Example:
# v = [1,1,0]; analyse_v(v)
##########################
data = {}
#monodromy
M = mon(v).expand()
data['mon'] = M
#zeros
sols = solve(M[1][0] == 0, l)
sol = list(map(lambda s: s.rhs(), sols))
sol.sort()
data['sol'] = sol
#trace
data['trace'] = list(map(lambda s: M.trace().subs(l=s), sol))
return data
#######Data Creation######
# input
Kmin = 1
Kmax = 8 #maximal period length, not larger than 9 to guarantee algebraic roots
sigma = [0,1] # alphabet
pots = {} #empty dictionary
for K in range(Kmin,Kmax + 1):
v_list = Tuples(sigma,K).list()
for v in v_list:
pots[str(v)] = deepcopy(analyse_v(v))
The cell below outputs all potentials, their corresponding monodromy matrices, the zeros of the entry M_{1,2}, and the corresponding value of the trace of M. For an interactive data analysis and further documentation, see the cells below.
for k in range(1,9):
show("Period Length K = ", k)
v_list = Tuples(sigma,k).list()
for v in v_list:
# skip the following potentials for readability
# they are treated separately at the the Section
# "Skipped potentials" of this notebook
if (v == [1,1,1,1,1,1,1] or v == [1,1,1,1,1,1,0]):
continue
show((str(v), pots[str(v)]))
The following cell analyses the previously skipped potentials
v = [1,1,1,1,1,1,0]
and
v = [1,1,1,1,1,1,1]
Both potentials lead to the monodromy matrix entry $M_{1,2}$.
$$ M_{1,2} = l^6 - 5l^4 + 6l^2 - 1 $$The algebraic roots produced with SageMath don't fully simplify with some remaining expressions that appear to have a non-vanishing imaginary part. The following calculations show that the imaginary parts add up, so that roots are only taken of positive real numbers.
Note that $$ 9 \cdot \left(\frac{7}{18} i \, \sqrt{3} + \frac{7}{54}\right)^{\frac{1}{3}} = 3 \, \sqrt{7} {\left(\cos\left(\frac{1}{3} \, \arctan\left(3 \, \sqrt{3}\right)\right) + i \, \sin\left(\frac{1}{3} \, \arctan\left(3 \, \sqrt{3}\right)\right)\right)} $$
and
$$ \frac{7}{{\left(\frac{7}{18} i \, \sqrt{3} + \frac{7}{54}\right)}^{\frac{1}{3}}} = \left(\cos\left(\frac{1}{3} \, \arctan\left(3 \, \sqrt{3}\right)\right) - i \, \sin\left(\frac{1}{3} \, \arctan\left(3 \, \sqrt{3}\right)\right)\right) $$This shows that $$ \mathrm{Im} \left( 9 \cdot \left(\frac{7}{18} i \, \sqrt{3} + \frac{7}{54}\right)^{\frac{1}{3}} + \frac{7}{{\left(\frac{7}{18} i \, \sqrt{3} + \frac{7}{54}\right)}^{\frac{1}{3}}} \right) = 0 $$
and
$$ \mathrm{Re} \left( 9 \cdot \left(\frac{7}{18} i \, \sqrt{3} + \frac{7}{54}\right)^{\frac{1}{3}} + \frac{7}{{\left(\frac{7}{18} i \, \sqrt{3} + \frac{7}{54}\right)}^{\frac{1}{3}}} \right) > 0 $$# Run this cell in order to analyse the roots of
# the monodromy matrices for v = [1,1,1,1,1,1,0]
# and v = [1,1,1,1,1,1,1]
v = [1,1,1,1,1,1,0]
print("Monodromy Matrix M for v = [1,1,1,1,1,1,0]")
show(pots[str(v)]['mon'])
v = [1,1,1,1,1,1,1]
print("Monodromy Matrix M for v = [1,1,1,1,1,1,1]")
show(pots[str(v)]['mon'])
print("Zeros of M_{2,1}")
for s in pots[str(v)]['sol']:
show(s)
The dictionary pots contains all possible periodic potentials. In order to filter the results we need to do 3 things
v_list = Tuples(sigma,2).list()
for v in v_list:
print([v, [s for s in pots[str(v)]['sol']], [s if s in QQ else s.n() for s in pots[str(v)]['trace']]])
gives as first line of output
[[0, 0], [r1], [-2]]This means that the entry $M_{2,1}$ of monodromy matrix for the potential $v = (0,0)$ is zero for all scalings (r1 corresponds to $\mathbb{R}$) and that the value of the trace of $M$ is always $-2$.
The line
[[1, 0], [0], [-2]]means that the entry $M_{2,1}$ of monodromy matrix for the potential $v = (1,0)$ is zero for the scaling l = 1 and that the value of the trace of $M$ is, in this case, $-2$.
Similarly to Example 1, we can look at potentials with period $K = 5$. A line like
[[1, 0, 1, 1, 0], [-1/2*sqrt(2), 1/2*sqrt(2)], [2.12132034355964, -2.12132034355964]]means that the entry $M_{2,1}$ of monodromy matrix for the potential $v = (1, 0, 1, 1, 0)$ is zero for the tuple of $\left( -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ and that the corresponding values of the trace of $M$ are $(2.12\dots, -2.12\dots)$.
Only list potentials such that $M_{2,1}$ has at least one rational zero.
v_list = Tuples(sigma,9).list()
for v in v_list:
isRational = [s in QQ for s in pots[str(v)]['sol']]
if True in isRational:
print([v, [s for s in pots[str(v)]['sol']], [s if s in QQ else s.n() for s in pots[str(v)]['trace']]])
Only list potentials such that $M_{2,1}$ has at least one rational zero and have absolute trace larger than 2.
v_list = Tuples(sigma,9).list()
for v in v_list:
isRational = [s in QQ for s in pots[str(v)]['sol']]
if True in isRational:
isTraceGt2 = [abs(s.n()) > 2 for s in pots[str(v)]['trace']]
if True in isTraceGt2:
print([v, [s for s in pots[str(v)]['sol']], [s if s in QQ else s.n() for s in pots[str(v)]['trace']]])
# example code, modify as indicated above
v_list = Tuples(sigma,5).list()
for v in v_list:
print([v, [s.full_simplify() for s in pots[str(v)]['sol']], [s if s in QQ else s.n() for s in pots[str(v)]['trace']]])
Export the pots dictionary in JSON file format.
You can read the input using a simple Python program
#!/bin/python3
import json
with open('pots_export.json') as json_file:
data = json.load(json_file)
for key in data:
print(key)
or an online JSON reader like https://jsonformatter.org/json-reader .
# run this cell to in order to save pots into json file
import json
def jsonify_dict(input_dict):
##########################
# create json object from dictionary,
# stringify everything for serialization
##########################
json_object ={}
for key in input_dict:
dictionary = {}
dictionary['mon'] = str(input_dict[key]['mon'])
dictionary['sol'] = str(input_dict[key]['sol'])
dictionary['trace'] = str(input_dict[key]['trace'])
json_object[str(key)] = deepcopy(dictionary)
return json_object
json_dict = jsonify_dict(pots)
with open('pots_export.json', 'w') as outfile:
json.dump(json_dict,outfile)