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Cell["Anharmonic Oscillator", "Subtitle"],

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Cell["Introduction and the simple harmonic oscillator", "Subsection"],

Cell[TextData[{
 "In this notebook we study some problems in quantum  mechanics using matrix \
methods. We know that we can solve quantum mechanics in any complete set of \
basis functions.  If we choose a particular basis, the Hamiltonian will not, \
in general, be diagonal, but the task is to diagonalize it to find the \
eigenvalues (which are the possible results of a measurement of the energy) \
and the eigenvectors. The reference for this material is Kinzel and Reents, \
p. 47-51.\n\nIn many cases this can not be done exactly and some numerical \
approximation is needed. A common approach, which is the basis of a lot of \
quantum chemistry is to take a ",
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 StyleBox["basis set and diagonalize it numerically. The ground state of this \
reduced basis state will not be  the exact ground state, but by increasing \
the size of the basis (up to a point) we can improve the accuracy and see if \
the energy converges as we increase the basis size. We will apply this \
approach here for an anharmonic oscillator. \n\nWe first discuss the exactly \
solvable case of the simple harmonic oscillator.  The Hamiltonian is given \
by",
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Cell["\<\
where p is the momentum, x the position, m the mass and \[Omega] the angular \
frequency of the classical oscillator.  This can be written in dimensionless \
form as \
\>", "Text"],

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Cell["\<\
In any textbook on quantum mechanics, it is shown that the energy levels are \
given by \
\>", "Text"],

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Cell["and the wavefunctions are given by", "Text"],

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 "where ",
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 "(x) is a Hermite polynomial.\n\nThe books also show that it is easier to \
determine the energy levels using operator methods rather than the Schr\
\[ODoubleDot]dinger equation, and that is the approach that we will take \
here.  In this approach, one introduces so-called raising and lowering \
operators, ",
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Cell["which have the commutation relations", "Text"],

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Cell["and in terms of which the Hamiltonian is written", "Text"],

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Cell["\<\
Denoting eigenstates of the Hamiltonian by |n\[RightAngleBracket], then one \
finds\
\>", "Text"],

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Cell["and so", "Text"],

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 "from which it follows that the energy ",
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Cell["\<\
Hence we can conveniently define the matrix elements for x as follows:\
\>", "Text"],

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Cell["\<\
and hence generate a matrix for x, if we keep a basis of \"basissize\" \
states:\
\>", "Text"],

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Cell["Lets check that this is correct for basissize = 4", "Text"],

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Cell["\<\
It is trivial to generate the Hamiltonian matrix of the simple harmonic \
oscillator, since it is diagonal, i.e.\
\>", "Text"],

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Cell["The anharmonic oscillator", "Subsection"],

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x and the convenient \"dot\" notation in ",
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Cell["\<\
The eigenvalues can also be obtained numerically and then sorted. Here we \
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\>", "Text"],

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Now we get some numbers. We start with a basis of size 15 and plot the \
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\>", "Text"],

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Cell["\<\
We see that the lowest levels appear to have converged well, but the higher \
levels have not. This is not surprising because the highest of the plotted \
levels are some of the highest possible for the size-10 basis, and so the \
effects of the truncation of the basis are naturally large in that region. \
\>", "Text",
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first element of ",
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Cell["\<\
We can also get a numerical value for a particular value of \[Lambda] (we \
take 0.2 here)\
\>", "Text",
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Cell["\<\
and plot it versus the inverse of the size of the basis set. Here are some \
results:\
\>", "Text",
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Cell["\<\
We see that for basis size greater than about 16, the energy has converged \
well. As you might expect, and you can verify yourself, a larger basis set is \
needed if \[Lambda] is larger. 

The eigenvectors can also be found, and these give the linear  coefficients \
of the Hermite polynomials which make up the coordinate space wavefunction, \
and hence the wavefunction of the anharmonic oscillator can also be obtained \
by this method.\
\>", "Text",
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Cell["\<\
In another handout we calculated the lowest three energy levels of this \
anharmonic oscillator in a comp;letely different way, from the Schrodinger \
equation. In that approach we solve a differential equation rather than \
diagonalize a matrix. We found the lowest three energy levels to be

      0.602405,      1.95054,      3.5363\
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Cell["\<\
We see that the lowest agrees very well with the ground state determined \
above from matrix methods. To compare all three numbers we now compute the \
lowest three levels from the matrix method, still with \[Lambda] = 0.2:\
\>", "Text",
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