- MATLAB Sandbox
- By "explicitly diagonalize" we
mean calculate something like V
^{-1}AV or VAV^{-1}and check if it is diagonal (check both!). Note that the inverse of A is`inv(A)`in MATLAB (actually, this is just one way to calculate it). Compare V^{-1}to the Hermitian conjugate of V (given by`V'`in MATLAB). What does this tell you? - The word "cyclic" should appear in your answer.
- To test whether A has to be complex or Hermitian, simply try out the different cases for a couple of examples each.
- We want to know how many decimal points are good for the
different numbers of terms. This presumably depends on the
matrix in detail, but you can try a few examples to see if
there is a consistent trend.
You may want
`format long`to get enough digits in the answers. - Make sure that T and V do not commute (unlikely if they
are chosen at random!). You are basically looking for how
many decimal digits improvement there is when epsilon is
reduced by factors of 10.
For this type of problem you should make you comparisons
with different epsilons using the
*same*T and V matrices so that it is clear that the differences are solely due to the numerical approximation. Then you can try another set to see if your results are robust (or just the result of some accident).

- By "explicitly diagonalize" we
mean calculate something like V
- SVM revisited.

Being stationary under arbitrary variations of the expansion coefficients just means that the partial derivative of the estimated energy with respect to any of the c_{i}coefficients is equal to zero. So just carry out this derivative, remembering that you have both a numerator and a denominator. You'll be able to cancel one sum from the denominator because it is positive definite and you will use that the H and B matrices are symmetric. - Symmetry factors.
- lambda
^{3}contribution to <xi^{2}>- This is just a review of the rules for diagrams (e.g., how is the number of vertices related to the power of lambda?). We're not looking for long answers!
- Follow the rules from the lecture notes. Remember when
calculating the vertex permutation factor (the third factor) that
the "external lines" for <xi
^{2}> are nailed down. There are 10 distinct diagrams. - Just plug in the numbers and add the fractions.
- If you get stuck or want to check your answers, look at the solution here (problem 3).

- partition function with xi
^{6}- The rule for lines should be the same as with xi
^{4}, because that comes from the xi^{2}term. For the vertex, before we had the -lambda/4 factor times the combinatorical factor for four derivatives hitting j^{4}. - How many lines should come from each vertex? Remember that the partition function (as opposed to its logarithm) includes ALL diagrams, connected and disconnected. The partition function has no external lines.
- <xi
^{2}> will have two external lines. Only connected diagrams contribute to - <xi
^{2}> but what about the partition function (the question asks for Z/Z0 rather than ln[Z/Z0]). - Look at the solution to the previous problem
here
(problem 3) to see how things work.
The exact answers via Mathematica are 1 - 5 alpha/(2 a^3)
+ 1155 alpha^2/(8 a^6) for Z/Z0 and 1/a - 15 alpha/a^4 + 1695
alpha^2/a^7 for <xi
^{2}>.

- The rule for lines should be the same as with xi
- replica method for <xi
^{2}>- This is a bit tricky: We are really
*continuing*the function to n=0. In the definition of O_{n}in the problem set, the first term is the numerator we want for <O> but then we want a factor of Z/Z_{0}in the*denominator*. You can get this by multiplying on top and bottom by Z/Z_{0}. How many factors of Z/Z_{0}are there now in the expression for O_{n}besides the part we want? Write them as Z/Z_{0}to that power. What happens if you then set ("continue") n to 0? - Where do the external legs come from? Can any index appear?
- If you have disconnected diagrams, how many are there (consider what indices can appear)? For the connected diagrams, how many different indices are there?
- The answer should be yes! How would the argument change
for another operator, such as <xi
^{4}>?

- This is a bit tricky: We are really

- lambda

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Last modified: 11:38 am, October 13, 2009.

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