What is the Partition Function for a Singular Exponential Hamiltonian?

[jarvGrad]

I need to find the integral as follows:

I am given a Hamiltonian of the form:

\<br /> <br /> H=\Sigma {(x_n+d y_n)^2}&lt; 2 m E<br /> <br />

(This should be a sum over n, but its not showing in the preview)


we integrate the exponential in n-space as
\<br /> \begin{equation}<br /> \int \exp{H} d^{3n}x d^{3n}y<br /> \end{equation}<br />
so that
\<br /> \begin{equation}<br /> \int \exp{(x+dy)^2} d^{3n}x d^{3n}y<br /> \end{equation}<br />
where (x+dy)^2 < E

I found a solution that tells me
\<br /> \begin{equation}<br /> \int \exp{(ax^2+bxy+cx^2)^2} d^{3n}x d^{3n}y<br /> \end{equation}<br />
which equals

\<br /> \begin{equation}<br /> \pi^{m/2}/{det[A]} <br /> \end{equation}<br />
where A is the 2-D matrix
A=[a b
b c]

However, the determinant is zero as I am given

\<br /> <br /> x^2+2mwxy+(mwy)^2<br /> <br /> [\tex]<br /> <br /> so this doesn&#039;t work. I found this solution at <a href="http://srikant.org/thesis/node13.html" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://srikant.org/thesis/node13.html</a> .<br /> There is a bit more work shown on the website. My professor assured me that the solution is closed form.

 

[hunt_mat]

Do you mean
<br /> \<br /> \begin{equation}<br /> \int \exp{(ax^2+bxy+cy^2)^2} d^{3n}x d^{3n}y<br /> \end{equation}<br /> <br />

Why not diagonalise A by doing a change of variables in you integral, or possibly a proof by induction?

 

[jarvGrad]

Oops sorry I miss wrote the integral in one of the lines above. It should read

\<br /> Z=\int exp[ax^2+bxy+cy^2] d^{3n}x d^{3n}y<br />

my problem happens to be specified that the exponential is of the form

\\x^2+(m w y)^2 + 2mwxyThis makes the matrix in the above solution singular and the integral cannot be performed.

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Note the problem that I was working on should not be solved like this. All I needed to do for that problem was do a simple variable change and integrate over a hypersphere described by the Hamiltonian. Not the exponential of the Hamiltonian. It is the usual integration one performs for multiplicity calculations. But another way to solve this problem is to calculate the partition function instead of the multiplicity. This calculation involves the partition function, which is denoted Z above.
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