Diagonalizing q1ˆ3q2ˆ3 with Degenerate Perturbation Theory

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The discussion focuses on the challenges of diagonalizing the perturbation q1ˆ3q2ˆ3 using degenerate perturbation theory. The first-order correction to the ground state is calculated using the expression $$\bra{GS} H_{int} \ket{GS}$$, where the ground state is defined as the vacuum state for two identical oscillators. The calculation involves evaluating the integral $$\bra{0}\bra{0} q_1^3 q_2^3 \ket{0} \ket{0}$$, which leads to integrals of the form $$\int dq_i q_i^3 e^{-2q_i^2}$$. A substitution of variables is suggested to simplify the integral, ultimately relating it to the Gamma function. This approach aims to resolve the difficulties encountered in finding the first-order correction.
ThiagoSantos
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Homework Statement
Determine the first order correction of a system of two identical harmonic oscilators
Relevant Equations
Hˆ =(p1ˆ2 + p2ˆ2+q1ˆ2 +q2ˆ2)/2+fq1ˆ3q2ˆ3. where f is the coupling constant
I tried to use the degenerated perturbation theory but I'm having problems when it comes to diagonalizing the perturbation q1ˆ3q2ˆ3 which I think I need to find the first order correction.
 
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I am rusty, but I try.

According to Wikipedia the first-order correction is
$$\bra{GS} H_{int} \ket{GS}$$
(assuming you want to calculate the correction to the ground state ##\ket{GS}##). Your ground state is the vacuum for both oscillators so ##\ket{GS} = \ket{0}\ket{0}##. Where ##\ket{0} \propto \exp(- \omega_i q_i^2)## with ##i = 1,2##.(here you have two identical oscillators so ##\omega_1 = \omega_2 = 1##). So you just have to calculate:
$$\bra{0}\bra{0} q_1^3 q_2^3 \ket{0} \ket{0}$$
which (if I am not mistaken) will result in integrals of the form
$$\int dq_i q_i^3 e^{-2q_i^2} $$
you can solve this by putting ##t=x^2## (##dt = 2xdx##), which will yield the Gamma function (https://en.wikipedia.org/wiki/Gamma_function).
 
So is there some elegant way to do this or am I just supposed to follow my nose and sub the Taylor expansions for terms in the two boost matrices under the assumption ##v,w\ll 1##, then do three ugly matrix multiplications and get some horrifying kludge for ##R## and show that the product of ##R## and its transpose is the identity matrix with det(R)=1? Without loss of generality I made ##\mathbf{v}## point along the x-axis and since ##\mathbf{v}\cdot\mathbf{w} = 0## I set ##w_1 = 0## to...

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