Find Ground State Energy of 3D Harmonic Oscillator

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The discussion focuses on finding the ground state energy of a 3D harmonic oscillator Hamiltonian. Participants suggest rewriting the equation using new variables, specifically introducing vectors R and r to facilitate the analysis. Clarification is requested regarding a potential typo in the original equation to ensure accurate communication. The goal of the proposed substitution is to separate the Hamiltonian into two distinct parts, simplifying the problem. Overall, the conversation emphasizes the importance of precise notation and effective variable transformation in solving the Hamiltonian.
koustav
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Summary:: I am trying to find the exact ground state energy of the hamiltonian.kindly help me with this

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I suggest rewriting your equation in terms of new variables:

##\vec{R} = \frac{1}{2} (\vec{r}_1 + \vec{r}_2)##
##\vec{r} = \vec{r}_1 - \vec{r}_2##
 
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To the OP, would you clarify the typo in the last term, just so we are all certain what we're working with.
 
Haborix said:
To the OP, would you clarify the typo in the last term, just so we are all certain what we're working with.
there will be vector sign and no square on the last term
 
Thank you. Stevendaryl's suggestion is a good one, give it a try and let us know how you get along. The goal of that kind of substitution is to get a Hamiltonian which separates into a sum two Hamiltonians, one in the ##R## coordinate and the other in the ##r## coordinate.
 
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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