Question:
What is the ground state energy of the potential $V\left(x_1,x_2,x_3\right)= \frac{M\omega^2}{2}\left(x_1^2+x_2^2+4x_3^2\right) $
Solution:
Following the same mindset in Question 4. The ground state energy is really just the sum of ground state energies, so that would be
$ \hbar\omega\frac{1}{2} + \hbar\omega\frac{1}{2} + \hbar 2\omega\frac{1}{2} = 2\hbar\omega $.
What is the ground state energy of the potential $V\left(x_1,x_2,x_3\right)= \frac{M\omega^2}{2}\left(x_1^2+x_2^2+4x_3^2\right) $
Solution:
Following the same mindset in Question 4. The ground state energy is really just the sum of ground state energies, so that would be
$ \hbar\omega\frac{1}{2} + \hbar\omega\frac{1}{2} + \hbar 2\omega\frac{1}{2} = 2\hbar\omega $.
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