Question:
Given an isotropic Quantum Harmonic Oscillator, what is the ground state energy.
$ \hat{H} = -\frac{\hbar^2}{2M} \nabla^2 + V\left(x_1,x_2,x_3\right)= -\frac{\hbar^2}{2M}\nabla^2 + \frac{M\omega^2}{2}\left(x_1^2+x_2^2+x_3^2\right). $
Solution:
The Hamilitonian can be written as a sum
$ \begin{eqnarray*} \hat{H} &=& -\frac{\hbar^2}{2M}\nabla^2 + \frac{M\omega^2}{2}\left(x_1^2+x_2^2+x_3^2\right) \\ &=& -\frac{\hbar^2}{2M}(\frac{\partial^2}{\partial x_1^2} + \frac{\partial^2}{\partial x_2^2} + \frac{\partial^2}{\partial x_3^2}) + \frac{M\omega^2}{2}\left(x_1^2+x_2^2+x_3^2\right) \\ &=& \sum\limits_{k=1}^{3}(-\frac{\hbar^2}{2M}\frac{\partial^2}{\partial x_k^2} + \frac{M\omega^2x_k^2}{2}) \end{eqnarray*} $
In this form, we can easily see the energy is just the sum of three harmonic oscillator, so the answer is $ \hbar\omega(n_1+n_2+n_3+\frac{3}{2}) $.
Phew - finally an easier question - blocked on question 3 for so long.
Given an isotropic Quantum Harmonic Oscillator, what is the ground state energy.
$ \hat{H} = -\frac{\hbar^2}{2M} \nabla^2 + V\left(x_1,x_2,x_3\right)= -\frac{\hbar^2}{2M}\nabla^2 + \frac{M\omega^2}{2}\left(x_1^2+x_2^2+x_3^2\right). $
Solution:
The Hamilitonian can be written as a sum
$ \begin{eqnarray*} \hat{H} &=& -\frac{\hbar^2}{2M}\nabla^2 + \frac{M\omega^2}{2}\left(x_1^2+x_2^2+x_3^2\right) \\ &=& -\frac{\hbar^2}{2M}(\frac{\partial^2}{\partial x_1^2} + \frac{\partial^2}{\partial x_2^2} + \frac{\partial^2}{\partial x_3^2}) + \frac{M\omega^2}{2}\left(x_1^2+x_2^2+x_3^2\right) \\ &=& \sum\limits_{k=1}^{3}(-\frac{\hbar^2}{2M}\frac{\partial^2}{\partial x_k^2} + \frac{M\omega^2x_k^2}{2}) \end{eqnarray*} $
In this form, we can easily see the energy is just the sum of three harmonic oscillator, so the answer is $ \hbar\omega(n_1+n_2+n_3+\frac{3}{2}) $.
Phew - finally an easier question - blocked on question 3 for so long.
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