Consider Schrödinger's Equation,
$$H=\sum^3_{i=1} \frac{p^2_i}{2m_i}+V(x_1,x_2,x_3).$$
In one dimensional case, we can analyse the shape of the potential, i.e
$$V(x)=\frac{1}{2}m_1 \omega^2_1 x^2$$
is the potential for quantum oscillator. The ground state of quantum oscillator looks like a Gaussian. For two dimensional oscillator we can write
$$V(x,y)=\frac{1}{2}m_1 \omega^2_1 x^2+ \frac{1}{2}m_2 \omega^2_2 y^2,$$
the ground state of this system is again looks like a Gaussian in two dimensions.
If we proceed further we can write
$$V(x,y,z)=\frac{1}{2} m_1 \omega^2_1 x^2+\frac{1}{2}m_2 \omega^2_2 y^2+\frac{1}{2}m_3 \omega^2_3 z^2$$
as the potential of thee dimensional harmonic oscillator.
I hope again the ground state of this system is a Gaussian, but in three dimensions I am unable to understand which shape it will get. What will happen if we further increase our dimensions say more than three?
Best Answer
The graphical representation of the probability density distributed over the three-dimensional space would be a four-dimensional plot--just like the plot of a probability density distribution over one dimension is two-dimensional and that of a probability density distribution over two dimensions is three-dimensional. There is no direct way to visualize a four-dimensional plot except via its projections onto lower dimensional spaces.
Now, for the specific theory of decoupled harmonic oscillators, the ground state would be the multiplication of Gaussians in each of the dimensions. Thus, the full probability density distribution over the three-dimensional space, when projected onto one (or two) dimension(s), would simply look like Gaussians in those lower dimensions. But, this simplification is owing to the decoupling of oscillators, such a simplification would not be generically possible.