Virial theorem as applied to hydrogen atom

[Dadface]

In the virial theorem the numerical value of the average potential energy within a system is exactly twice that of the average kinetic energy. I know the theorem is proved mathematically but to me it seems a coincidence that one value is exactly twice the other value. I find that interesting.
I find it more interesting and more of a coincidence when I take into account the fact that the potential energy term is not an absolute value but the value it would have if the potential energy at the chosen separation of infinity is given the chosen value of zero.
Can anyone explain, without the maths, why one value is twice the other. I'm trying to get an intuitive feeling of why this should be the case.
Also, does the Schrödinger analysis of the hydrogen atom give a better proof of the virial theorem than the Bohr treatment? I ask this because I think the simple Bohr analysis ignores the kinetic energy of the proton.

 

[PeterDonis]

In the virial theorem the numerical value of the average potential energy within a system is exactly twice that of the average kinetic energy.

More precisely, in the virial theorem for a potential which goes like 1/r, we have $2 \langle T \rangle = - \langle V \rangle$, where $\langle T \rangle$ is the time average of the total kinetic energy and $\langle V \rangle$ is the time average of the total potential energy. The minus sign comes from the $1 / r$ potential; it is a special case of a more general theorem which says that, for a potential that goes like $r^n$, we have $2 \langle T \rangle = n \langle V \rangle$. The $1 / r$ potential is just the case $n = -1$.

See here:

https://en.wikipedia.org/wiki/Virial_theorem

the potential energy term is not an absolute value but the value it would have if the potential energy at the chosen separation of infinity is given the chosen value of zero

Again, this is for a potential that goes like $1 / r$; it obviously vanishes as $r \rightarrow \infty$. More generally, a potential that goes like $r^n$ will vanish as $r \rightarrow \infty$ for any $n < 0$. This makes physical sense since we expect two particles at infinite separation to have zero interaction with each other.

 

[Dadface]

Thank you PeterDonis. The thing that interests is the fact that one value is numerically twice the other value. Without looking into the theory in more detail it seems to be a nice coincidence.