# Virial theorem hydrogen atom

I calculated $$\langle T \rangle$$ and $$\langle V \rangle$$ as a function of time of a given state of the hydrogen atom $$|\psi\rangle=a|1,0,0\rangle+b|2,0,0\rangle$$ and I found that

$$\langle V \rangle =-c+e\cos(dt)$$ $$\langle T \rangle=\frac{c}{2}-e\cos(dt)$$

where $$a,b,c,d,e$$ are constants. I see then that for $$t=0$$ the virial theorem for the potential $$\frac{-e^2}{r}$$, that is, $$\langle V \rangle=-2\langle T \rangle$$ holds. However, for other times it is not true.

Why is the virial theorem only valid here for the initial time

PS: I calculated the expected values as a function of time applying the time evolution operator to $$|\psi\rangle$$

• Please reproduce all of the steps of your derivation so they can be checked. – my2cts Mar 6 '19 at 22:32
• Please check $cos(0) = 1$, so it does not hold at $t=0$. – wang1908 Mar 6 '19 at 23:12
• It is kind of a long precedure, but I got the cosine term from the expected values $<1,0,0|V|2,0,0>$ and $<2,0,0|V|1,0,0>$, which are equal. factorizing these constants I end up having $e^{iu}+e^{-iu}$, and that's a cosine function. I got the $<T>$ result by computing $<H>-<V>$ and knowing that $H|n,l,m>=h_{n}|n,l,m>$. I used too the fact that $<n,l,m|V|n,l,m>=\frac{1}{a_{0}n^2}$ with $a_{0}$ the Bohr radius.\\ on the other hand, yes, it doesn't hold at $t=0$, thanks for bringing that up! – Juan Pablo Arcila Mar 7 '19 at 1:58
• I suspect you threw out the baby with the bathwater, and did not calculate the full expression for the general nonstationary state, ie the one not discarding the "velocity". – Cosmas Zachos Mar 8 '19 at 15:00

Virial theorem relates time averaged energies. Thus there should be no time in the expressions of $$V$$ and $$T$$ that you are relating.
• @Jaun Yes. Part of the problem may be notation. The angle-brackets $\langle \cdot \rangle$ are often used in discussions of the Virial theorem where they mean "a suitable time-average" and they are also often used in quantum mehcanics where they mean "a suitable expectation value". If the quantum system is expressed in the position basis, then that expectation value represents a position average so it's not the same thing usually meant when the brackets are used in discussions of the Virial theorem. – dmckee --- ex-moderator kitten Mar 8 '19 at 21:08