0
$\begingroup$

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 $|\psi\rangle$

$\endgroup$
  • 3
    $\begingroup$ Please reproduce all of the steps of your derivation so they can be checked. $\endgroup$ – my2cts Mar 6 at 22:32
  • $\begingroup$ Please check $cos(0) = 1$, so it does not hold at $t=0$. $\endgroup$ – wang1908 Mar 6 at 23:12
  • $\begingroup$ 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! $\endgroup$ – Juan Pablo Arcila Mar 7 at 1:58
  • $\begingroup$ 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". $\endgroup$ – Cosmas Zachos Mar 8 at 15:00
0
$\begingroup$

Virial theorem only holds for stationary/eigen state.

$\endgroup$
  • 2
    $\begingroup$ Er ... the Virial system can be applied to any system that is (A) bound and (B) either periodic or endures for a long time. But the system has to be averaged over the time period in which it qualifies for (B). Eigen-states are special because they qualify at every instant, but they are not the only states for which you can apply the theorem. $\endgroup$ – dmckee Mar 7 at 1:28
  • 1
    $\begingroup$ Isn't what you're saying the answer to the problem dmckee? the average of a cosine function is 0. In that case, Virial theorem is working! $\endgroup$ – Juan Pablo Arcila Mar 7 at 1:59
  • $\begingroup$ @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. $\endgroup$ – dmckee Mar 8 at 21:08
  • $\begingroup$ Considering the molecular systems where Virial theorem only holds at its geometrical equilibrium, the two conditions clearly insufficient, even being eigenstate is necessary but not suffcient. $\endgroup$ – wang1908 Mar 9 at 12:57
  • $\begingroup$ @wang Those are precisely the pre-conditions for deriving the Virial theorem in the first place. Whether applying the theorem to a qualifying system is useful or not is a distinct problem from it being meaningful to invoke it. $\endgroup$ – dmckee Mar 9 at 17:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.