Discussion
A particle of mass m in three dimensions is subjected to a potential \(V(\vec{r})\) = A ln(r/ro) where \(r = |\vec{r}|\) and A and ro are constants. Then, in some normalized stationary state say \(\psi(\vec{r})\), the mean square velocity, \(\langle \vec{v}^{2} \rangle\), is equal to :
Answer: D
The mean square velocity is \(A/m\).
In any stationary state the quantum virial theorem holds:
\[ \begin{aligned} 2\langle T \rangle &= \langle \vec{r}\cdot\vec{\nabla}V \rangle \\ \vec{r}\cdot\vec{\nabla}V &= r\frac{d}{dr}A\ln\frac{r}{r_{o}} = A \end{aligned} \]
so \(\langle T \rangle = A/2\). With \(\langle T \rangle = \frac{1}{2}m\langle v^{2} \rangle\), this gives \(\langle v^{2} \rangle = A/m\), the same in every bound state of the logarithmic potential.
\(2A/m\) and \(A/2m\) are off by a factor of 2 in one direction or the other (from dropping the 2 in the virial theorem or the 1/2 in the kinetic energy), and \(3A/m\) wrongly multiplies by the number of dimensions.
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