In mechanics, the ''virial theorem'' provides a general equation relating the average over time of the total kinetic energy $T$ of a stable system consisting of N particles, bound by potential forces, with that of the total potential energy, $\langle V_{tot}\rangle$, where angle brackets represent the average over time of the enclosed quantity. Mathematically, the theorem states $$2 \left\langle T \right\rangle = -\sum_{k=1}^N \left\langle \mathbf{F}_k \cdot \mathbf{r}_k \right\rangle$$ where $F_k$ represents the force on the $k$-th particle, which is located at position $r_k$. In Quantum Mechanics, we have similar results. **Theorem** Let $H=\boldsymbol p^2/2m + V(r)$ be the Hamiltonian for a three-dimensional particle in a spherically-symmetric potential. Show that $$2\langle \psi|T|\psi\rangle =\big\langle \psi|r\f{\p V}{\p r}|\psi\big\rangle,$$ where $T$ is the kinetic energy and $|\psi\rangle$ is an energy eigenstate. **Remark** If $V(r)$ is proportional to $r^k$, then for the energy eigenstate $|\psi\rangle$ with energy $E$, we have $$ E=(1+\f{k}{2})\langle \psi|V|\psi\rangle.$$ //Proof// Let $|\psi\rangle$ be a normalized bound state with definite energy $E$ (i.e. $H|\psi\rangle = E|\psi\rangle$), and $A$ be any well-defined Hermitean operator, then it is easy to show that $$(1)\hspace{25pt}\langle \psi|[A,H]|\psi\rangle=0.$$ By the fundamental relation between $\boldsymbol x$ and $\boldsymbol p$, we derive that $$(2)\hspace{25pt}[H,\boldsymbol x]=-i\hbar \f{\boldsymbol p}{m},\hspace{25pt}[H,\boldsymbol p]=i\hbar\grad V(r).$$ Let $A=\boldsymbol x\cdot\boldsymbol p$, we compute by using (2): \begin{equation} \begin{split} [H,A]&=[H,\boldsymbol x]\cdot p +\boldsymbol x\cdot[H,\boldsymbol p]~\\ &=-i\hbar \f{\boldsymbol p^2}{m} + \boldsymbol x\cdot i\hbar\grad V(r)~\\ &=-i\hbar\, 2T + i\hbar r\f{\p V}{\p r}. \end{split} \end{equation} Using (1) gives $$2\langle \psi|T|\psi\rangle =\big\langle \psi|r\f{\p V}{\p r}|\psi\big\rangle.$$ For $V(r)$ is proportional to $r^k$, it is easy to verify that $$r\f{\p V}{\p r} = k V(r).$$ Therefore, \begin{equation} \begin{split} E=\langle\psi|H|\psi\rangle&=\langle \psi|T|\psi\rangle + \langle \psi|V|\psi\rangle~\\ &=\f{1}{2}\langle \psi|r\f{\p V}{\p r}|\psi\rangle + \langle \psi|V|\psi\rangle~\\ &=(1+\f{k}{2}) \langle \psi|V|\psi\rangle, \end{split} \end{equation} which proves the **Remark**.