The potential is given by $V\left(\left\|(x,y,z)\right\|\right)$, so $[\hat{L}, \hat{H}] = 0$.
Using the definition of $\langle \hat{L} \rangle$ and the time-dependent Schrödinger equation, show that the expectation value of angular momentum does not change with time.
Expectation value of angular momentum, $\langle \hat{L} \rangle$ = $\langle \psi \mid \hat{L} \mid \psi \rangle $ where $\psi$ is a generic wavefunction.
Best Answer
The TDSE is: $$ \hat{H}\Psi = i \hbar \frac{\partial \Psi}{\partial t} $$
Taking the complex conjugate (note that $H=H^{*}$ since the Hamiltonian is Hermitian):
$$ -i \hbar \frac{\partial \Psi^{*}}{\partial t} = (\hat{H}\Psi)^{*} = \Psi^{*}H^{*} = \Psi^{*}H $$
By definition:
$$ \langle\hat{L}\rangle=\langle\Psi|\hat{L}|\Psi\rangle = \int_{\mathbf{R^3}}\Psi^{*}\hat{L}\Psi \,\mathbf{dr}^3 $$
Therefore, since $\hat{L}$ is time-independent:
$$ \frac{\partial}{\partial t} \langle\hat{L}\rangle= \int_{\mathbf{R^3}}\frac{\partial \Psi^{*}}{\partial t}\hat{L}\Psi \,\mathbf{dr}^3 + \int_{\mathbf{R^3}}\Psi^{*}\hat{L}\frac{\partial \Psi}{\partial t} \,\mathbf{dr}^3 $$
Sub in the first two equations and multiply through by $i \hbar$:
$$ i\hbar \frac{\partial}{\partial t} \langle\hat{L}\rangle= -\int_{\mathbf{R^3}}\Psi^{*}\hat{H}\hat{L}\Psi \,\mathbf{dr}^3 + \int_{\mathbf{R^3}}\Psi^{*}\hat{L}\hat{H}\Psi \,\mathbf{dr}^3 = \int_{\mathbf{R^3}}\Psi^{*}(\hat{L}\hat{H}-\hat{H}\hat{L})\Psi^{*} $$
$$ i\hbar \frac{\partial}{\partial t} \langle\hat{L}\rangle = \int_{\mathbf{R^3}}\Psi^{*}[\hat{H},\hat{L}]\Psi^{*} = 0 $$ Therefore, $\frac{\partial}{\partial t}\langle\hat{L}\rangle=0$, which means that $\langle\hat{L}\rangle$ is a constant, as we wanted to show.