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.

Just apply Hermitian conjugation to both sides of the equality: $C^+=(i[A,B])^+=-i[AB-BC]^+=-i[B^+A^+-A^+B^+]$= ... (sorry, had to use + sign instead of the dagger).

## Best Answer

Try to answer the following question as a warm-up. What can you say about the expectation value of the $1D$ momentum operator $p = -i \hbar \tfrac{d}{dx}$ of a real wavefunction? Note that the expectation value itself should be real, because momentum is a real quantity (and the operator is hermitian).