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.


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    $\begingroup$ Can you add a little detail for those in the audiences who are not physicists, as well as what you're doing to attack the problem? $\endgroup$ – Ron Gordon Mar 24 '13 at 23:58
  • $\begingroup$ I've tried to add a little more detail, but I don't really understand it much myself, I'm not a physicist either. $\endgroup$ – Tom Mar 25 '13 at 0:08
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    $\begingroup$ @Tom: Would you like this question moved to physics.SE? $\endgroup$ – Thomas Mar 25 '13 at 0:11
  • $\begingroup$ I was actually just on there and found that someone has asked the same question, so maybe just have it deleted here? $\endgroup$ – Tom Mar 25 '13 at 0:18
  • $\begingroup$ @Tom, so you didn't like the answer that was given? $\endgroup$ – Thomas Mar 25 '13 at 0:18

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.


One elegant proof of this that clearly displays why commutation with the Hamiltonian tells you something about time dependence is most easily seen recalling the Heisenberg equation of motion $$ \dot A = \frac{i}{\hbar}[H, A(t)]+ \partial_t A, \qquad A(t) = e^{i H t/\hbar}Ae^{-iHt/\hbar} $$ whose derivation from the Schrodinger equation is given in the Wiki link. For any operator $A$ with $\partial_t A = 0$, the second term on the right vanishes. Moreover, if the operator commutes with the Hamiltonian, then the first time on the right vanishes as well. Therefore, for any state $|\psi(t)\rangle = e^{-iHt/\hbar}|\psi(0)\rangle$ one has $$ \frac{d}{dt}\langle\psi(t)|A|\psi(t)\rangle=\frac{d}{dt}\langle \psi(0)|e^{iHt/\hbar}Ae^{-iHt/\hbar}|\psi(0)\rangle = \langle\psi(0)|\dot A|\psi(0)\rangle = 0 $$ In this case, $L$, the angular momentum operator, has both of the properties that cause the right hand side of the Heisenberg equation of motion to vanish, so we are done!

By the way, I've used some abuses of notation that are common in physics when switching between the Heisenberg and Schrodinger pictures; let me know if this is confusing as a result, and I can make things more notationally precise.


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