Shankar 14.4.1 using Ehrenfest's Theorem

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SUMMARY

This discussion focuses on the application of Ehrenfest's theorem to derive the time evolution of the expectation value of angular momentum, specifically in the context of a Hamiltonian defined as H = -γL·B, where B is a position-independent magnetic field. The key equations include the angular momentum commutation relations [L_i, L_j] = iħε_{ijk}L_k and the expression for the magnetic moment μ = γL. The main conclusion reached is that the correct expression for the commutator [H, L]_x leads to a factor of 2 in the time derivative of the expectation value of angular momentum, which is resolved by correctly applying the commutation relations.

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Homework Statement


Show that if
[tex]H = -\gamma\mathbf{L\cdot B}[/tex], and B is position independent,

[tex]\frac{\mathrm{d} \left \langle \mathbf{L} \right \rangle}{\mathrm{d} t}=\left \langle \boldsymbol\mu \times\mathbf{B} \right \rangle=\left \langle \boldsymbol\mu \right \rangle\times\mathbf{B}[/tex]

Here H is the Hamiltonian, B is the magnetic field, L is the angular momentum operator, mu is the magnetic moment, and gamma is the gyromagnetic ratio.

I want to solve this using Ehrenfest's theorem.

Homework Equations


[tex]\boldsymbol\mu=\gamma\mathbf{L}[/tex]

[tex][L_i,L_j]=i\hbar\epsilon_{ijk}L_k[/tex]
(the fancy way of writing the usual angular momentum commutation relations)

and Ehrenfest's theorem:
[tex]\frac{\mathrm{d} \left \langle Q \right \rangle}{\mathrm{d} t} = \frac{i}{\hbar}\left \langle [H,Q] \right \rangle+\left \langle \frac{\partial Q}{\partial t} \right \rangle[/tex]

The Attempt at a Solution


First of all, L has no explicit time dependence so the last term in Ehrenfest's theorem cancels. I seem to be getting a factor of 2 in [H,L] that shouldn't be there:

The x component of [H,L] is:
[tex][H,L]_x=\gamma\left ( L_xL_xB_x+L_xL_yB_y+L_xL_zB_z - L_xB_xL_x-L_yB_yL_x-L_zB_zL_x \right )[/tex]

[tex]=\gamma i\hbar\left ( L_zB_y-L_yB_z +L_zB_y-L_yB_z \right )[/tex]
Edit: ^^^^ This is my mistake.[tex]= 2i\hbar\left ( \mu_zB_y-\mu_yB_z \right )=\frac{2\hbar}{i}(\boldsymbol\mu\times\mathbf{B})_x[/tex]
and likewise for the other components.

Putting it together I get
[tex]\frac{\mathrm{d} \left \langle \mathbf{L} \right \rangle}{\mathrm{d} t}=\frac{i}{\hbar}\left \langle [H,L]\right \rangle=\frac{i}{\hbar}\left \langle \frac{2\gamma\hbar}{i}\mathbf{L}\times\mathbf{B} \right \rangle=\left \langle 2\boldsymbol\mu\times\mathbf{B} \right \rangle[/tex]

and I'm left with that pesky two.

Where did I go wrong?Edit: I found my error and pointed out where I went wrong above.

That line should be
[tex][H,\mathbf{L}]_x=\gamma\left ((L_xL_y-L_yL_x)B_y+(L_xL_z-L_zL_x)B_z \right )=\gamma \left (i\hbar L_zB_y-i\hbar L_yB_z \right )=\frac{\hbar}{i}(\boldsymbol\mu \times\mathbf{B})_x[/tex]
then everything works.
 
Last edited:
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Your mistake is in the first step where you wrote [H,L]_x. The correct expression should be [H,L]_x=\gamma\left ( L_xL_xB_x+L_xL_yB_y+L_xL_zB_z - L_xB_xL_x-L_yB_yL_x-L_zB_zL_x \right )=\gamma i\hbar\left ( L_zB_y-L_yB_z +L_zB_y-L_yB_z \right )= 2\gamma i\hbar\left ( \mu_zB_y-\mu_yB_z \right )=2\gamma\hbar(\boldsymbol\mu\times\mathbf{B})_x
The extra factor of 2 comes from the fact that [L_i,L_j]=i\hbar\epsilon_{ijk}L_k is a sum over k, so when you expand the commutator [H,L]_x, you will have two terms with the same coefficient.
 

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