Derivation of the Lorentz force, QM

[ag_swe]

Studying in the Heisenberg picture, we have

\frac{\text{d}x_i}{\text{d}t}=-\frac{\text{i}}{\hbar}[x_i,H]=\frac{1}{m}\left(p_i-\frac{q}{c}A_i\right)

where the last bracket is known as the kinematic momentum \pi. Now, to find \frac{\text{d}^2{\bf{x}}}{\text{d}t^2} I do the following:

\frac{\text{d}^2{\bf{x}}}{\text{d}t^2} = [\frac{\text{d}{\bf{x}}}{\text{d}t},H]<br /> =[\frac{\pi}{m},\frac{\pi^2}{2m}+q\Phi]=

\frac{\pi}{m}(\frac{\pi^2}{2m}+q\Phi)-(\frac{\pi^2}{2m}+q\Phi)\frac{\pi}{m}=

\frac{\pi}{m}q\Phi-q\Phi\frac{\pi}{m}=

\frac{q}{m}(-\text{i}\hbar\nabla-\frac{q}{c}{\bf{A}})\Phi-\frac{q}{m}\Phi(-\text{i}\hbar\nabla-\frac{q}{c}{\bf{A}})

\frac{\text{i}\hbar q}{m}{\bf{E}}-\frac{q^2}{mc}{\bf{A}}\Phi+\frac{\text{i}\hbar q}{m}\Phi \nabla+\frac{q^2}{mc}\Phi {\bf{A}}=?

The correct expression should be

\frac{\text{d}^2{\bf{x}}}{\text{d}t^2} =\frac{q}{m}[{\bf{E}}+\frac{1}{2c}(\frac{\text{d}{\bf{x}}}{t} \times {\bf{B}}-{\bf{B}} \times \frac{\text{d}{\bf{x}}}{t})]

My expression should be multiplied with the constant -\text{i}/\hbar, so at least the electric field term in my expression is correct. But how do I transform the rest into the wanted expression?

 

[samalkhaiat]

Use
m\frac{dv_{i}}{dt} = im[H , v_{i}] + m\frac{\partial v_{i}}{\partial t}

then write
H = \frac{1}{2}m \sum_{k}v_{k}v_{k} + q \phi

Now
<br /> [H , v_{i}] = \frac{m}{2}\sum_{k} \{v_{k}[v_{k}, v_{i}] + [v_{k} , v_{i}] v_{k}\} + \frac{q}{m}[\phi , p_{i}]<br />

Next use the operator equation (try to prove it!)
[v_{i},v_{k}] = i \frac{q}{m^{2}}\epsilon_{ikj}B_{j}
where
B_{j}= (\vec{\nabla}\times \vec{A})_{j}
Notice that
m\frac{\partial v_{i}}{\partial t} = - q \frac{\partial A_{i}}{\partial t}
Put every thing together and you will get the Lorentz force operator. Can you now see your mistakes?

Sam

 

[ag_swe]

Thank you very much. Very appreciated! However, I didn't get how to find the two vector products without "looking into the future", i.e., knowing what I was looking for.