- #1
ag_swe
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Studying in the Heisenberg picture, we have
[itex] \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) [/itex]
where the last bracket is known as the kinematic momentum [itex] \pi [/itex]. Now, to find [itex] \frac{\text{d}^2{\bf{x}}}{\text{d}t^2} [/itex] I do the following:
[itex] \frac{\text{d}^2{\bf{x}}}{\text{d}t^2} = [\frac{\text{d}{\bf{x}}}{\text{d}t},H]
=[\frac{\pi}{m},\frac{\pi^2}{2m}+q\Phi]= [/itex]
[itex] \frac{\pi}{m}(\frac{\pi^2}{2m}+q\Phi)-(\frac{\pi^2}{2m}+q\Phi)\frac{\pi}{m}= [/itex]
[itex] \frac{\pi}{m}q\Phi-q\Phi\frac{\pi}{m}= [/itex]
[itex] \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}}) [/itex]
[itex] \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}}=?[/itex]
The correct expression should be
[itex] \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})][/itex]
My expression should be multiplied with the constant [itex] -\text{i}/\hbar[/itex], so at least the electric field term in my expression is correct. But how do I transform the rest into the wanted expression?
[itex] \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) [/itex]
where the last bracket is known as the kinematic momentum [itex] \pi [/itex]. Now, to find [itex] \frac{\text{d}^2{\bf{x}}}{\text{d}t^2} [/itex] I do the following:
[itex] \frac{\text{d}^2{\bf{x}}}{\text{d}t^2} = [\frac{\text{d}{\bf{x}}}{\text{d}t},H]
=[\frac{\pi}{m},\frac{\pi^2}{2m}+q\Phi]= [/itex]
[itex] \frac{\pi}{m}(\frac{\pi^2}{2m}+q\Phi)-(\frac{\pi^2}{2m}+q\Phi)\frac{\pi}{m}= [/itex]
[itex] \frac{\pi}{m}q\Phi-q\Phi\frac{\pi}{m}= [/itex]
[itex] \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}}) [/itex]
[itex] \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}}=?[/itex]
The correct expression should be
[itex] \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})][/itex]
My expression should be multiplied with the constant [itex] -\text{i}/\hbar[/itex], so at least the electric field term in my expression is correct. But how do I transform the rest into the wanted expression?