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 Exercise [22.09] 
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Joined: 12 Jul 2010, 07:44
Posts: 154
Post Exercise [22.09]
Taylor series expansion of \psi(t) is:


\psi(t) = \psi(0) + t\frac{d}{dt}\psi(0) + t^2\frac{1}{2}\frac{d^2}{dt^2}\psi(0) + \cdots = e^{\left(t\frac{d}{dt}\right)}\psi(0)


Therefore the time evolution operator \boldsymbol{U_t} is actually:


\boldsymbol{U_t} = e^{t\frac{d}{dt}} = e^{-t\frac{i}{\hbar}\mathcal{H}}



This has inverse:


\boldsymbol{U_t}^{-1} = \boldsymbol{U_t}^* = e^{t\frac{i}{\hbar}\mathcal{H}}


Therefore, the Heisenberg version \boldsymbol{Q_H} of an operator \boldsymbol{Q} is:


\boldsymbol{Q_H} = \boldsymbol{U_t}^{-1}\boldsymbol{Q}\boldsymbol{U_t} = e^{t\frac{i}{\hbar}\mathcal{H}}\boldsymbol{Q}e^{-t\frac{i}{\hbar}\mathcal{H}}



Taking the derivative of this, assuming neither \mathcal{H} nor \boldsymbol{Q} has any explicit time-dependence we get:


\frac{d}{dt}\boldsymbol{Q_H} = \frac{i}{\hbar}\left(\mathcal{H}e^{t\frac{i}{\hbar}\mathcal{H}}\boldsymbol{Q}e^{-t\frac{i}{\hbar}\mathcal{H}} - e^{t\frac{i}{\hbar}\mathcal{H}}\boldsymbol{Q}\mathcal{H}e^{-t\frac{i}{\hbar}\mathcal{H}}\right) = \frac{i}{\hbar}\left(\mathcal{H}\boldsymbol{Q_H}-\boldsymbol{Q_H}\mathcal{H}\right)


and therefore


i\hbar\frac{d}{dt}\boldsymbol{Q_H} = \boldsymbol{Q_H}\mathcal{H} - \mathcal{H}\boldsymbol{Q_H}


as required.

(The book has the two terms on the RHS swapped around, but that's an error).


07 Oct 2012, 14:23
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