Exercise [22.09]

deant · 07 Oct 2012, 14:23

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).