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(Shankar 12.2.4)

Let $U[R(\epsilon_z\hat k)] = I - {i\over\hbar}\epsilon_z L_z$ be the infinitesimal generator for rotation operators, and $T(\vec\epsilon) = I - {i\over\hbar}\vec\epsilon\cdot\vec P$ the generator for translations in the x-y plane, where $\vec\epsilon=(\epsilon_x,\epsilon_y)$ and $\vec P=(P_x,P_y)$. Then consider $\bar U = U[R(-\epsilon_z\hat k)]\:T(-\vec\epsilon)\:U[R(\epsilon_z\hat k)]\:T(\vec\epsilon)$ which infinitesimally translates a quantum state by $\vec\epsilon$, rotates it by $\epsilon_z$, translates it by $-\vec\epsilon$, and then rotates it by $-\epsilon_z$. Given a coordinate $(x,y)$, we can follow it through the transformations:

$$ \left(\begin{array}{c} x \\ y \end{array}\right) \overset{T(\vec\epsilon)}{\longrightarrow} \left(\begin{array}{c} x + \epsilon_x \\ y + \epsilon_y \end{array}\right) \overset{U[R(\epsilon_z\hat k)]}{\longrightarrow} \left(\begin{array}{c} x + \epsilon_x - \epsilon_z(y + \epsilon_y) \\ y + \epsilon_y + \epsilon_z(x + \epsilon_x) \end{array}\right) \\ \overset{T(-\vec\epsilon)}{\longrightarrow} \left(\begin{array}{c} x - \epsilon_z(y + \epsilon_y) \\ y + \epsilon_z(x + \epsilon_x) \end{array}\right) \overset{U[R(-\epsilon_z\hat k)]}{\longrightarrow} \left(\begin{array}{c} x - \epsilon_z(y + \epsilon_y) + \epsilon_z(y + \epsilon_z(x + \epsilon_x)) \\ y + \epsilon_z(x + \epsilon_x) - \epsilon_z(x - \epsilon_z(y + \epsilon_y)) \end{array}\right) \\ = \left(\begin{array}{c} x + x\epsilon_z^2 + \epsilon_x\epsilon_z^2 - \epsilon_y\epsilon_z \\ y + y\epsilon_z^2 + \epsilon_y\epsilon_z^2 + \epsilon_x\epsilon_z \end{array}\right) $$

Then I think $\bar U = T \left(\begin{array}{c} \epsilon_x\epsilon_z^2 - \epsilon_y\epsilon_z \\ \epsilon_y\epsilon_z^2 + \epsilon_x\epsilon_z \end{array}\right) = I - {i\over\hbar}(\epsilon_x\epsilon_z^2 - \epsilon_y\epsilon_z)P_x - {i\over\hbar}(\epsilon_y\epsilon_z^2 + \epsilon_x\epsilon_z)P_x$. Expanding the original form gives the following equality:

$$ \bar U = \left(I + {i\over\hbar}\epsilon_z L_z\right) \left(I + {i\over\hbar}\epsilon_x P_x + {i\over\hbar}\epsilon_y P_y\right) \left(I - {i\over\hbar}\epsilon_z L_z\right) \left(I - {i\over\hbar}\epsilon_x P_x - {i\over\hbar}\epsilon_y P_y\right) $$

Looking at the terms of order $\epsilon_x\epsilon_z^2$, I get the following constraint:

$$ -L_zP_xL_z+L_z^2P_x = \hbar^2 P_x $$

But I am told I should conclude

$$ -2L_zP_xL_z+P_xL_z^2+L_z^2P_x = [L_z,[L_z,P_x]] = -i\hbar[L_z,P_y] = \hbar^2P_x $$

What happened to the other part of the commutator form?

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You could use a little outside knowledge and observe that for any Lie group with lie algebra $\mathfrak{g}$ with $X\,Y\in\mathfrak{g}$:

$$\exp(Y)\,\exp(X)\,\exp(-Y) = \exp(Z);\\\\ Z= X + [Y,\,X] + \frac{1}{2!} [Y,\,[Y,\,X]]+\frac{1}{3!}[Y,\,[Y,\,[Y,\,X]]]+\cdots$$

This is another form of the so called braiding formula $\mathrm{Ad}(e^Y) = \exp(\mathrm{ad}(Y))$. You will need to learn this for the study of QM.

Now use it to simplfy the $U[R(-\epsilon_z\hat k)]\:T(-\vec\epsilon)\:U[R(\epsilon_z\hat k)]$ in your product with $Y$ as the "infinitessimal generator" for $U[R(-\epsilon_z\hat k)]$ i.e. $Y=+{i\over\hbar}\epsilon_z L_z$ and $X$ as the "infinitessimal generator" for $T(-\vec\epsilon)$ i.e. $X=+ {i\over\hbar}\vec\epsilon\cdot\vec P$. We get $Z=+\frac{\epsilon_z}{\hbar^2}\,[L_z,\,\epsilon_x\,P_x+\epsilon_y\,P_y]$. So you get, to first order:

$$U[R(-\epsilon_z\hat k)]\:T(-\vec\epsilon)\:U[R(\epsilon_z\hat k)]\approx \exp\left(+ {i\over\hbar}(\epsilon_x\,P_x+\epsilon_y\,P_y) - \frac{\epsilon_z}{\hbar^2}\,[L_z,\,\epsilon_x\,P_x+\epsilon_y\,P_y]+ 3^{rd}\text{ order terms & higher}\right)$$

Now we add in the last term $T(+\vec\epsilon)$ of the product on the right to get:

\begin{array}{cl}&U[R(-\epsilon_z\hat k)]\:T(-\vec\epsilon)\:U[R(\epsilon_z\hat k)]\: T(+\vec\epsilon)\\\\=&\left(\mathrm{id}+{i\over\hbar}(\epsilon_x\,P_x+\epsilon_y\,P_y)-\frac{\epsilon_z}{2!\,\hbar^2}\,[L_z,\,\epsilon_x\,P_x+\epsilon_y\,P_y]+\frac{1}{2!}\left({i\over\hbar}(\epsilon_x\,P_x+\epsilon_y\,P_y)\right)^2+\cdots\right)\times\left(\mathrm{id}- {i\over\hbar}(\epsilon_x\,P_x+\epsilon_y\,P_y)+\cdots\right)\\\\=&\mathrm{id}+\frac{\epsilon_z}{\hbar^2}\,[L_z,\,\epsilon_x\,P_x+\epsilon_y\,P_y]+\cdots\end{array}

which will hopefully give you the expression you need. The second order terms all cancel out if you expand correctly

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  • $\begingroup$ Ah, I see where I've made the error originally now: I didn't expand the rotation operators out to second order. I didn't realize the meaning of I+iεL/h as an expansion of exp(iεL/h) up to linearity. Expanding up to quadraticity gives the full commutator form. $\endgroup$ – Alan Jun 21 '15 at 8:04
  • $\begingroup$ Though thanks for the info. Getting the commutator relation through the braiding formula is a neater way to the constraint. $\endgroup$ – Alan Jun 21 '15 at 8:10

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