(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?


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

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.