# Rotational transformations using passive transformations

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.2.2.

We can also derive the generator of rotations ${L_{z}}$ by considering passive transformations of the position and momentum operators, in a way similar to that used for deriving the generator of translations. In a passive transformation, the operators are modified while the state vectors remain the same. For an infinitesimal rotation ${\varepsilon_{z}\hat{\mathbf{z}}}$ about the ${z}$ axis in two dimensions, the unitary operator has the form

$\displaystyle U\left[R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)\right]=I-\frac{i\varepsilon_{z}L_{z}}{\hbar} \ \ \ \ \ (1)$

For a finite rotation by ${\phi_{0}\hat{\mathbf{z}}}$ the transformations are given by

 $\displaystyle \left\langle X\right\rangle _{R}$ $\displaystyle =$ $\displaystyle \left\langle X\right\rangle \cos\phi_{0}-\left\langle Y\right\rangle \sin\phi_{0}\ \ \ \ \ (2)$ $\displaystyle \left\langle Y\right\rangle _{R}$ $\displaystyle =$ $\displaystyle \left\langle X\right\rangle \sin\phi_{0}+\left\langle Y\right\rangle \cos\phi_{0}\ \ \ \ \ (3)$ $\displaystyle \left\langle P_{x}\right\rangle _{R}$ $\displaystyle =$ $\displaystyle \left\langle P_{x}\right\rangle \cos\phi_{0}-\left\langle P_{y}\right\rangle \sin\phi_{0}\ \ \ \ \ (4)$ $\displaystyle \left\langle P_{y}\right\rangle _{R}$ $\displaystyle =$ $\displaystyle \left\langle P_{x}\right\rangle \sin\phi_{0}+\left\langle P_{y}\right\rangle \cos\phi_{0} \ \ \ \ \ (5)$

For the infinitesimal transformation, ${\phi_{0}=\varepsilon_{z}}$ and these equations reduce to

 $\displaystyle \left\langle X\right\rangle _{R}$ $\displaystyle =$ $\displaystyle \left\langle X\right\rangle -\left\langle Y\right\rangle \varepsilon_{z}\ \ \ \ \ (6)$ $\displaystyle \left\langle Y\right\rangle _{R}$ $\displaystyle =$ $\displaystyle \left\langle X\right\rangle \varepsilon_{z}+\left\langle Y\right\rangle \ \ \ \ \ (7)$ $\displaystyle \left\langle P_{x}\right\rangle _{R}$ $\displaystyle =$ $\displaystyle \left\langle P_{x}\right\rangle -\left\langle P_{y}\right\rangle \varepsilon_{z}\ \ \ \ \ (8)$ $\displaystyle \left\langle P_{y}\right\rangle _{R}$ $\displaystyle =$ $\displaystyle \left\langle P_{x}\right\rangle \varepsilon_{z}+\left\langle P_{y}\right\rangle \ \ \ \ \ (9)$

In the passive transformation scheme, we move the transformation to the operators to get

 $\displaystyle U^{\dagger}\left[R\right]XU\left[R\right]$ $\displaystyle =$ $\displaystyle X-Y\varepsilon_{z}\ \ \ \ \ (10)$ $\displaystyle U^{\dagger}\left[R\right]YU\left[R\right]$ $\displaystyle =$ $\displaystyle X\varepsilon_{z}+Y\ \ \ \ \ (11)$ $\displaystyle U^{\dagger}\left[R\right]P_{x}U\left[R\right]$ $\displaystyle =$ $\displaystyle P_{x}-P_{y}\varepsilon_{z}\ \ \ \ \ (12)$ $\displaystyle U^{\dagger}\left[R\right]P_{y}U\left[R\right]$ $\displaystyle =$ $\displaystyle P_{x}\varepsilon_{z}+P_{y} \ \ \ \ \ (13)$

Substituting 1 into these equations gives us the commutation relations satisfied by ${L_{z}}$. For example, in the first equation we have

 $\displaystyle U^{\dagger}\left[R\right]XU\left[R\right]$ $\displaystyle =$ $\displaystyle \left(I+\frac{i\varepsilon_{z}L_{z}}{\hbar}\right)X\left(I-\frac{i\varepsilon_{z}L_{z}}{\hbar}\right)\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle X+\frac{i\varepsilon_{z}}{\hbar}\left(L_{z}X-XL_{z}\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle X-Y\varepsilon_{z} \ \ \ \ \ (16)$

Equating the last two lines, we get

$\displaystyle \left[X,L_{z}\right]=-i\hbar Y \ \ \ \ \ (17)$

Similarly, for the other three equations we get

 $\displaystyle \left[Y,L_{z}\right]$ $\displaystyle =$ $\displaystyle i\hbar X\ \ \ \ \ (18)$ $\displaystyle \left[P_{x},L_{z}\right]$ $\displaystyle =$ $\displaystyle -i\hbar P_{y}\ \ \ \ \ (19)$ $\displaystyle \left[P_{y},L_{z}\right]$ $\displaystyle =$ $\displaystyle i\hbar P_{x} \ \ \ \ \ (20)$

We can use these commutation relations to derive the form of ${L_{z}}$ by using the commutation relations for coordinates and momenta:

$\displaystyle \left[X,P_{x}\right]=\left[Y,P_{y}\right]=i\hbar \ \ \ \ \ (21)$

with all other commutators involving ${X,Y,P_{x}}$ and ${P_{y}}$ being zero. Starting with 17, we see that

$\displaystyle \left[X,L_{z}\right]=-\left[X,P_{x}\right]Y \ \ \ \ \ (22)$

We can therefore deduce that

$\displaystyle L_{z}=-P_{x}Y+f\left(X,Y,P_{y}\right) \ \ \ \ \ (23)$

where ${f}$ is some unknown function. We must include ${f}$ since the commutators of ${X}$ with ${X,Y}$ and ${P_{y}}$ are all zero, so adding on ${f}$ still satisfies 17. (You can think of it as similar to adding on the constant in an indefinite integral.)

Now from 18, we have

$\displaystyle \left[Y,L_{z}\right]=\left[Y,P_{y}\right]X \ \ \ \ \ (24)$

so combining this with 23 we have

$\displaystyle L_{z}=-P_{x}Y+P_{y}X+g\left(X,Y\right) \ \ \ \ \ (25)$

The undetermined function is now a function only of ${X}$ and ${Y}$, since the dependence of ${L_{z}}$ on ${P_{x}}$ and ${P_{y}}$ has been determined uniquely by the commutators 17 and 18.

From 19 we have

$\displaystyle \left[P_{x},L_{z}\right]=\left[P_{x},X\right]P_{y} \ \ \ \ \ (26)$

We can see that this is satisfied already by 25, except that we now know that the function ${g}$ cannot depend on ${X}$, since then ${\left[P_{x},g\right]\ne0}$. Thus we have narrowed down ${L_{z}}$ to

$\displaystyle L_{z}=-P_{x}Y+P_{y}X+h\left(Y\right) \ \ \ \ \ (27)$

Finally, from 20 we have

$\displaystyle \left[P_{y},L_{z}\right]=-\left[P_{y},Y\right]P_{x} \ \ \ \ \ (28)$

This is satisfied by 27 if we take ${h=0}$ (well, technically, we could take ${h}$ to be some constant, but we might as well take the constant to be zero), giving us the final form for ${L_{z}}$:

$\displaystyle L_{z}=-P_{x}Y+P_{y}X \ \ \ \ \ (29)$