Daily Archives: Tue, 18 April 2017

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)