1.   Rotations Y Changes in frame of reference or point of view
involve transformations
of coordinate axes
(or, more generally, basis set) Y´  X´  X
= Z´ Z

2. Rotations Y Y´  x´  y´ y X´    x X
= Z´ Z

3. v´ = R v
cos sin 0
-sin cos 0
R =

4. r´ r r´ = r - a a Translations parallel translationY´ Y
Translations
parallel translation
(no rotation) of axes r´ r
r´ = r - a X´ a X Z´
Vectors (and functions) are translated in the
“opposite direction” as the coordinate system. Z
How can we possibly express an operator like this as a matrix?

5. The trick involves using
to cast matrix operators as exponentials
where H is an
operator…or matrix
the unit matrix
···

6. and we’ll make that connection through
Taylor Series (in 1-dimension)
…and this useful limit

7.  δax= δay= δaz= f (r) f (r) f f (x0+δx) = f (x0) + δx x ax N
For an infinitesimal translation
x=x0 3 
f (r)
i=1
Ok…but how can any matrix represent this? ax N
δax=
δay=
δaz=
Imagine dividing the entire translation a into ay N
and applying this little step N times N az N
f (r)

8. making this a continuous smooth translation
lim
N∞ i ħ
(-iħ )

9. For homework you will be asked to do the same thing for rotations
i.e.,
show you can cast in the same form.
cos sin 0
-sin cos 0
R=  -
You should start from:
R=
Later we will generalize this result to:

10. Rotation of coordinate axes by  about any arbitrary axis ^
Rotation of coordinate axes by  about any arbitrary axis 
Rotation of the physical system within fixed coordinate axes
Recall, even more fundamentally, the QM relation:
Time evolution of an initial state,
generated by the Hamiltonian

11. p a J  H t Amount of transformation Nature of the transformation
Operator
“Generator” p a
Translation:
moving linearly
through space
rotating
through space J  H t
translation
through time

12. The Silver Surfer, Marvel Comics Group, 1969