Quantum Mechanics for Engineers 

© Leon van Dommelen 

D.68 Awkward questions about spin
Now of course you ask: how do you know how the mathematical
expressions for spin states change when the coordinate system is rotated around some
axis? Darn.
If you did a basic course in linear algebra, they will have told you
how the components of normal vectors change when the coordinate system
is rotated, but not spin vectors, or spinors, which are
twodimensional vectors in threedimensional space.
You need to go back to the fundamental meaning of angular momentum.
The effect of rotations of the coordinate system around the
axis was discussed in addendum {A.19}.
The expressions given there can be straightforwardly generalized to
rotations around a line in the direction of an arbitrary unit vector
. Rotation by an angle multiplies the
direction angular momentum eigenstates by
if is the angular momentum in the
direction. For electron spin, the values for are
, so, using the Euler formula (2.5) for
the exponential, the eigenstates change by a factor
For arbitrary combinations of the eigenstates, the first of the two
terms above still represents multiplication by the number
.
The second term may be compared to the effect of the
direction angular momentum operator , which
multiplies the angular momentum eigenstates by
; it is seen to be
. So the
operator that describes rotation of the coordinate system over an
angle around the axis is

(D.44) 
Further, in terms of the , , and angular
momentum operators, the angular momentum in the direction is
If you put it in terms of the Pauli spin matrices, drops out:
Using this operator, you can find out how the spinup and spindown
states are described in terms of correspondingly defined basis states
along the  or axis, and then deduce these
correspondingly defined basis states in terms of the ones.
Note however that the very idea of defining the positive and
angular momentum states from the ones by rotating the coordinate
system over 90 is somewhat specious. If you rotate the
coordinate system over 450 instead, you get a different
answer! Off by a factor 1, to be precise. But that is as bad as
the indeterminacy gets; whatever way you rotate the axis system to the new
position, the basis vectors you get will either be the same or only a
factor 1 different {D.69}.
More awkwardly, the negative momentum states obtained by rotation do
not lead to real positive numerical factors for the corresponding
ladder operators. Presumably, this reflects the fact that at the wave
function level, nature does not have the rotational symmetry that it
has for observable quantities. Anyway, if nature does not bother to
obey such symmetry, then there seems no point in pretending it does.
Especially since the nonpositive ladder factors would mess up various
formulae. The negative spin states found by rotation go out of the
window. Bye, bye.