Download Exercise [22.29] - Road to Reality forum

Exercise [22.29]
This problem asks us to compare the polarization ellipse obtained via the procedure described by fig.
22.13 to that obtained “directly” from the quantum state w   z  , and verify that they are
the same (up to a scale factor).
We first consider the ellipse obtained in fig. 22.13.
Suppose q is a point in
  . Then the 3-dimensional coordinates of its stereographic
projection onto the Riemann sphere q (where the x and y axes are along the real and imaginary
axes of the original complex plane, and the z axis points upwards through the north pole) are:
qx

 q 

 2 Re 
2
1  q 


qy

 q 

 2 Im 
2
1  q 


qz

1 q
2
1 q
2
(1)
Now consider the great circle perpendicular to this vector, projected back down to the equatorial
plane. This will be an ellipse, whose narrow axis is along the direction of q ; we can regard this as a
circle that has been squashed along this direction. Along its widest axis (perpendicular to q in the
complex plane), the ellipse contacts the Riemann sphere, so it has a radius of 1 in this direction.
Along q its radius is in fact just qz :
z axis
perpendicular
great circle
qz
equatorial plane
along direction of q
projected
ellipse
(cross-section)
qz
If q 
z
w
then the narrow axis of the ellipse points along q  12  z  w , and the “squashing
factor” qz along this direction is
1
z
w
1
z
w
wz
.
wz

Im
Re
1
When z  w , q  1 and so q is in the northern hemisphere; the direction of rotation is
anticlockwise. Conversely, when z  w
q is in the southern hemisphere, the ellipse is “flipped
over”, and the direction of rotation is clockwise. (When z  w the ellipse flattens into a line,
indicating linear polarization, and there is no “direction of rotation”).
Note that taking q  
z
w
instead of 
z
w
simply rotates q by 180°. This in turn rotates the
projected ellipse by 180° also… which has no effect, since 180° rotations are a symmetry of ellipses!
That’s why there’s no need to worry about the sign of q .
The ellipse above can be given explicitly in the complex plane by the formula:
z    ie 2 
i 1 z w 
cos   i

  sin  
w z
w z
(2)
Where the direction of rotation is given by increasing  . (The i in the front is an extra 90° rotation
because inside the […] we are squashing the vertical rather than the horizontal axis).
We now turn our attention to the polarization ellipse obtained “directly” from the quantum state
w   z  . We know that the polarization ellipse is the path traced out by the electric field
vector, as we move along the photon’s trajectory. We know that for pure 
and 
state
photons moving up the z-axis, this will be a circle in the x-y plane, rotating anticlockwise and
clockwise respectively. If we identify the x-y plane and the complex plane as in fig. 22.13, then we
can associate the electric field of the  and  states with complex numbers eit and e  i t
respectively, where  is the angular frequency (   2 f ). But note that if we take w  z  1, the
“ellipse” we obtain by our previously described method is a line along the (imaginary) y-axis, not
along the (real) x-axis. So, if (as we may reasonably assume) adding quantum states together
linearly adds their corresponding electric field vectors, then we should associate the  and 
states with complex numbers ieit and ie  it respectively, to match the results obtained from this
method. This is really just an adjustment of the phase (position along the wave at t  0 ), so this
choice is just as valid as our previous one.
It remains to determine how multiplying a quantum state  or  by a complex number affects
the corresponding electric field. We might (correctly) guess that the electric field strength is
increased by the magnitude of this multiplier, and the phase (angle) changes in proportion to its
argument. However the intuitive identification:
w
 iweit
z
 izeit
does not work: Taking w  z  ei would in that case rotate both electric vectors by  , thereby
rotating the entire polarization ellipse through this angle. But the orientation of the polarization
ellipse must depend only on the ratio z : w , which is unchanged by choice of  .
The real problem here is that we are only guessing at the relationship between electric field
direction and the actual complex-component representation of the states 
and  , which
Penrose has not even hinted at (so far in the book, anyway). A full description or derivation of this
representation is far beyond the scope of the current exercise, but interested readers might wish to
consult this online paper[1]. In any case, given the considerations above, we might instead propose
the (correct) identification:
w
 iweit
z
For which we can see that taking w  z  ei simply gives us
 izeit
(3)
w
 ieit  
z 
 ieit  
…and so clearly produces exactly the same ellipse as when w  z  1, as time rotates t through a
complete 2 revolution.
[Reading the linked paper, or at least its conclusions, we discover that the components of  and
 are in fact complex 3-vectors, the real and imaginary parts of which give the (perpendicular)
electric and magnetic field vectors respectively. Time dependence is due to a simple multiplying
factor of e  i t , rotating the “real” and “imaginary” vectors into one another; so multiplying by a
constant factor ei is equivalent (as above) to simply subtracting a constant time interval; it
therefore rotates  and  in opposite directions. Our initially incorrect guess was due to our
artificial identification of the x-y plane and the complex plane, which is a construction introduced for
the purpose of simplifying this particular problem, but which is not an intrinsic feature of the 
and  wavefunctions; they in fact have the same time dependence, but mirror-image geometry
(rather than the other way around).]
Having obtained the expressions for the individual electric field vectors of w  and z  (whether
by deductive guesswork or further reading), we may now add them to obtain the electric field of the
arbitrary linear superposition of states w   z  :
w
 z

iweit
 izeit
(4)
…and then we have:
iweit  izeit
 i  w eiweit  z eiz e  it 
 ie 2
we
 i 12  z w  it
 ie 2 

i t  12  z w  
i 1 z w 
i 1 z w 
we
e
i 1 z w   it

 i t  12  z w  

 z e 2
 ze
e
Substituting   t  12  w  z  gives:
iweit  izeit
 ie 2 
we
 ie 2 
 w  ei  ei  

i 1 z w
i 1 z w
i
 z e i 
 z  w e
 i


And further substituting ei  cos   i sin  gives:
iweit  ize it
i 1 z w 
 z  w   cos   i sin  
 ie 2 
 2 w cos  
 ie 2 
 w  z  cos   i  w  z  sin  
i 1 z w 
 i  w  z  e 2
i 1 z w 
cos   i

  sin  
w z
w z
And so finally we obtain:
w

z 

 w  z  . ie
i 12  z w 
…which we can see differs from (2) only by the scale factor
cos   i

 w  z .
  sin  
w z
w z
We have therefore obtained
the same ellipse (up to this scale factor) in both cases, as required.
∎
[1]
(5)
“The single Photon”, J. Dressel 2010, http://www.justindressel.com/jdressel/static/pdf/singlephoton.pdf