Question Regarding Commutator of two incompatible Hermitian Operators

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I have two questions that are based on the following example involving the Hermitian operator i[A,B]=iAB-iBA for the case of a plane polarized photon.


The observable (Hermitian Matrix) for the plane polarized photon, which Professor Susskind gave in his quantum mechanics lecture, lecture number 6 available on the web, is given by the Hermitian matrix


A(theta) = [cos(2 theta) , sin(2 theta) ; sin(2 theta) , -cos(2 theta) ]. The
polarization angle, theta, corresponds to a plane polarizer oriented with
an angle theta made with the x axis in, say, the x, y plane.

Note: in the matrix notation above, the row elements are separated by commas,
and the columns are separated by a semicolon. So the above matrix is a
two-by-two matrix.


Professor Susskind gave as an example, also in lecture 6 of his quantum mechanics video lectures, that one way to make such a polarizer is to have a circular ring with thin highly conductive wires strung close together to form parallel chords across the 2-d
region enclosed by the ring.


Let A(theta), and A(gamma) be two observables associated with two different
plane polarization angles, theta and gamma; respectively.

Then the matrices commute iff theta-gamma = n*pi/2 for integers n=0,
+-1,+-2,...etc, as can be seen by direct calculation.


Let zeta = theta-gamma.

Then a direct calculation shows that


G(theta, gamma) = i[A(theta),A(gamma)]
= [0 , -2i sin(2 zeta) ; 2i sin(2 zeta), 0],

which has eigenvalues 2 sin(2 zeta) and -2 sin(2 zeta) with corresponding


eigenstate vectors

| 2 sin(2 zeta) > = [ 1/sqrt(2) ; i/sqrt(2) ]

and

| -2 sin(2 zeta) > = [ -i/sqrt(2) ; 1/sqrt(2) ]; respectively,

where zeta = theta - gamma is the difference in the polarization angles theta
and gamma.


The eigenvectors are constants, and only the eigenvalues
2 sin(2 zeta), and -2 sin(2 zeta) vary as the angle between the two polarizers vary (for this particular example).


It turns out, in this case, the incompatibility of the two observables A(theta) and
A(gamma) are in some sense dual to the output values of the observable
G(theta, gamma) in the following sense: when ever A(theta) and A(gamma) are least compatible, i.e. when the polarization angle is pi/4 radians, (or 45 degrees), then the eigenvalues--which are the two possible measurement output values of a measurement of the observable G(theta, gamma)--are largest in magnitude.

However, if A(theta) and A(gamma) are compatible, that is if they commute,
which occurs only when zeta=theta - gamma = an integer multiple pi/2 radians, then the eigenvalues of the observable G(theta, gamma) are both equal to zero.

Now here is the first of two questions I have about this example:

Is this saying that the value of any measurement of the observable G(theta, gamma) is equal to 0 if the experiment is set up such that zeta is an integer multiple of pi/2 radians?

If the answer is yes, then this leads me to ask the second question: suppose that zeta is close to say, pi/2, then the eigenvalues of G(theta, gamma) are
2 sin(2 zeta), and -2 sin(2 zeta), and are close to zero, and G still has the same two mutually orthogonal eigenstate vectors


| 2 sin(2 zeta) > = [ 1/sqrt(2) ; i/sqrt(2) ]

and

| -2 sin(2 zeta) > = [ -i/sqrt(2) ; 1/sqrt(2) ].

So, 2 sin(2 zeta) is approximately equal to -2 sin(2 zeta), and so the two eigenvalues, which are the two possible outputs of the measurement corresponding to G, are very close to being equal. That is to say, can the two possible measurement output values be made to be necessarily virtually indistinguishable by setting up the experiment with zeta being made close enough to being 90 degrees (pi/2 radians)?

If so, this seems like a different kind of uncertainty: it is an uncertainty where a measurement is constrained to yield one of two possible nearly-equal output values since the eigenvalues 2 sin(2 zeta), and -2 sin(2 zeta), are nearly equal.


Is this accurate? I am just learning the subject, and may have made a mistake above. Any thoughts, suggested readings, or insights are most appreciated! Thanks for your help!
 
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