1 Quantum Mechanics Lecture 16B Preliminaries: Spin 1 operators 1 Recall in the discussion of the Stern-Gerlach measurement of spin that for a spin 2 particle only two outcomes ~ ~ result (as opposed to the classical result of a smooth distribution from + 2 2 ). It was also mentioned that higher spin objects exist and the outcomes are always 2S + 1 in number. For→ the − following discussion we want to examine spin 1 particles. They have outcomes of +~, 0, ~. In what follows we will ignore the factor of ~ and just call the eigenvalues +1, 0, 1. Just as before we can make− spin measurements in three perpendicular directions. − 1 The observables associated with these three bases will be complementary, just as was the case for spin 2 . We will take the z basis as our first and define the eigenvectors in terms of column matrices as follows,
1 0 0 1, +1 z = 0 1, 0 z = 1 1, 1 z = 0 | i | i | − i 0 0 1
With these eigenvectors and setting the eigenvalues (the second term in the ket vectors) we can easily construct the spin 1 operator in the z basis. It should be obvious it takes the matrix form,
10 0 Sˆz = 00 0 0 0 1 − We could spend some time constructing the two other complementary bases but in what follows we will only be concerned with the operators. They are,
0 1 0 0 i 0 1 1 − Sˆx = 1 0 1 Sˆy = i 0 i √2 √2 − 0 1 0 0 i 0 1 ˆ ˆ ˆ These satisfy an operator algebra much the same as for spin 2 , [Sx, Sy] = iSz. The square of each of these operators is found to be,
0 1 0 0 1 0 1 0 1 1 1 Sˆ2 = 1 0 1 1 0 1 = 0 2 0 x 2 2 0 1 0 0 1 0 1 0 1 0 i 0 0 i 0 1 0 1 2 1 − − 1 − Sˆ = i 0 i i 0 i = 0 2 0 y 2 − − 2 0 i 0 0 i 0 10 1 − 10 0 10 0 1 0 0 ˆ2 Sz = 00 0 00 0 = 0 0 0 0 0 1 0 0 1 0 0 1 − − ˆ2 ˆ2 ˆ2 ˆ2 ˆ Notice that if we define the squared spin operator, S = Sx + Sy + Sz the result is 21 (2 times the identity matrix). In addition the squared spin in each direction is an observable that can be measured as well. Are the squared spin observables complementary? We can check by evaluating the following commutators (and using [Sˆx, Sˆy]= iSˆz),
1 0 1 1 0 1 1 0 1 1 0 1 1 1 [Sˆ2, Sˆ2] = 0 2 0 0 2− 0 0 2− 0 0 2 0 x y 4 − 4 1 0 1 10 1 10 1 1 0 1 − − 0 0 0 0 0 0 1 1 = 0 4 0 0 4 0 =0 4 − 4 0 0 0 0 0 0
You can quickly check that all commutator involving the squared spins vanish. Thus, you can measure the square of the spin in any combination of the three directions simultaneously.
1 The Kochen-Specker Theorem The spin of a particle with spin 1 can be measured in one of three perpendicular directions, defined by the operators Sx,Sy,Sz. The outcomes of any one measurement is only one of the three possible values: -1, 0, or +1. We know from our discussion above that we should not expect these operators to commute (they won’t) but the squares of these operators do commute. The result of any measurement of the spin squared is 1 or 0. Lastly, as can be determined by explicitly calculate from the matrix forms of these operators, or can be taken from the theory of quantum mechanical spin is that, ˆ2 ˆ2 ˆ2 ˆ2 I S = Sx + Sy + Sz =2 . (1)
Since this is proportional to the identity, Sˆ2 commutes with all of these spin operators (it is a Casimir invariant) and again, the three squared operators commute amongst themselves. ˆ2 ˆ2 ˆ2 Since the set of operators Sx, Sy , Sz commute, they are compatible observables - they can be measured simultaneously. We see that since the only two values these operators can take are 0 and 1, and that their sum is 2, it must be that in any simultaneous measurement of these observables, two are 1 and one is zero. This must be in order to satisfy (1) above. Lastly, since the directions x, y, and z were chosen arbitrarily (the system is spherically symmetric) any other choice of three perpendicular axes must also satisfy these constraints. The question is: can we assign values to every triad (set of perpendicular operators) whose directions cross the unit sphere? That is, can we find a region that is one-third of the area of a unit sphere where only the 0 outcome lies and the rest of the sphere belonging to results of 1? First, why the restriction to 1/3 of the area? Consider if the area defining 0 outcomes was less than 1/3 of 4π the surface area (i.e. 3 ). There would then be triads that all had 1 outcomes -in violation of our criterion. 4π Obviously, if the area were larger than 3 , then there would be outcomes with two 0s, also violating our criterion. So the problem simplifies to whether we can draw regions adding up to one third of the area of the unit sphere such that only 0 result axes cross. Since the axes are constrained to be orthogonal, this is going to require a little thought. Let’s define the area resulting in 0 outcomes to be A and those for 1 outcomes to be B. First some (hopefully) obvious conditions.
1) If any point is defined as a 0 outcome (Ai) then the opposite point on the sphere must also be defined as a 0 outcome (and likewise for 1 outcomes). The reason for this will be clear after the next condition. 2) For any point on the sphere defined as a 0 outcome, the equatorial radius perpendicular to this point must be a B region.
Let’s draw this and see why it must be true.
Measured outcome 0 at this point ×
× Measured outcome 0 at this point This ring must be outcome 1
Since the two 1 axes are perpendicular to the 0 axis, they cross the unit sphere along the ring that is at an angle of 90o from the 0 axis. Since we demand that these assignments be unique we see that the red ring can not contain any 0 outcomes. Now, let’s make an argument to see what the maximal area of A can be. From condition 2 above, we see that we can not consider a single contiguous area. Thus, consider 2 contiguous areas that must be symmetric about the equator. Consider the regions of A to be two spherical caps of angle up to but not including 45o from the z axis (see figure).
2 zˆ zˆ 0 0 Measured outcome 0
45o 45o 45o 45o 1 1
0 Measured outcome 1 0
Notice in the right hand figure that if the 0 axis (blue unit vector) were to be rotated by an angle more than 45o from the z axis then one of the other 1 axes (red unit vectors) would necessarily lie in the upper A region. Thus the spherical caps can not extend 45o and beyond. 4π 2 Can we reach the requirement of 3 for the area of A? The formula for a spherical cap is Acap =2πR (1 cos θ) and since this is the unit sphere we have, −
o 1 A2caps = 2 2π(1 cos45 )=4π 0.293 < 4π × − × 3 Thus we see with this method you can not uniquely define every point on the sphere as being 0 and 1 following our conditions. 4π To explicitly show that the area of A must be 3 and that the distribution above allows results that are all 1, consider the following traid. zˆ 0
1
0
It is possible to arrange the three perpendicular unit vectors to lie just on a circular cone such that all three vectors lie in region B. In the figure above, the blue circle is the intersection of a circular cone, just like the on defined for the upper cap, with surface of the unit sphere closest towards us. This arrangement would yield a simultaneous measurement of all 1s, violating the quantum mechanical prediction (and experimentally observed result). There are full, formal proofs of this result and thus far we have only given a hand waving argument. To embolden this argument you might try to discretize the caps - have many infinitesimal area elements that together form the two caps above. One might then try adding more infinitesimal area elements in the B region. Recall however for every point we add that is 0, we add a ring perpendicular to it that must be 1. Thus adding a small area element dA = dx2 outside of the caps will create a band within both caps of macroscopic length and width approximately dx, that must be region B. Since we would be subtracting much more area from the caps than is added outside of the caps, we see that we can never increase the area of A in this manner. This should be sufficient to convince you of the result that no distribution of 0s and 1s on the unit sphere can satisfy our requirement that one axis is 0 and the other two 1. This is the result of the Kochen-Specker theorem (though they proved it in a much more formal way). ˆ2 ˆ2 ˆ2 There is no possible way to ascribe values to every possible measurement of Sx, Sy , Sz for all orientations of the three axes while conforming to the rules of quantum mechanics.
3 Another proof of the KS theorem
(0,0,0,1) (0,0,0,1) (1,-1,1,-1) (1,-1,1,-1) (0,0,1,0) (1,-1,-1,1) (1,1,-1,1) (1,1,-1,1) (1,1,1,-1) (0,0,1,0) (0,1,0,0) (1,-1,-1,1) (1,1,1,1) (0,1,0,0) (1,1,1,1) (1,1,1,-1) (-1,1,1,1) (-1,1,1,1) (1,1,0,0) (1,0,1,0) (1,1,0,0) (1,0,-1,0) (1,0,0,1) (1,0,0,-1) (1,-1,0,0) (1,0,1,0) (1,0,0,1) (1,-1,0,0) (1,0,-1,0) (0,0,1,1) (0,1,0,-1) (1,0,0,-1) (0,1,-1,0) (0,0,1,1) (0,1,0,-1) (0,1,-1,0)
4