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TFY4250/FY2045 Tillegg <strong>13</strong> - <strong>Addition</strong> <strong>of</strong> <strong>angular</strong> <strong>momenta</strong> 5<br />
Conclusion and comments<br />
(i) By measuring both S 1z and S 2z we can prepare the two-spin system in one <strong>of</strong> the<br />
“old” states, |↑ 1 ↑ 2 〉, |↑ 1 ↓ 2 〉, |↓ 1 ↑ 2 〉, |↓ 1 ↓ 2 〉. In these states S 1z and S 2z have sharp values,<br />
and so does S z .<br />
(ii) For these “old” states S 2 is not sharp, except for |↑ 1 ↑ 2 〉 and |↓ 1 ↓ 2 〉, which happen<br />
to be the upper and lower rungs in the triplet ladder. The reason that S 2 is not sharp for<br />
the other “old” states (|↑ 1 ↓ 2 〉 and |↓ 1 ↑ 2 〉) is that Ŝ2 does not commute with Ŝ1z and Ŝ2z.<br />
(iii) If we choose instead to measure S z and |S|, this spin system is prepared in one <strong>of</strong><br />
the “new” states, either in one <strong>of</strong> the triplet states<br />
|1, 1〉 = |↑ 1 ↑ 2 〉,<br />
|1, 0〉 = 1 √<br />
2<br />
( |↑ 1 ↓ 2 〉 + |↓ 1 ↑ 2 〉 ) ,<br />
|1, −1〉 = |↓ 1 ↓ 2 〉,<br />
(triplet)<br />
(T<strong>13</strong>.<strong>13</strong>)<br />
or in the singlet,<br />
|0, 0〉 = 1 √<br />
2<br />
( |↑ 1 ↓ 2 〉 − |↓ 1 ↑ 2 〉 ) . (singlet) (T<strong>13</strong>.14)<br />
Here S 1z and S 2z are unsharp, except in the two states |↑ 1 ↑ 2 〉 and |↓ 1 ↓ 2 〉. The figure below<br />
gives an illustration <strong>of</strong> both the old and the new states.<br />
Remember: The “old” states on the left are prepared by measuring S 1z and S 2z separately.<br />
The “new” states on the right are prepared by measuring the size |S| and the z-component<br />
S z <strong>of</strong> the total spin.<br />
(iv) The singlet corresponds to S = 0, which means that the √ two spins S 1 and S 2 are<br />
antiparallel. The triplet corresponds to an angle α = 2 arccos 2/3 ≈ 70.5 ◦ between the<br />
two vectors S 1 and S 2 . This is the closest the two spins come to being parallel.