hw202a - Physics 512 Homework Set#2 Solutions Winter 2003 1...

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Physics 512 Winter 2003 Homework Set #2 – Solutions 1. We may add three angular momenta, ~ J 1 , ~ J 2 and ~ J 3 , by frst adding the frst two, ~ J 12 = ~ J 1 + ~ J 2 , and then by adding this result to the last one, ~ J 123 = ~ J 12 + ~ J 3 . Using this (or any other suitable method), add three spin-1 / 2 states and write down the resulting coupled | jm i states in terms oF the uncoupled basis. Give a complete set oF commuting operators that are diagonal in the coupled basis. [Note that | 3 2 3 2 i = |↑↑↑i should be an obvious state.] We start by adding the frst two angular momenta as suggested: | 11 i = |↑↑i | 10 i = 1 2 ( |↑↓i + |↓↑i ) | 00 i = 1 2 ( |↑↓i − |↓↑i ) | 1 1 i = |↓↓i (where it should be understood that these apply to the frst two spins). Now, when we add the third spin to this, we have to make a choice. Either we can add it to the triplet or the singlet. Adding it to the singlet is trivial, and gives | 1 2 1 2 i = | 00 i⊗|↑ i = 1 2 ( |↑↓↑i − |↓↑↑i ) | 1 2 1 2 i = | 00 i⊗|↓ i = 1 2 ( |↑↓↓i − |↓↑↓i ) (1) We now turn to the triplet. When adding the third spin to the triplet, we can end up with either j = 1 2 or j = 3 2 . Looking up the Clebsch-Gordan coefficients, we fnd | 3 2 3 2 i = | 11 i = | 3 2 1 2 i = 1 3 ( | 11 i + 2 | 10 i ) = 1 3 ( |↑↑↓i + |↑↓↑i + |↓↑↑i ) | 3 2 1 2 i = 1 3 ( 2 | 10 i + | 1 1 i ) = 1 3 ( |↑↓↓i + |↓↑↓i + |↓↓↑i ) | 3 2 3 2 i = | 1 1 i = |↓↓↓i (2) For resulting angular momentum 3 2 , and | 1 2 1 2 i 0 = 1 3 ( 2 | 11 i−| 10 i ) = 1 6 ( 2 |↑↑↓i − |↑↓↑i − |↓↑↑i ) | 1 2 1 2 i 0 = 1 3 ( | 10 i− 2 | 1 1 i ) = 1 6 ( |↑↓↓i − |↓↑↓i − 2 ) (3) For resulting angular momentum 1 2 . Note that the prime is to indicate that this set oF j = 1 2 states is di±erent From the frst one Found above. This type oF degeneracy is quite common whenever we add three or more angular momenta together. One way to express this is that when we combine the frst two spins, we have 1 2 1 2 = 0 1 1
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Next we combine either 0 1 2 = 1 2 or 1 1 2 = 1 2 3 2 . So the fnal result is 1 2 1 2 1 2 = 1 2 1 2 3 2 This decomposition also shows that you start with eight states (oF spins up or down in the uncoupled basis), and end up with eight states as well (in the coupled basis), which is a good thing. While the two 1 2 states are linearly independent, the explicit answers (1) and (3) are not necessarily unique. One could take appropriately normalized linear combinations oF them (with the same m values, oF course). This is just a symptom oF a degeneracy in the states. OF course, the 3 2 states, (2) , are unique. This degeneracy between the two 1 2 states indicates that by simplying saying “total spin- 1 / 2 ” we have not Fully accounted For a complete set oF quantum numbers in the coupled basis. OF course, looking back at how we arrived at the coupled basis, we see that the frst spin- 1 / 2 state that we derived, (1) , came From the singlet oF ~ J 12 , while the second spin- 1 / 2 state, (3) , came From the triplet oF ~ J 12 . This suggests that we may look at the eigenvalue oF J 2 12 to distinguish among the coupled states. To be more
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hw202a - Physics 512 Homework Set#2 Solutions Winter 2003 1...

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