The three matrix operators for spin one satisfy sz Sy – Sy 8z = isz andcyclic permutations. Show thats = sz, (sz tisy )³ = 0.For the sane in, m, can have the values im, m – 1, .., -m, whileA12 has eigenvalue m(m + 1). ThusM? = m(m + 1)±25 x 1 = 5 timeseach once6154 x 8 = 32 times+3/2±1/23/2each 8 times4±13 x 27 = 81 times1each 27 times32 x 18 = 96 times1/2±1/2 each 48 times1 × 12 = 42 times0, each 42 timesTotal 256 eigenvaluesA certain state | 4) is an eigenstate of L? and L,:L'|v) = 1(l+ 1) h² |»),mh|v) .For this state calculate (La) and (L²).

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The three matrix operators for spin one satisfy sz Sy – Sy 8z = is, and cyclic permutations. Show that s = sz, (sz tisy)3 = 0. For the same in, m, can have the valucs im, m – 1, .., -m, while A12 has eigenvalue m(m + 1). Thus M2 = m(m + 1) +2 土1 each once 6 5 x1 = 5 times 15 +3/2 ±1/2 3/2 each 8 times 4 x 8 = 32 times 4 1 each 27 times 3 x 27 = 81 times 3 1/2 ±1/2 each 48 times 2 x 18 = 96 times 0, each 42 timcs 1x 12 = 42 times Total 256 eigenvalues A certain state | ) is an eigenstate of L2 and L,: L'|v) = 1(l+ 1) h² |), mh| v). For this state calculate (La) and (L2).

The three matrix operators for spin one satisfy sz Sy – Sy 8z = is, and cyclic permutations. Show that s = sz, (sz tisy)3 = 0. For the same in, m, can have the valucs im, m – 1, .., -m, while A12 has eigenvalue m(m + 1). Thus M2 = m(m + 1) +2 土1 each once 6 5 x1 = 5 times 15 +3/2 ±1/2 3/2 each 8 times 4 x 8 = 32 times 4 1 each 27 times 3 x 27 = 81 times 3 1/2 ±1/2 each 48 times 2 x 18 = 96 times 0, each 42 timcs 1x 12 = 42 times Total 256 eigenvalues A certain state | ) is an eigenstate of L2 and L,: L'|v) = 1(l+ 1) h² |), mh| v). For this state calculate (La) and (L2).