solution7 - 24 Electron-positron pairs are emitted in a...

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24. Electron-positron pairs are emitted in a decay process (moving in opposite directions) in a spin state | 0 , 0 i spin = | s, m i of zero total spin angular momentum ~ S = ~ S 1 + ~ S 2 . (We focus here only on the spin state of the pair). (a) What is the mean value that will be obtained in a measurement performed on an ensemble of such electron-positron pairs of the operator S 1 z S 2 z ? Is the value the same as that which one would obtain by independently measuring the mean value of S 1 z and S 2 z and forming the product of the mean values of each? Explain. From an earlier problem we know that this spin state can be written in terms of the direct product states in the form | 0 , 0 i = 1 2 [ | , i | , i ]= 1 2 | 1 2 , 1 2 i | 1 2 , 1 2 i ¸ . We thus can calculate h S 1 z S 2 z i = 1 2 h 1 2 , 1 2 | h 1 2 , 1 2 | ¸ S 1 z S 2 z | 1 2 , 1 2 i | 1 2 , 1 2 i ¸ = μ 1 4 1 2 h 1 2 , 1 2 | h 1 2 , 1 2 | ¸ | 1 2 , 1 2 i | 1 2 , 1 2 i ¸ = 1 4 . On the other hand h S 1 z i = 1 2 h 1 2 , 1 2 | h 1 2 , 1 2 | ¸ S 1 z | 1 2 , 1 2 i | 1 2 , 1 2 i ¸ = μ 1 2 1 2 h 1 2 , 1 2 | h 1 2 , 1 2 | ¸ | 1 2 , 1 2 i + | 1 2 , 1 2 i ¸ =0 and h S 2 z i = 1 2 h 1 2 , 1 2 | h 1 2 , 1 2 | ¸ S 2 z | 1 2 , 1 2 i | 1 2 , 1 2 i ¸ = μ 1 2 1 2 h 1 2 , 1 2 | h 1 2 , 1 2 | ¸ | 1 2 , 1 2 i + | 1 2 , 1 2 i ¸ so that h S 1 z ih S 2 z i , which shows that that average of the product is not generally the same as the product of the averages, which is a general statistical result. (b) What is the mean value that will be obtained in a measurement performed on an ensemble of such electron-positron pairs of the operator S 1 x S 2 x ? Using the fact that S 1 x = 1 2 [ S 1 + S 1+ ] S 2 x = 1 2 [ S 2 + S 2+ ]
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we f nd that h S 1 x S 2 x i = 1 2 h 1 2 , 1 2 | h 1 2 , 1 2 | ¸ S 1 x S 2 x | 1 2 , 1 2 i | 1 2 , 1 2 i ¸ = 1 8 h 1 2 , 1 2 | h 1 2 , 1 2 | ¸ [ S 1 + S 1+ ][ S 2 + S 2+ ] | 1 2 , 1 2 i | 1 2 , 1 2 i ¸ = 1 8 h 1 2 , 1 2 | h 1 2 , 1 2 | ¸ [ S 1 + S 1+ ] | 1 2 , 1 2 i | 1 2 , 1 2 i ¸ = 1 8 h 1 2 , 1 2 | h 1 2 , 1 2 | ¸ | 1 2 , 1 2 i | 1 2 , 1 2 i ¸ = 1 8 [0 1 1+0]= 1 4 which is the same as for h S 1 z S 2 z i . This actually follows from symmetry. Since the combined state has zero total angular momentum it is rotationally invariant. So | 0 , 0 i = U R | 0 , 0 i ,
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solution7 - 24 Electron-positron pairs are emitted in a...

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