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Unformatted text preview: Solutions to Problems in Peskin and Schroeder, An Introduction To Quantum Field Theory Homer Reid March 13, 2003 Chapter 3 Problem 3.1 Part a. The commutation relations are [ J μν , J ρσ ] = i parenleftBig g νρ J μ σ g μρ J νσ g νσ J μρ + g μσ J νρ parenrightBig . Then we have [ K i , K j ] = [ J i , J j ] = i parenleftBig g i J j g 00 J ij g ij J 00 + g j J i parenrightBig We can replace g 00 = 1 , J 00 = 0 , and g i = g j = 0 since we know i and j are space indices. Then [ K i , K j ] = iJ ij = epsilon1 ijk L k . Next, [ L i , K n ] = 1 2 epsilon1 ijk [ J jk , J n ] This is really just a fancy way of using the properties of the epsilon1 tensor to write = 1 2 braceleftBig [ J pq , J n ] [ J qp , J n ] bracerightBig 1 Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 3 2 where p = 1 + (( i + 1) mod 3) and q = 1 + (( i + 2) mod 3) . In the second commutator we can switch q and p and simultaneously flip the sign since J is antisymmetric to obtain = [ J pq , J n ] = i parenleftBig g q J pn g p J qn g qn J p + g pn J q parenrightBig = i parenleftBig g pn J q g qn J p parenrightBig . But only one of g pn and g qn can be nonzero, and whichever one is nonzero has the value 1 since p and q are space indices. So we get [ L i , K n ] = braceleftBigg iJ i +2 , , ( n = i + 1) + iJ i +1 , , ( n = i + 2) (where all addition is to be carried out modulo 3). = iepsilon1 ink J k = iepsilon1 ink J k = iepsilon1 ink K k . Finally, the last commutator is calculated through a calculation exactly anal ogous to the one we just went through, and actually the answer is given to us in the problem, namely, [ L i , L j ] = iepsilon1 ijk L k . Next, we have [ J + , J ] = 1 4 bracketleftBig ( L + i K ) , ( L i K ) bracketrightBig = 1 4 braceleftBig [ L , L ] + i [ K , L ] i [ L , K ] + [ K , K ] bracerightBig = 2 i [ K , L ] = 2 i braceleftBig [ K 1 , L 1 ] ˆ i + [ K 2 , L 2 ] ˆ j + [ K 3 , L 3 ] ˆ k bracerightBig = 0 since [ K i , L j ] negationslash = 0 only for i negationslash = j as we saw above. Next, bracketleftBig J i + , J j + bracketrightBig = 1 4 bracketleftBig ( L i + iK i ) , ( L j + iK j ) bracketrightBig = 1 4 braceleftbig [ L i , L j ] + i [ K i , L j ] + i [ L i , K j ] [ K i , K j ] bracerightbig = 1 4 braceleftbig iepsilon1 ijk L k + iepsilon1 ijk + iepsilon1 ijk K k + iepsilon1 ijk L k bracerightbig = 1 2 epsilon1 ijk J j + Homer Reid’s Solutions to Peskin and Schroeder Problems: Chapter 3 3 so the J + operators satisfy the angular momentum commutation relations among themselves. The demonstration that the same is true for the J operators is just too similar to what we just did to warrant repeating. Part b. For the ( 1 2 , 0) representation of the Lorentz group we want to take J + = 1 2 ( L + i K ) = 1 2 vectorσ, J = 1 2 ( L i K ) = 0 ....
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This note was uploaded on 02/07/2011 for the course PHYS 1002 taught by Professor Imanpour during the Spring '11 term at Alabama.
 Spring '11
 Imanpour
 Quantum Field Theory

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