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Physics 745 - Group Theory Solution Set 22 1. [10] It is rare we will actually use the representation matrices Γ( j ) ( R ) , but occasionally it is useful. We want to work out explicitly Γ( 2 ) ( R ( x ) ) , generated 1 by the generators Ta = 12 σ a , ⎛0 1⎞ ⎛ 0 −i ⎞ ⎛1 0 ⎞ where σ 1 = ⎜ ⎟, σ2 = ⎜ ⎟, σ3 = ⎜ ⎟. ⎝1 0⎠ ⎝i 0 ⎠ ⎝ 0 −1 ⎠ We will write x in the form x = xrˆ , where r̂ is a unit vector, and x is the magnitude of x. 2 (a) [3] Show that ( rˆ ⋅ σ ) = 1 , where 1 is the unit matrix, for any unit vector r̂ . Furthermore, find a simplification for ( rˆ ⋅ σ ) for arbitrary positive integer n n when n is even or odd. We simply write out the expression explicitly, writing rˆ = ( r1 , r2 , r3 ) , so that ⎛ r 2 ( rˆ ⋅ σ ) = ⎜ 3 ⎝ r1 + ir2 r1 − ir2 ⎞⎛ r3 ⎟⎜ − r3 ⎠⎝ r1 + ir2 r1 − ir2 ⎞ ⎛ r32 + r12 + r22 ⎟=⎜ − r3 ⎠ ⎝ 0 ⎞ = 1 ( rˆ 2 ) = 1 2 2 2⎟ r3 + r1 + r2 ⎠ 0 Now, if you have it raised to an even power, you will simply get ( rˆ ⋅ σ ) = 1n = 1 , while 2n for an odd power, you will have ( rˆ ⋅ σ ) 2 n +1 = 1n ( rˆ ⋅ σ ) = rˆ ⋅ σ . That’s pretty simple. (b) [4] Expand out the representation Γ ( 12 ) ∞ ( R ( x ) ) = exp ( −iT ⋅ x ) = ∑ n=0 ( −i T ⋅ x ) n n! dividing it into even and odd terms. Simplify as much as possible 1 (c) [3] Show that Γ( 2 ) ( R ( x ) ) = 1 cos ( 12 x ) − i ( rˆ ⋅ σ ) sin ( 12 x ) We simply follow the instructions, and then divide it into even and odd terms: Γ ( 12 ) ∞ ( R ( x)) = ∑ ( −iT ⋅ x ) n ∞ 1 n ( −i 12 rˆ ⋅ σx ) n! n =0 n =0 n ! ∞ ∞ 1 1 n n n n = ∑ ( −i 12 x ) ( rˆ ⋅ σ ) + ∑ ( −i 12 x ) ( rˆ ⋅ σ ) n even n ! n odd n ! =∑ 2 4 3 5 = 1 ⎡1 − 2!1 ( 12 x ) + 4!1 ( 12 x ) − + "⎤ + ( rˆ ⋅ σ ) ⎡ −i ( 12 x ) + i 3!1 ( 12 x ) − i 5!1 ( 12 x ) + −"⎤ ⎣ ⎦ ⎣ ⎦ 1 1 = 1 cos ( 2 x ) − i ( rˆ ⋅ σ ) sin ( 2 x )