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Unformatted text preview: PHY 5667 Problem Set no. 1 solution Problem 1 ( a ) [ A,BC ] = ABC − BCA = ABC − BAC + BAC − BCA = ( AB − BA ) C + B ( AC − CA ) = [ A,B ] C + B [ A,C ] ( b ) [ A,BC ] = ABC − BCA = ABC + BAC − BAC − BCA = ( AB + BA ) C − B ( AC + CA ) = { A,B } C + B { A,C } Problem 2 This identity is straightforward to show by repeated use of ψ ( x i ) ψ † ( x j ) = [ ψ ( x i ) ,ψ † ( x j )] + ψ † ( x j ) ψ ( x i ) = δ ( x i − x j ) + ψ † ( x j ) ψ ( x i ) . ψ ( x ) ψ ( y ) ψ † ( x 1 ) ψ † ( x 2 )  ) = ψ ( x ) parenleftBig δ ( y − x 1 ) + ψ † ( x 1 ) ψ ( y ) parenrightBig ψ † ( x 2 )  ) = parenleftBig ψ ( x ) δ ( y − x 1 ) ψ † ( x 2 ) + ψ ( x ) ψ † ( x 1 ) δ ( y − x 2) + ψ ( x ) ψ † ( x 1 ) ψ † ( x 2 ) ψ ( y ) parenrightBig  ) = parenleftBig δ ( x − x 2 ) δ ( y − x 1 ) + δ ( y − x 1 ) ψ † ( x 2 ) ψ ( x ) + δ ( x − x 1 ) δ ( y − x 2 ) + δ ( y − x 2 ) ψ † ( x 1 ) ψ ( x ) + ψ ( x ) ψ † ( x 1 ) ψ † ( x 2 ) ψ ( y ) parenrightBig  ) = ( δ ( x − x 2 ) δ ( y − x 1 ) + δ ( x − x 1 ) δ ( y − x 2 ))  ) From the third line, the second, fourth and fifth term have vanished as consequence of ψ ( x i )  ) = 0. 1 Problem 3 (a) The two particlestate is defined as:  φ ) = integraldisplay d x 1 d x 2 φ ( t,x 1 ,x 2 ) ψ † ( x 1 ) ψ † ( x 2 )  ) We can write the function φ as sum of its symmetric and antisymmetric...
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 Fall '10
 Okui
 symmetric functions, Brown dwarf, Symmetric polynomial, symmetric function

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