HW1_soln

# HW1_soln - PHY 5667 Problem Set no 1 solution Problem 1 a...

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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 particle-state 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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