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852HW2_Solutions - PHYS852 Quantum Mechanics II Spring 2009...

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Unformatted text preview: PHYS852 Quantum Mechanics II, Spring 2009 HOMEWORK ASSIGNMENT 2: Solutions 1. Consider the most general normalized spin-1/2 state: | ψ ) = c + | + ) + c- |−) , where S z |±) = ± planckover2pi1 2 |±) . a.) Compute ( S x ) , ( S y ) and ( S z ) . b.) Compute the variances Δ S x , Δ S y , and Δ S z . c.) Prove that Δ S x = planckover2pi1 2 | c 2 + − c 2- | , Δ S y = planckover2pi1 2 | c 2 + + c 2- | and Δ S z = planckover2pi1 | c + || c- | . Hint: Use the fact that ( | c + | 2 + | c- | 2 ) 2 = 1 2 = 1. Answer: a.) ( S x ) = planckover2pi1 2 ( c * + , c *- ) parenleftbigg 0 1 1 0 parenrightbiggparenleftbigg c + c- parenrightbigg = planckover2pi1 2 ( c * + , c *- ) parenleftbigg c- c + parenrightbigg = planckover2pi1 2 ( c * + c- + c *- c + ) ( S y ) = planckover2pi1 2 ( c * + , c *- ) parenleftbigg − i i parenrightbiggparenleftbigg c + c- parenrightbigg = planckover2pi1 2 ( c * + , c *- ) parenleftbigg − ic- ic + parenrightbigg = planckover2pi1 2 i ( c * + c- − c *- c + ) ( S z ) = planckover2pi1 2 ( c * + , c *- ) parenleftbigg 1 − 1 parenrightbiggparenleftbigg c + c- parenrightbigg = planckover2pi1 2 ( c * + , c *- ) parenleftbigg c + − c- parenrightbigg = planckover2pi1 2 ( c * + c + − c *- c- ) b.) We know that ( S 2 x ) = ( S 2 y ) = ( S 2 z ) = planckover2pi1 2 4 . So that Δ S x = radicalbig ( S 2 x ) − ( S x ) 2 = planckover2pi1 2 radicalBig 1 − ( c * + c- + c *- c + ) 2 Δ S y = radicalBig ( S 2 y ) − ( S y ) 2 = planckover2pi1 2 radicalBig 1 + ( c * + c- − c *- c + ) 2 Δ S z = radicalbig ( S 2 z ) − ( S z ) 2 = planckover2pi1 2 radicalBig 1 − ( c * + c + − c *- c- ) 2 c.) With ( c * + c + + c *- c- ) 2 = 1 2 = 1 we can write Δ S x as Δ S x = planckover2pi1 2 radicalBig ( c * + c + + c *- c- ) 2 − ( c * + c- + c *- c + ) 2 = planckover2pi1 2 radicalBig | c + | 4 + 2 | c + | 2 | c- | 2 + | c- | 4 − ( c * + ) 2 c 2- − 2 | c + | 2 | c- | 2 − ( c *- ) 2 c 2 + = planckover2pi1 2 radicalBig ( c * + ) 2 c 2 + − ( c * + ) 2 c 2- − ( c *- ) 2 c 2 + + ( c *- ) 2 c 2- = planckover2pi1 2 radicalBig (( c * + ) 2 − ( c *- ) 2 )( c 2 + − c 2- ) = planckover2pi1 2 | c 2 + − c 2- | Similarly, we obtain Δ S y = planckover2pi1 2 radicalBig ( c * + ) 2 c 2 + + 2 c * + c *- c + c- + ( c *- ) 2 c 2- + ( c * + ) 2 c 2- − 2 c * + c *- c + c- + ( c *- ) 2 c 2 + = planckover2pi1 2 radicalBig ( c * + ) 2 c 2 + + ( c *- ) 2 c 2- + ( c * + ) 2 c 2- + ( c *- ) 2 c 2 + = planckover2pi1 2 | c 2 + + c 2- | 1 and Δ S z = planckover2pi1 2 radicalbig ( | c + | 2 + | c- | 2 ) 2 − ( | c + | 2 − | c- | 2 ) 2 = planckover2pi1 2 radicalbig 4 | c + | 2 | c- | 2 = planckover2pi1 | c + || c- | 2 2. The unitary rotation operator for spin Hilbert space is U R ( vector θ ) = e- i planckover2pi1 vector θ · vector S . The component of spin along the direction given by θ, φ in spherical polar coordinates is found by taking S z and rotating first by θ about the y-axis, then by φ about the z-axis, giving S θφ = U R (...
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852HW2_Solutions - PHYS852 Quantum Mechanics II Spring 2009...

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