Solutions to Assignment 2

Solutions to Assignment 2 - Physics 681-481 CS 483...

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February 16, 2006 Physics 681-481; CS 483: Discussion of #2 I. Constructing a spooky 2-Qbit state We can write the state | Ψ i as | Ψ i = 1 3 ( | 00 i + | 01 i + | 10 i ) = ( 1 H )( q 2 3 | 0 i| 0 i + q 1 3 | 1 i H | 0 i ) = ( 1 H ) C H h ( q 2 3 | 0 i + q 1 3 | 1 i ) | 0 i i = ( 1 H ) C H ( w 1 ) | 0 i| 0 i , (1) where w is any one-Qbit unitary transformation that takes | 0 i into q 2 3 | 0 i + q 1 3 | 1 i . To construct a controlled-Hadamard C H from a controlled-NOT C , note that the NOT operation X is x · σ while the Hadamard transformation is H = 1 2 ( X + Z ) = 1 2 ( x + z ) · σ. It follows from Section A2 of the appendix to chapter 1 that H = uXu , (2) where u is the one-Qbit unitary transformation associated with any rotation that takes x into 1 2 ( x + z ). Since we also have 1 = uu , it follows that C H = ( 1 u ) C ( 1 u ) . (3) So (1) reduces to the compact form | Ψ i = ( 1 Hu ) C ( w u ) | 0 i| 0 i , (4) which produces the state | Ψ i by acting on | 0 i| 0 i with three one-Qbit unitaries ( Hu , w , and u ) and one (two-Qbit) cNOT gate.
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  • Spring '08
  • Ginsparg
  • Linear Algebra, unitary transformation, Qbits

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