In the basis we can repeat this reasoning for

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Unformatted text preview: escribed by a qubit. Very roughly speaking, the spin is a quantum description of the magnetic moment of an electron which behaves like a spinning charge. The two allowed states can roughly be thought of as clockwise rotations (“spin-up”) and counter clockwise rotations (“spin-down”). We will say much more about the spin of an elementary particle later in the course. 1.7. QUBITS 15 Measurement Example I: Phase Estimation Now that we have discussed qubits in some detail, we can are prepared to look more closesly at the measurement principle. Consider the quantum state, eiθ 1 | ψ ￿ = √ | 0￿ + √ | 1￿ . 2 2 If we were to measure this qubit in the standard basis, the outcome would be 0 with probability 1/2 and 1 with probability 1/2. This measurement tells us only about the norms of the state amplitudes. Is there any measurement that yields information about the phase, θ? To see if we can gather any phase information, let us consider a measurement in a basis other than the standard basis, namely 1 |+￿ ≡ √ (|0￿ + |1￿) 2 and 1 |−￿ ≡ √ (|0￿ − |1￿). 2 What does |φ￿ look like in this new basis? This can be expressed by first writing, 1 |0￿ = √ (|+￿ + |−￿) 2 and 1 |1￿ = √ (|+￿ − |−￿). 2 Now we are equipped to rewrite |ψ ￿ in the {|+￿ , |−￿}-basis, 1 eiθ |ψ ￿ = √ |0￿ + √ |1￿) 2 2 1 e iθ = (|+￿ + |−￿) + (|+￿ − |−￿) 2 2 1 + eiθ 1 − e iθ = |+￿ + |−￿ . 2 2 Recalling the Euler relation, eiθ = cos θ + i sin θ, we see that the probability of measuring |+￿ is 1 ((1 + cos θ)2 + sin2 θ) = cos2 (θ/2). A similar calcula4 tion reveals that the probability of measuring |−￿ is sin2 (θ/2). Measuring in the (|+￿ , |−￿)-basis therefore reveals some information about the phase θ. Later we shall show how to analyze the measurement of a qubit in a general basis. 16 CHAPTER 1. INTRODUCTION Measurement example II: General Qubit Bases What is the result of measuring a￿ general qubit state |ψ ￿ = α |0￿ + β |1￿, in ￿ a general orthonormal basis |v ￿ , ￿v ⊥ , where |v ￿ =￿a|0￿ + b|1￿ and |v ⊥ ￿ = ￿ b∗ |0￿ − a∗ |1￿? ￿You should also check that |v ￿ and ￿v ⊥ are orthogonal by ￿ showing that v ⊥ |v = 0. To answer this question, let us make use of our recently acquired bra￿￿ ket notation. We first show that the states |v ￿ and ￿v ⊥ are orthogonal, that is, that their inner product is zero: ￿ ￿ v ⊥ |v = (b∗ |0￿ − a∗ |1￿)† (a |0￿ + b |1￿) = (b ￿0| − a ￿1|)† (a |0￿ + b |1￿) = ba ￿0|0￿ − a2 ￿1|0￿ + b2 ￿0|1￿ − ab ￿1|1￿ = ba − 0 + 0 − ab =0 Here we have used the fact that ￿i|j ￿ = δij . Now, the probability of measuring the state |ψ ￿ and getting |v ￿ as a result is, Pψ ( v ) = | ￿ v | ψ ￿ | 2 = | ( a ∗ ￿ 0| + b ∗ ￿ 1| ) ( α | 0￿ + β | 1￿ ) | 2 = | a ∗ α + b∗ β | 2 Similarly, ￿￿ ￿￿ 2 ￿ ￿ Pψ ( v ⊥ ) = ￿ v ⊥ | ψ ￿ = | ( b ￿ 0| − a ￿ 1| ) ( α | 0￿ + β | 1￿ ) | 2 = | bα − a β | 2...
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This document was uploaded on 09/22/2013.

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