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Assignment 2 Solutions 1. the General State of a Spin Half Particle

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Assignment 2 Solutions 1. the General State of a Spin Half Particle PHYSICS 301 QUANTUM PHYSICS I (2007) Assignment 2 Solutions 1 1. The general state of a spin half particle with spin component S n = S · nˆ = 2 ~ can be shown to be given by 1 1 1 iφ 1 1 |S n = 2 ~i = cos( 2 θ)|S z = 2 ~i + e sin( 2 θ)|S z = − 2 ~i where nˆ is a unit vector nˆ = sin θ cos φ ˆi + sin θ sin φ jˆ + cos θ kˆ, with θ and φ the usual angles for spherical polar coordinates. 1 1 (a) Determine the expression for the the states |S x = 2 ~i and |S y = 2 ~i. 1 (b) Suppose that a measurement of S z is carried out on a particle in the state |S n = 2 ~i. 1 What is the probability that the measurement yields each of ± 2 ~? 1 (c) Determine the expression for the state for which S n = − 2 ~. 1 (d) Show that the pair of states |S n = ± 2 ~i are orthonormal. SOLUTION 1 (a) For the state |S x = 2 ~i, the unit vector nˆ must be pointing in the direction of the X axis, i.e. θ = π/2, φ = 0, so that 1 1 1 1 |S x = ~i = √ |S z = ~i + |S z = − ~i 2 2 2 2 1 For the state |S y = 2 ~i, the unit vector nˆ must be pointed in the direction of the Y axis, i.e. θ = π/2 and φ = π/2. Thus 1 1 1 1 |S y = ~i = √ |S z = ~i + i|S z = − ~i 2 2 2 2 1 1 2 (b) The probabilities will be given by |hS z = ± 2 ~|S n = 2 ~i| . These probabilities will be 1 1 2 2 1 |hS z = 2 ~|S n = 2 ~i| = cos ( 2 θ) 1 1 2 iφ 1 2 2 1 |hS z = − 2 ~|S n = 2 ~i| = |e sin( 2 θ)| = sin ( 2 θ). 1 1 1 (c) The state for which S n = − 2 ~ is the state for which S · nˆ = − 2 ~, i.e. S · (−nˆ) = 2 ~. Thus the required result can be obtained by making the replacement nˆ → −nˆ, which is achieved through the replacements θ → π − θ and φ → φ + π. Thus, the required expression is 1 1 1 i(φ+π) 1 1 |S n = − 2 ~i = cos( 2 (π − θ))|S z = 2 ~i + e sin( 2 (π − θ))|S z = − 2 ~i 1 1 iφ 1 1 = sin( 2 θ)|S z = 2 ~i − e cos( 2 θ)|S z = − 2 ~i 1 (d) The required inner products are, using the orthonormality of the basis states |S z = 1 ± 2 ~i: 1 1 h 1 1 −iφ 1 1 i hS n = − 2 ~|S n = 2 ~i = sin( 2 θ)hS z = 2 ~| − e cos( 2 θ)hS z = − 2 ~| h 1 1 iφ 1 1 i × cos( 2 θ)|S z = 2 ~i + e sin( 2 θ)|S z = − 2 ~i 1 1 1 1 = sin( 2 θ) cos( 2 θ) − cos( 2 θ) sin( 2 θ) = 0 1 1 h 1 1 −iφ 1 1 i hS n = 2 ~|S n = 2 ~i = cos( 2 θ)hS z = 2 ~| + e sin( 2 θ)hS z = − 2 ~| h 1 1 iφ 1 1 i × cos( 2 θ)|S z = 2 ~i + e sin( 2 θ)|S z = − 2 ~i 2 1 2 1 = cos ( 2 θ) + sin ( 2 θ) = 1. 1 1 h 1 1 −iφ 1 1 i hS n = − 2 ~|S n = − 2 ~i = sin( 2 θ)hS z = 2 ~| − e cos( 2 θ)hS z = − 2 ~| h 1 1 iφ 1 1 i × sin( 2 θ)|S z = 2 ~i − e cos( 2 θ)|S z = − 2 ~i 2 1 2 1 = sin ( 2 θ) + cos ( 2 θ) = 1. 2. The state of a system is given by √ 1 −iπ/4 |ψi = 3 (1 − i)|1i + 2e |2i − (2 + i)|3i where the states |ni, n = 1, 2, 3 form a complete set of orthonormal basis states. (a) What is the probability amplitude of finding the system, in each of the states |1i and |2i? What are the probabilities of finding the system in each of these states? (b) What is the associated bra vector hψ|? (c) Another state of the system is given by 1 |φi = √ (1 − i)|1i − 2i|2i. 6 Calculate the inner product hψ|φi and determine the probability of observing the sys- tem in the state |ψi given that it was in the state |φi. SOLUTION (a) The probability amplitude of finding the system in the state |1i is the coefficient of |1i, i.e. h1|ψi = (1 − i)/3. The probability ampliude of finding the system in the state |2i is √ h2|ψi = 2e−iπ/4/3. Finally, the probability amplitude of finding the system in the state |3i will be h2|ψi = −(2 + i)/3. The corresponding probabilities will then be 1 − i 2 2 |h1|ψi|2 = = 3 9 2 and √ 2e−iπ/4 2 2 |h2|ψi|2 = = 3 9 and 2 + i 2 5 |h3|ψi|2 = − = 3 9 Note that these probability amplitudes add to give unity, i.e. the initial state was cor- rectly normalized. (b) The bra vector corresponding to the ket √ 1 −iπ/4 |ψi = 3 (1 − i)|1i + 2e |2i − (2 + i)|3i is obtained by taking the complex conjugate of the coefficients, and replacing the ket vectors |1i, |2i and |3i by the corresponding bras i.e. 1 √ hψ| = (1 + i)h1| + 2eiπ/4h2| − (2 − i)h3|. 3 (c) The inner product hψ|φi is then 1 √ 1 hψ|φi = (1 + i)h1| + 2eiπ/4h2| − (2 − i)h3| · √ (1 − i)|1i − 2i|2i 3 6 1 h √ i = √ (1 + i)(1 − i) − 2i 2eiπ/4 3 6 1 = √ (4 − 2i). 3 6 where we have used 1 + i eiπ/4 = cos(π/4) + i sin(π/4) = √ . 2 The probability of observing the system in the state |ψi given that it was in the state |φi will is 2 2 1 10 |hψ|φi| = √ (4 − 2i) = . 3 6 27 For this to be a true probability, the final state |φi must be normalized to unity. To check this, the inner product hφ|φi must be evaluated. This is just 1 hφ|φi = |1 − i|2 + |2i|2 = 1. 6 Hence the probability |hφ|ψi|2 is the true probability. 3. (a) Consider a system consisting of two spin half atoms (or qubits), usually referred to as a ‘bipartite system’. Each of these atoms can be put into a spin state for which 1 1 1 S z = ± 2 ~. Using the notation 0 ↔ S z = − 2 ~ and 1 ↔ S z = + 2 ~ the four possible states for this system are |00i, |01i and two others. (i) What are the other possible states? The state of this bipartite system can be manipulated by the use of magnetic fields. Suppose this bipartite system is initially prepared in the state |ψi. 3 (ii) What is the symbol in bra-ket notation for the probability amplitude of finding the system in state hmn|, m, n = 0 or 1, given that it was initially prepared in the state |ψi? (iii) What is the symbol in bra-ket notation for the probability amplitude of finding the system in the final state hφ| given that it was in one of the intermediate states |mni? (iv) What is the symbol in bra-ket notation for the probability amplitude of finding the system in the final state hφ| given that it ‘passed through’ the intermediate state |mni? (Recall that the system was initially prepared in the state |ψi.) (v) Expressed as a sum over appropriate terms, write down an expression for the total probability amplitude of finding the atom in the final state hφ| if the intermediate state through which it passed is not observed. (b) Suppose the initial and final states are such√ that the probability amplitudes of part 1 m+n (a)(iv) above have the values 2 i , i = −1, if the system has passed through the intermediate state |mni, m, n = 0 or 1. (i) What is the total probability of finding the system in the final state hφ| if the intermediate state through which it passed is not observed? (ii) Decoherence (as produced by external ‘noise’ affecting the atoms) is a phyical process which is equivalent to an observer determining which state each atom is in by some suitable measuring process. What is the total probability of finding the system in the final state hφ| if the intermediate state through which it passed is ‘observed’ through decoherence? SOLUTION (a) (i) hmn|ψi (ii) hφ|mni (iii) hφ|mnihmn|ψi (iv) Since the intermediate state is not observed, the total probability amplitude of observing the atom in the hφ| is the sum of the probability amplitudes of arriving in this final state via each of the possible intermediate states. Since the initial state is |ψi, the required probabilitiy amplitude is hφ|ψi, and this is X hφ|ψi = hφ|mnihmn|ψi mn where the sum is over the four possible values of mn. (b) Given that the probability amplitudes hφ|mnihmn|ψi = (i)m+n/2, then (i) The total probability amplitude of finding the atom in the final state hφ| if the intermediate state through which it passed is not observed is given in answer 3(a)iv i.e. X hφ|ψi = hφ|iihi|ψi mn 1 0 1 1 2 = 2 i + i + i + i 1 = 2 1 + i + i − 1 =i.
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