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Introduction to Quantum Computing (Instructor Solution Manual, Solutions) PDF
Preview Introduction to Quantum Computing (Instructor Solution Manual, Solutions)
Solutions Manual Introduction to Quantum Computing R.R. LaPierre Chapter 1 ______________________________________________________________________________ Exercise 1.1. Derive Eq. (1.11): I | E + E |2 = I + I + 2√I I cos 12 1 2 1 2 1 2 E = E ei(kx−t) 1 01 E = E ei(kx−t+) 2 02 E E * = (E ei(kx−t))(E ei(kx−t+)* 1 2 01 02 = E E e−i , assuming E is parallel to E (same polarization) so E E = E E 01 02 01 02 01 02 01 02 E E * = (E ei(kx−t+)(E ei(kx−t))* 2 1 02 01 = E E e+i, assuming E is parallel to E (same polarization) 01 02 01 02 E E * + E E * = E E (e+i + e−i) 1 2 2 1 01 02 = 2E E cos 01 02 2√I I cos 1 2 | E + E |2 = (E + E )( E + E )* 1 2 1 2 1 2 = E E * + E E * + E E * + E E * I 1 1 2 2 1 2 2 1 12 I1 I2 2√I1I2cos I = I +I +2√I I cos 12 1 2 1 2 ______________________________________________________________________________ Chapter 2 ______________________________________________________________________________ Exercise 2.1. Show that the first two states (two lowest energy levels) of the infinite quantum well are orthonormal. 2 π (x) = √ sin( x) 1 L L 2 2π (x) = √ sin( x) 2 L L +∞ ∫ ∗ dx 1 2 −∞ 2 L π 2π = ∫ sin( x)sin( x) dx L L L 0 4 L π π π = ∫ sin( x)sin( x)cos( x) dx, using sin(2x)=2sinxcosx L 0 L L L 4 L π π = ∫ sin2( x)cos( x) dx L L L 0 4 π L = sin3( x)| 3π L 0 4 = [sin3(π)−sin3(0)] 3π = 0 +∞ ∫ ∗ dx 1 1 −∞ 2 L π π = ∫ sin( x)sin( x) dx L L L 0 2 L π = ∫ sin2( x)dx L L 0 1 L 2π = ∫ [1−cos( x)]dx, using 2sin2(x) = 1−cos(2x) L 0 L 1 L 1 2π L = x| − sin( x)| L 0 2π L 0 = 1 Similarly, +∞ ∫ ∗ dx = 1 2 2 −∞ ∫+∞∗ dx = δ where i, j = {0, 1} −∞ i j ij (x) and (x) are orthonormal. 1 2 ______________________________________________________________________________ Exercise 2.2. Prove Eq. (2.49): <A> = |c |2 a + |c |2 a + … + |c |2 a 1 1 2 2 n n = c + c + … c 1 1 2 2 n n ∫+∞∗Â dx <A> = −∞ ∫+∞∗ dx −∞ The numerator is: +∞ ∫ ∗Â dx −∞ +∞ = ∫ (c∗∗ +c∗∗ +c∗∗ )Â(c +c +...c ) dx 1 1 2 2 n n 1 1 2 2 n n −∞ +∞ = ∫ (c∗∗ +c∗∗ +c∗∗ )(a c +a c +...a c ) dx 1 1 2 2 n n 1 1 1 2 2 2 n n n −∞ = c∗c a +c∗c a +...c∗c a 1 1 1 2 2 2 n n n = |c |2a +|c |2a +...|c |2a 1 1 2 2 n n All other terms in the numerator, like a c∗c ∫+∞∗ dx, are zero since and are 2 1 2 −∞ 1 2 1 2 orthonormal. Similarly, the denominator is: +∞ ∫ ∗ dx −∞ +∞ = ∫ (c∗∗ +c∗∗ +c∗∗ )(c +c +...c ) dx 1 1 2 2 n n 1 1 2 2 n n −∞ +∞ +∞ +∞ = c∗c ∫ ∗ dx+c∗c ∫ ∗ dx+...c∗c ∫ ∗ dx 1 1 1 1 2 2 2 2 n n n n −∞ −∞ −∞ = |c |2 +|c |2+...|c |2 , assuming each is normalized 1 2 n i = 1 , since all probabilities must sum to 1 All other terms in the denominator, like c∗c ∫+∞∗ dx, are zero. 1 2 −∞ 1 2 ∫+∞∗Â dx <A> = −∞ = |c |2a +|c |2a +...|c |2a ∫+∞∗ dx 1 1 2 2 n n −∞ Chapter 3 ______________________________________________________________________________ Exercise 3.1. How might you use the SG experiment to build a random number generator? Pass atoms from an oven with random spin through a Stern-Gerlach apparatus. If the S-G apparatus passes spin up, then assign bit 0 to the atom. If the S-G apparatus passes spin down, then assign bit 1 to the atom. The result is a random sequence of 0’s and 1’s. A string of n bits can represent any number from 0 to 2n-1. Therefore, the random string of 0’s and 1’s can be converted to a random number. ______________________________________________________________________________ Exercise 3.2. Show that Eq. (3.38) and (3.39) satisfy Eq. (3.33) to (3.36). Ŝ x ⟩ = +ħx ⟩ x + + 2 ħ(0 1) 1 (1) = ħ 1 (1) = +ħx ⟩ 2 1 0 √2 1 2√2 1 2 + Ŝ x ⟩ = −ħx ⟩ x − − 2 ħ(0 1) 1 ( 1) = ħ 1 (−1) = −ħx ⟩ 2 1 0 √2 −1 2√2 1 2 − Ŝ y ⟩ = +ħy ⟩ y + + 2 ħ(0 −i) 1 (1) = ħ 1 (1) = +ħy ⟩ 2 i 0 √2 i 2√2 i 2 + Ŝ y ⟩ = −ħy ⟩ y − − 2 ħ(0 −i) 1 ( 1) = ħ 1 (−1) = −ħy ⟩ 2 i 0 √2 −i 2√2 i 2 − Ŝ z ⟩ = +ħz ⟩ z + + 2 ħ(1 0)(1) = ħ(1) = +ħz ⟩ 2 0 −1 0 2 0 2 + Ŝ z ⟩ = −ħz ⟩ z − − 2 ħ(1 0)(0) = ħ( 0) = −ħz ⟩ 2 0 −1 1 2 −1 2 − ______________________________________________________________________________ Exercise 3.3. Prove that any complex 2x2 matrix can be written as I + σ̂ + σ̂ + σ̂ where , x y z , and are complex numbers. I + σ̂ + σ̂ + σ̂ x y z 1 0 0 1 0 −i 1 0 = ( ) + ( ) + ( ) + ( ) 0 1 1 0 i 0 0 −1 α+γ β−δi = ( ) β+δi α−γ Equate this to a general 2x2 complex matrix: α+γ β−δi a c ( ) = ( ) where a, b, c, d are complex numbers β+δi α−γ b d a = α+γ b = β+δi c = β−δi d = α−γ To generate the complex 2x2 matrix, set the following parameters: = (a+d)/2 = (b+c)/2 = (b−c)/2i = (a−d)/2 ______________________________________________________________________________ Exercise 3.4. Prove the following relations: X2 = Y2 = Z2 = − i XYZ = I Z = − i XY ZY = − i X ZYX = − i I YX = − i Z 0 1 0 1 1 0 X2 = ( )( ) = ( ) = I 1 0 1 0 0 1 0 −i 0 −i 1 0 Y2 = ( )( ) = ( ) = I i 0 i 0 0 1 1 0 1 0 1 0 Z2 = ( )( ) = ( ) = I 0 −1 0 −1 0 1 0 1 0 −i 1 0 0 1 0 i i 0 1 0 − i XYZ = − i ( )( )( ) = − i ( )( ) = − i ( ) = ( ) = I 1 0 i 0 0 −1 1 0 i 0 0 i 0 1 0 1 0 −i i 0 1 0 − i XY = − i ( )( ) = − i ( ) = ( ) = Z 1 0 i 0 0 −i 0 −1 1 0 0 −i 0 −i 0 1 ZY = ( )( ) = ( ) = − i ( ) = − i X 0 −1 i 0 −i 0 1 0 ZYX = (ZY)X = (− i X)X = − i (XX) = − i I 0 −i 0 1 −i 0 1 0 YX = ( )( ) = ( ) = − i ( ) = − i Z i 0 1 0 0 i 0 −1 ______________________________________________________________________________ Exercise 3.5. Prove Eq. (3.50) from Eq. (3.49). Show that the Ŝ operator has eigenvalues ± ħ and n 2 θ θ cos sin corresponding eigenvectors ( 2 ) and ( 2 ) eisinθ − eicosθ 2 2 n = cossinθ x n = sinsinθ y n = cosθ z Ŝ = n Ŝ + n Ŝ + n Ŝ n x x y y z z ħ 0 1 ħ 0 −i ħ 1 0 = cossinθ ( )+ sinsinθ( )+ cosθ( ) 2 1 0 2 i 0 2 0 −1 cosθ cossinθ−isinsinθ ħ = ( ) 2 cossinθ+isinsinθ −cosθ cosθ (cos−isin)sinθ ħ = ( ) 2 (cos+isin)sinθ −cosθ ħ cosθ e−isinθ = ( ) 2 eisinθ −cosθ Ŝ ( cosθ2 ) = ħ( cosθ e−isinθ)( cosθ2 ) = ħ( cosθcosθ2 + sinθsinθ2 ) = ħ( cosθ2 ) n eisinθ 2 eisinθ −cosθ eisinθ 2 ei(sinθcosθ − cosθsinθ) 2 eisinθ 2 2 2 2 2 where we have used the identities cox(x−y) = cos(x)cos(y) +sin(x)sin(y), and sin(x−y) = sin(x)cos(y) – cos(x)sin(y) Ŝ ( sinθ2 ) = ħ( cosθ e−isinθ)( sinθ2 ) = ħ( cosθsinθ2− sinθcosθ2 ) = ħ(sin(−θ2)) = n − eicosθ 2 eisinθ −cosθ − eicosθ 2 ei(sinθsinθ + cosθcosθ) 2 eicosθ 2 2 2 2 2 θ ħ sin − ( 2 ) 2 −eicosθ 2 Chapter 4 ______________________________________________________________________________ 1 1 1 Exercise 4.1. Show that the following state is normalized: ( + i)− 2 2 √2 1 1 2 1 2 1 1 | + i| +| | = + = 1 2 2 √2 2 2 ______________________________________________________________________________ Exercise 4.2. Rewrite 0> and 1> in terms of |+⟩ and |−⟩. Show that |+⟩ and |−⟩ are orthonormal. 1 |+⟩ = ( |0⟩ + |1⟩ ) √2 1 |−⟩ = ( |0⟩ − |1⟩ ) √2 Rearranging gives: 1 |0⟩ = ( |+⟩+|−⟩ ) √2 1 |1⟩ = ( |+⟩−|−⟩ ) √2 The bras are: 1 ⟨+| = (⟨0|+⟨1|) √2 1 ⟨−| = (⟨0|−⟨1|) √2 The inner products are: 1 1 ⟨+|−⟩ = (⟨0|+⟨1|) ( |0⟩ − |1⟩ ) √2 √2 1 = (⟨0|0⟩−⟨0|1⟩+⟨1|0⟩−⟨1|1⟩) 2 1 = (1−0+0−1) 2 = 0 1 1 ⟨+|+⟩ = (⟨0|+⟨1|) ( |0⟩+ |1⟩ ) √2 √2 1 = (⟨0|0⟩+⟨0|1⟩+⟨1|0⟩+⟨1|1⟩) 2 1 = (1+0+0+1) 2 = 1 1 1 ⟨−|−⟩ = (⟨0|−⟨1|) ( |0⟩− |1⟩ ) √2 √2 1 = (⟨0|0⟩−⟨0|1⟩−⟨1|0⟩+⟨1|1⟩) 2 1 = (1−0−0+1) 2 = 1 ______________________________________________________________________________ Exercise 4.3. Prove the following: (a) < >* = < > 1 2 2 1 (b) < > can be a complex number, but < > is real and positive 1 2 1 1 (c) <c > = c*< > 1 2 1 2 (d) < c > = c< > 1 2 1 2 α δ Suppose | > = ( ) and | > = ( ) 1 β 2 γ Then < | = (α∗β∗) and < | = (δ∗γ∗) 1 2 (a) ∗ δ < >* = [(α∗β∗)( )] = [α∗δ+β∗γ]∗ = αδ∗ + βγ∗ 1 2 γ α < > = [(δ∗γ∗)( )] = δ∗α+γ∗β = αδ∗ + βγ∗ 2 1 β < >* = < > 1 2 2 1 (b) δ < > = (α∗β∗)( ) = α∗δ+β∗γ , which in general is a complex number 1 2 γ α < > = (α∗β∗)( ) = α∗α+β∗β = |α|2 +|β|2 , which is real and positive 1 1 β (c) <c | = (|c >)† = (c| >)† = c*< | 1 1 1 1 <c > = c*< > 1 2 1 2 (d) c > = c > 2 2 < c > = < c > = c< > 1 2 1 2 1 2 ______________________________________________________________________________