Question { 1652 }

QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - EXAMPLE 31.2 : (a) Let | - > = 1 | x > + 0 | y >, | | > = 0 | x > + 1 | y >. Find the value of 2 | x > + 3 | y > in term of | - > and | | >. (b) Let m to be the reduced mass. Find the value of m in term of Ma and Mb where 1 / m = 1 / Ma + 1 / Mb.

Answer

QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - ANSWER 31.2 : (a) For | - > = 1 | x > + 0 | y >, multiply it with 2 to produce 2 | - > = 2 | x > + 0 | y >, then 2 | x > = 2 | - > - 0 | y > as first equation. For | | > = 0 | x > + 1 | y >, multiply it with 3 to produce 3 | | > = 0 | x > + 3 | y >, then 3 | y > = 3 | | > - 0 | x > as second equation. Finally first equation plus second equation to produce 2 | x > + 3 | y > = 2 | - > + 3 | | >. (b) Let 1 / m = 1 / Ma + 1 / Mb = (Ma + Mb) / (Ma x Mb). Then m = (Ma x Mb) / (Ma + Mb). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1711 }

QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - EXAMPLE 31.3 : In photoelectrical effect analysis of quantum chemistry, let E = kinetic energy of electron, p = intensity of UV light, f = frequency of UV light. According to Classical Theory, E = c for all values of f, E = mp. According to Quantum Theory, E = c for all values of p, E = mf + c. In a graph, m and c are constants where m is slope and c is y intercept. If m = 2 and c = 3 with similar value of E : (a) find the value of p according to Classical Theory; (b) find the value of f according to Quantum Theory.

Answer

QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - ANSWER 31.3 : (a) E = c = 3, then E = mp = 2p = 3, then p = 3 / 2. (b) E = c = 3, then E = mf + c = 2f + 3 = 3, then f = (3 - 3) / 2 = 0. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1405 }

QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - EXAMPLE 31.4 : In a rigid rotor model in quantum chemistry, the moment of inertia I is given by an Equation E as I = Ma x La x La + Mc x Lc x Lc = m x L x L, where m = (Ma x Mc) / (Ma + Mc) and L = La + Lc, m is the reduced mass, Ma is the mass of a, Mc is the mass of c, La is the radius of a from point O, Lc is the radius of c from point O. Prove by simplest method that Equation E is wrong.

Answer

QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - ANSWER 31.4 : Let Ma = 0 and Mc = 1 as assumption. Substitute them into Equation E as I = Ma x La x La + Mc x Lc x Lc = 0 + Lc x Lc = Lc x Lc. However, m = (0 x 1) / (0 + 1) = 0, then I = Lc x Lc, which is not equal to 0 but I = m x L x L = 0 x L x L = 0. Equation E is proven wrong. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1537 }

QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - EXAMPLE 31.5 : In a wavefunction, let P(x) = A cos kx + B sin kx. By using the boundary conditions of x = 0 and x = l, where P(0) = P(l) = 0, prove by mathematical calculation that P(x) = B sin (npx / l) where p = 22 / 7 approximately, n is a rounded number. A, B and k are constants.

Answer

QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - ANSWER 31.5 : When x = 0, P(0) = A cos (k x 0) + B sin (k x 0) = A = 0. When x = l, P(l) = A cos (k x l) + B sin (k x l) = 0 = B sin (k x l), then k x l = arc sin 0 = np, k = np / l. Finally P(x) = B sin kx = B sin (npx / l) as proven. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1446 }

QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - EXAMPLE 31.6 : In N + 1 Rule in Quantum Chemistry, whenever a spin 1 / 2 nucleus is adjacent to N other nuclei, it is split into N + 1 distinct peaks. In 1 peak or singlet, there is only 1 magnitude. In 2 peaks or doublet, the ratio of magnitude of each peak is 1 : 1. In 3 peaks or triplet, the ratio of magnitude of each peak is 1 : 2 : 1. In 4 peaks or quartet, the ratio of magnitude of each peak is 1 : 3 : 3 : 1. In 5 peaks or quintet, the ratio of magnitude of each peak is 1 : 4 : 6 : 4 : 1. (a) By using binomial coefficients or Triangle of Pascal find the ratio of magnitude of each peak if 6 peaks exists. (b) How many adjacent nuclei are available in a spin 1 / 2 nucleus in such situation of 6 peaks?

Answer

QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - ANSWER 31.6 : (a) 1 : 5 : 10 : 10 : 5 : 1, since (0 + 1) : (1 + 4) : (4 + 6) : (6 + 4) : (4 + 1) : (1 + 0) with reference to the ratio of magnitude of each peak in quintet. (b) Number of adjacent nuclei = Number of peaks - 1 = 6 - 1 = 5. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1876 }

QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - EXAMPLE 31.7 : (a) The correct statement about both the average value of position () and momentum (

) of a 1-dimensional harmonic oscillator wavefunction is =

= 1 - x. Find the value of x. (b) The probabilities of finding a particle around points A, B and C in the wavefunction y = f(x) are P(A), P(B) and P(C) respectively. Coordinates are A (3,5), B (4,-10) and C (6,7). Arrange P(A), P(B) and P(C) in term of a < b < c, when | y-coordinate | signifies the probability.

Answer

QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - ANSWER 31.7 : (a) Let =

= 1 - x = 0, then x = 1 - 0 = 1. (b) With reference to the | y-coordinate | for each point in A, B and C coordinates, if 5 < 7 < |-10| then a < b < c = P(A) < P(C) < P(B). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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Question { 1586 }

QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - EXAMPLE 31.8 : (a) Acceptable wavefunction in quantum mechanics in the range of : negative infinity < x < positive infinity, vanishes at least at one boundary. Which of the following is the wavefunction or are the wavefunctions of acceptable theory : P = x, P = | x |, P = sin x, P = exp (-x), P = exp (-| x |)? State the reason. (b) Let linear momentum operator P = -ih d / dz. The wavefunction is S = exp (-ikz) where i x i = -1, k and h are constants. Find the linear momentum of such wavefunction by using the term P x S.

Answer

QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - ANSWER 31.8 : (a) Acceptable wavefunctions are P = sin x as the boundaries are P = +1 and -1, and P = exp (-| x |) as the boundaries are P = 1 and 0. (b) P x S = -ih d / dz [ exp (-ikz) ] = -ik x (-ih) x exp (-ikz) = -i (-i) kh x S = -kh x S. Then P = -kh = linear momentum. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1540 }

Question 112 - In quantum computing, let the amplitude A = a | 0 > + b | 1 >, | a | | a | + | b | | b | = 1. Find the values of b if A = 0.8 | 0 > + b | 1 >.

Answer

Answer 112 - Let A = a | 0 > + b | 1 > = 0.8 | 0 > + b | 1 >, then a = 0.8. When | a | | a | + | b | | b | = 1, | b | | b | = 1 - | a | | a | = 1 - 0.8 x 0.8 = 1 - 0.64 = 0.36, | b | = 0.6. Finally b = +0.6 or -0.6. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1800 }

QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - EXAMPLE 31.9 : When an algebraic product is defined on the space, the Lie bracket is the commutator [x,y] = xy - yx according to Lie algebra in mathematics. If [p,x] f(x) = px f(x) - xp f(x), p = -ih d / dx, find the value of [p,x] in term of i and h.

Answer

QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - ANSWER 31.9 : Let [p,x] f(x) = px f(x) - xp f(x) = (-ih d / dx) x f(x) - x (-ih d / dx) f(x) = -ih { d / dx [ x f(x) ] - x d f(x) / dx } = -ih [ f(x) ] + x d f(x) / dx - x d f(x) / dx = -ih [ f(x) ]. Then [p,x] = -ih. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1429 }

QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - EXAMPLE 31.10 : There are 6 spin orbitals in p subshell in a ground state carbon atom. Only 2 electrons fill the p subshell. Number of different ways for n electrons to occupy the k spin orbitals are k! / [ (n!) (k-n)! ]. Find the number of different configurations of electrons to occupy the p subshell in a carbon atom.

Answer

QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - ANSWER 31.10 : Let k = 6, n = 2. Number of ways = k! / [ (n!) (k-n)! ] = 6! / [ (2!) (6-2)! ] = 6! / (2! 4!) = 15. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1535 }

QUANTUM COMPUTING - EXAMPLE 32.1 : In quantum computing, let the amplitude A = a | 0 > + b | 1 >, | a | | a | + | b | | b | = 1. Find the values of b if A = 0.8 | 0 > + b | 1 >.

Answer

QUANTUM COMPUTING - ANSWER 32.1 : Let A = a | 0 > + b | 1 > = 0.8 | 0 > + b | 1 >, then a = 0.8. When | a | | a | + | b | | b | = 1, | b | | b | = 1 - | a | | a | = 1 - 0.8 x 0.8 = 1 - 0.64 = 0.36, | b | = 0.6. Finally b = +0.6 or -0.6. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1432 }

QUANTUM COMPUTING - EXAMPLE 32.2 : (a) If | 001 > = | 1 >, | 111 > = | 7 >, find the 2 possible values of ( | 001 > + | 1 > + | 7 > ) ( | 111 > ). (b) In quantum money, a duplicate will have probability P of passing the verification test of a bank, if the total number of photons on the bank note is N. The would be counterfeiter has a probability p of success in duplicating the quantum money correctly for each photon. Guess the relationship of P, p and N as a mathematical formula involving natural logarithm ln.

Answer

QUANTUM COMPUTING - ANSWER 32.2 : (a) First answer : ( | 001 > + | 1 > + | 7 > ) ( | 111 > ) = ( | 1 > + | 1 > + | 7 > ) ( | 111 > ) = ( | 9 > ) ( | 111 > ) = | 999 >. Second answer : ( | 001 > + | 1 > + | 7 > ) ( | 111 > ) = ( | 1 > + | 1 > + | 7 > ) ( | 7 > ) = ( | 9 > ) ( | 7 > ) = | 63 >. (b) If N = 1, then P = p. If N = 2, then P = p power 2. If N = 3, then P = p power 3. For N, P = p power N, ln P = ln (p power N), then ln P = N ln p. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1494 }

QUANTUM COMPUTING - EXAMPLE 32.3 : A system of linear congruences consists of 3 equations : X ≡ 1 (mod 3), X ≡ 3 (mod 5), X ≡ 4 (mod 6). X has positive values. (a) List the values of these equations from 1 to 35. Then find the minimum value of X. (b)(i) Find the least common multiple (LCM) of b = 3, 5 and 6 where X ≡ a (mod b). (ii) If b - a has the same value of all equations above, then X + (b - a) is divisible by LCM. Find the value of minimum value of X via LCM division.

Answer

QUANTUM COMPUTING - ANSWER 32.3 : (a) X ≡ 1 (mod 3) = 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34. X ≡ 3 (mod 5) = 3, 8, 13, 18, 23, 28, 33. X ≡ 4 (mod 6) = 4, 10, 16, 22, 28, 34. All equations have minimum value of X = 28. (b)(i) LCM for b = 3, 5 and 6 = (30 / 10, 30 / 6, 30 / 5) is 30. (ii) Since b - a = 3 - 1 = 5 - 3 = 6 - 4 = 2, then X + (b - a) is divisible by LCM. X + 2 is divisible by 30. X = 30 - 2 = 28. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1564 }

QUANTUM COMPUTING - EXAMPLE 32.4 : A system of linear congruences consists of 3 equations : X ≡ 1 (mod 2), X ≡ 3 (mod 3), X ≡ 4 (mod 5). X has positive values. (a)(i) List the values of these equations from 1 to approximately 40. (ii) Find the first smallest value and second smallest value of X. (iii) Guess the third smallest value of X. (b) Let X ≡ Aa (mod Ma), X ≡ Ab (mod Mb), X ≡ Ac (mod Mc). According to Chinese remainder theorem, X ≡ (Aa x Ya x Md + Ab x Yb x Me + Ac x Yc x Mf) [ mod (Ma x Mb x Mc) ]. (i) Show that Ma, Mb and Mc have the greatest common divisor of Ma x Mb x Mc. (ii) Find the values of Md, Me and Mf if Md = Mb x Mc, Me = Ma x Mc and Mf = Ma x Mb. (iii) Find the values of Ya, Yb and Yc if Ya = Remainder of (Md / Ma), Yb = Remainder of (Me / Mb) and Yc = Remainder of (Mf / Mc). (iv) Use Chinese remainder theorem to find X.

Answer

QUANTUM COMPUTING - ANSWER 32.4 : (a)(i) 1 (mod 2) = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39. 3 (mod 3) = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39. 4 (mod 5) = 4, 9, 14, 19, 24, 29, 34, 39. (ii) By observation on X, 3 equations have common values of 9 and 39. First smallest value = 9, second smallest value = 39. (iii) Third smallest value = second smallest value + (second smallest value - first smallest value) = 39 + (39 - 9) = 69. (b)(i) Let Ma = 2, Mb = 3, Mc = 5 where they are prime numbers. Their greatest common divisor is 2 x 3 x 5 = Ma x Mb x Mc (shown). (ii) Md = Mb x Mc = 3 x 5 = 15, Me = Ma x Mc = 2 x 5 = 10, Mf = Ma x Mb = 2 x 3 = 6. (iii) Ma = 2, Mb = 3, Mc = 5, Md = 15, Me = 10, Mf = 6. Md / Ma = 15 / 2 = 7 remain 1, Ya = 1. Me / Mb = 10 / 3 = 3 remain 1, Yb = 1. Mf / Mc = 6 / 5 = 1 remain 1, Yc = 1. (iv) Let Aa = 1, Ab = 3, Ac = 4, Ya = 1, Yb = 1, Yc = 1, Ma = 2, Mb = 3, Mc = 5, Md = 15, Me = 10, Mf = 6. X ≡ (Aa x Ya x Md + Ab x Yb x Me + Ac x Yc x Mf) [ mod (Ma x Mb x Mc) ] = (1 x 1 x 15 + 3 x 1 x 10 + 4 x 1 x 6) [ mod (2 x 3 x 5) ] = 69 mod 30 = 39 mod 30 = 9 mod 30. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question { 1831 }

QUANTUM COMPUTING - EXAMPLE 32.5 : In quantum teleportation, let (C0 + D1) (00 + 11) = C000 + C011 + D100 + D111. Ba = 00 + 11, Bc = 10 + 01, Be = 00 - 11, Bm = 10 - 01. (a) Find the values of 00, 01, 10 and 11 in term of Ba, Bc, Be and Bm. (b) Prove by calculation that (C0 + D1) (00 + 11) = 0.5 [ Ba (C0 + D1) + Bc (C1 + D0) + Be (C0 - D1) + Bm (-C1 + D0) ].

Answer

QUANTUM COMPUTING - ANSWER 32.5 : (a) Ba + Be = (00 + 11) + (00 - 11) = 2 (00), 00 = (Ba + Be) / 2. Bc - Bm = (10 + 01) - (10 - 01) = 2 (01), 01 = (Bc - Bm) / 2. Bc + Bm = (10 + 01) + (10 - 01) = 2 (10), 10 = (Bc + Bm) / 2. Ba - Be = (00 + 11) - (00 - 11) = 2 (11), 11 = (Ba - Be) / 2. (b) (C0 + D1) (00 + 11) = C000 + C011 + D100 + D111 = (00) C0 + (01) C1 + (10) D0 + (11) D1 = (1/2) (Ba + Be) C0 + (1/2) (Bc - Bm) C1 + (1/2) (Bc + Bm) D0 + (1/2) (Ba - Be) D1 = (1/2) [ (Ba + Be) C0 + (Bc - Bm) C1 + (Bc + Bm) D0 + (Ba - Be) D1 ] = 0.5 ( Ba C0 + Be C0 + Bc C1 - Bm C1 + Bc D0 + Bm D0 + Ba D1 - Be D1) = 0.5 [ Ba C0 + Ba D1 + Bc C1 + Bc D0 + Be C0 - Be D1 + Bm (-C1) + Bm D0 ] = 0.5 [ Ba (C0 + D1) + Bc (C1 + D0) + Be (C0 - D1) + Bm (-C1 + D0) ] (Proven). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.