Question { 1573 }

QUANTUM COMPUTING - EXAMPLE 32.6 : (a) Let H | 0 > = 0.707 ( | 0 > + | 1 > ), H | 1 > = 0.707 ( | 0 > - | 1 > ). Find the values for H | 0 > + H | 1 > and H | 0 > - H | 1 >. (b) In quantum computing, a qubyte is a quantum byte, or 8 quantum bits, a sequence processed as a unit. A qubit is a quantum bit. According to Alexander Holevo in his theorem, n qubits cannot carry more than n classical bits of information. What is the maximum amount of classical bits of information that can be carried by 1 qubyte.

Answer

QUANTUM COMPUTING - ANSWER 32.6 : (a) H | 0 > + H | 1 > = 0.707 ( | 0 > + | 1 > ) + 0.707 ( | 0 > - | 1 > ) = 0.707 ( | 0 > + | 1 > + | 0 > - | 1 > ) = 0.707 ( 2 | 0 > ) = 1.414 | 0 >. H | 0 > - H | 1 > = 0.707 ( | 0 > + | 1 > ) - 0.707 ( | 0 > - | 1 > ) = 0.707 [ | 0 > + | 1 > - ( | 0 > - | 1 > ) ] = 0.707 ( 2 | 1 > ) = 1.414 | 1 >. (b) A qubyte = 8 qubits. When n = 8, maximum amount of classical bits = 8. 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 { 1600 }

QUANTUM COMPUTING - EXAMPLE 32.7 : If | ± > = 0.707 ( | 0 > ± | 1 > ), prove that | Ψ (t = 0) > = | 0 > = 0.707 ( | + > + | - > ).

Answer

QUANTUM COMPUTING - ANSWER 32.7 : | + > = 0.707 ( | 0 > + | 1 > ), | - > = 0.707 ( | 0 > - | 1 > ), | + > + | - > = 0.707 ( | 0 > + | 1 > ) + 0.707 ( | 0 > - | 1 > ) = 0.707 ( | 0 > + | 1 > + | 0 > - | 1 > ) = 0.707 ( 2 | 0 > ). | Ψ (t = 0) > = | 0 > = [ 1 / (0.707 x 2) ] ( | + > + | - > ) = 0.707 ( | + > + | - > ) (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 { 1509 }

QUANTUM COMPUTING - EXAMPLE 32.8 : In quantum computing, a quantum state is given by S = a | 00 > + b | 01 > + g | 10 > + d | 11 >. (a) Find S in term of | 0 > and | 1 > etc. (b) The probability of getting x is P(x). For S = 0.5 | 00 > + 0.5 | 01 > + 0.5 | 10 > + 0.5 | 11 >, find P(0) and P(1). Hint : P(00) + P(01) = P(0) = a x a + b x b, P(10) + P(11) = P(1) = g x g + d x d.

Answer

QUANTUM COMPUTING - ANSWER 32.8 : (a) S = a | 00 > + b | 01 > + g | 10 > + d | 11 > = | 0 > ( a | 0 > + b | 1 > ) + | 1 > ( g | 0 > + d | 1 > ). (b) By comparison of S, a = b = g = d = 0.5. P(0) = a x a + b x b = 0.5 x 0.5 + 0.5 x 0.5 = 0.5, P(1) = g x g + d x d = 0.5 x 0.5 + 0.5 x 0.5 = 0.5. 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 { 1410 }

QUANTUM COMPUTING - EXAMPLE 32.9 : In quantum computing, find the equations of S = (a | 0 > + b | 1 >) (g | 0 > + d | 1 >) in term of | 00 >, | 01 >, | 10 > and / or | 11 > when ad = 0.

Answer

QUANTUM COMPUTING - ANSWER 32.9 : If ad = 0, then a = 0 but d ≠ 0, or d = 0 but a ≠ 0, or a = d = 0. When a = 0 but d ≠ 0, S = (b | 1 >) (g | 0 > + d | 1 >) = bg | 10 > + bd | 11 >. When d = 0 but a ≠ 0, S = (a | 0 > + b | 1 >) (g | 0 >) = ag | 00 > + bg | 10 >. When a = d = 0, S = (b | 1 >) (g | 0 >) = bg | 10 >. 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 { 1329 }

QUANTUM COMPUTING - EXAMPLE 32.10 : In quantum computing, the conversion of Control Not (CNOT) gate in two input quantum bit gate could be decribed as : | 00 > --> | 00 >, | 01 > --> | 01 >, | 10 > --> | 11 >, | 11 > --> | 10 >. If | P > = 0.707 ( | 01 > - | 11 > ), find the value of CNOT | P >.

Answer

QUANTUM COMPUTING - ANSWER 32.10 : Since | 01 > --> | 01 > and | 11 > --> | 10 >, then CNOT | P > = CNOT [ 0.707 ( | 01 > - | 11 > ) ] = 0.707 ( | 01 > - | 10 > ). 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 { 1478 }

QUANTUM BIOLOGY - EXAMPLE 33.1 : In quantum biology, conditional entropy is given by H (S | O) = H (SO) - H (O). With reference to such equation, find another equation for H (SA).

Answer

QUANTUM BIOLOGY - ANSWER 33.1 : If H (S | O) = H (SO) - H (O), then H (SO) = H (S | O) + H (O). Substitute O term with A will give H (SA) = H (S | A) + H (A). 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 { 1396 }

QUANTUM BIOLOGY - EXAMPLE 33.2 : (a) In a biomolecule, ATP hydrolysis will produce 3 kcal / mol to 60 kcal / mol of energy, according to energy scale. Find the range of such energy generation if 2 moles of molecules involved. (b) A formula is given by ln P = ln a + b ln W where P = metabolic rate, W = body size, a = dependent of taxa, b = scaling exponent. If b is approximately 1 for plant, and b is approximately 0.75 for animal, find the relationship of P as a function of a and W in plant.

Answer

QUANTUM BIOLOGY - ANSWER 33.2 : (a) Range of energy, E = 3 kcal / mol to 60 kcal / mol. For 2 moles, multiply the range by 2 so that E = (3 x 2) to (60 x 2) kcal or E = 6 kcal to 120 kcal. (b) Let ln P = 1n a + b ln W. For plant, b = 1 approximately. Then ln P = 1n a + ln W = ln (aW), P = aW. 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 { 1343 }

QUANTUM BIOLOGY - EXAMPLE 33.3 : In quantum biology, microtubule is used to store information in a cell. At temperature of T = 300 K, measured current I in a probe is directly proportional to the supplied voltage V, when passing through a microtubule with resistance R. (a) Form an equation of V as a function of I involving k as a constant. (b) If the microtubule has R = 1 ohm at such condition, find the value of V when I = 2 A. Hint : Ohm's law. (c) Find the relationship of k as a function of R.

Answer

QUANTUM BIOLOGY - ANSWER 33.3 : (a) Let I = kV, V = I / k. (b ) By Ohm's law, when I = 2 A, V = IR = 2 A x 1 ohm = 2 V. (c) From answers in (a) and (b), V = I / k = IR. Then k = 1 / R. 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 { 1344 }

QUANTUM BIOLOGY - EXAMPLE 33.4 : (a) According to Landauer (1986), the capacity of human memory is approximately X bits. Assume that a human retains 2 bits / second of visual, verbal, tactile and musical memory, find the value of X if a human lifetime is approximately 2.5 billion seconds. (b) The total power consumption of the human brain is about 25 Watts. The bread of 100 grams will produce 1000 kilojoules of energy. How much bread is needed to run a human brain for 1 day?

Answer

QUANTUM BIOLOGY - ANSWER 33.4 : (a) Let X = human memory / second x human lifetime, then X = 2 bits / second x 2.5 billion seconds = 5 billion bits. (b) One day has 24 hours or 24 hours x 3600 seconds / hour = 86400 seconds. Let Watt = Joule / second. Total energy consumption of human brain per day = 25 Joules / second x 86400 seconds = 2160000 Joules = A. Energy generation of bread = 1000000 Joules / 100 grams = 10000 Joules / gram = B. Mass of bread needed to run a 25 Watt human brain for 1 day = A / B = 216 grams. 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 { 1297 }

QUANTUM BIOLOGY - EXAMPLE 33.5 : (a) In the measurement of the number of photons received by a plant during photosynthesis, quantum meter is used. At 8 AM, 11 AM, 12 PM, 2 PM and 4 PM of the same day, the measured readings of the meter are 10, 70, 60, 120 and 120 units. Find the mode, median and min of the readings of the quantum meter. (b) Microtubule is used to carry information in a cell. If the cross-sectional area of the microtubule has a diameter of 25 nanometers, find the volume of the microtubule of 1 nanometer in height. State the assumption of calculation.

Answer

QUANTUM BIOLOGY - ANSWER 33.5 : (a) Mode = 120 (2 readings, greater than others with 1 reading). Median = 70 (since 70 is the middle, when the values are rearranged orderly as 10, 60, 70, 120, 120). Min = (sum of readings) / (number of readings) = (10 + 70 + 60 + 120 + 120) / 5 = 380 / 5 = 76. (b) Assumption : Microtubule is a cylinder. Let V = 3.142 x h x d x d / 4 where V = volume, d = diameter, h = height. Then V = 3.142 x h x d x d / 4 = 3.142 x n x 25 n x 25 n / 4 = 490.9375 n x n x n = 4.909375 x 10^2 x 10^(-9) x 10^(-9) x 10^(-9) = 4.909375 x 10^(-25) meters, where ^ is the symbol of power, n is nano or 10^(-9). 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 { 1390 }

QUANTUM BIOLOGY - EXAMPLE 33.6 : The time to maintain quantum coherence in a biological system is t. Let t = k / (MT) where k = constant, M = mass, T = temperature. For living farm animals, normal rectal temperatures ranges for pig and goat are 38.7 - 39.8 and 38.5 - 39.7 in degree Celsius respectively. Let a pig has a mass of 100 kg and a goat has a mass of 300 pounds. (a) Find the mass of goat in the unit of kilogram, when 1 pound = 0.4536 kg. (b) Find t (p) / t (g) for living animals when t (p) = t for pig and t (g) = t for goat. (c) State the assumption of your calculation in question (b).

Answer

QUANTUM BIOLOGY - ANSWER 33.6 : (a) M for goat = 300 pounds x 0.4536 kg / pound = 136.08 kg. (b) Let t (p) = k / (100 T), t (g) = k / (136.08 T). Then t (p) / t (g) = 136.08 / 100 = 1.3608. (c) Assumption : Both pig and goat have the same temperature T, times to lose their quantum information t are approximations. 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 { 1326 }

QUANTUM BIOLOGY - EXAMPLE 33.7 : (a) In a DNA of a living cell, the quantum information available in the bases guanine (G) and thymine (T) are | G > = | 110 > and | T > = | 010 > respectively. Calculate | G > - | T >. (b) In a living biological cell, the step time for random walk of an electron is t. The localization time of an electron is T. If i is the geometric average of T and t, find log T as a function of t and i.

Answer

QUANTUM BIOLOGY - ANSWER 33.7 : (a) | G > - | T > = | 110 > - | 010 > = | 100 >. (b) Geometric average of T and t is the square root of (T x t) = i. Then log i = 0.5 log (Tt), log (Tt) = log T + log t = 2 log i, log T = 2 log i - log t. 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 { 1384 }

QUANTUM BIOLOGY - EXAMPLE 33.8 : (a) Let ^ be the symbol of power where 1 ^ 2 = 1 x 1 = 1, 2 ^ 2 = 2 x 2 = 4. Let the number of electrons in a human body to be 10 ^ 28 = A, the number of all of the grains of sand on Earth planet to be 7 x (10 ^ 20) = B, the number of all the stars in the visible sky to be 8 x (10 ^ 3) = C. By assuming that every star in the visible sky has the same number of grains of sand as on Earth planet, prove by mathematical calculations that there are more electrons in one human body compared to the number of all of the grains of sand on the stars in the visible sky. (b) The incoming solar radiation to the Earth's surface is mainly from sun. Around 51 % of the radiation is absorbed by Earth's surface. Around 19 % is absorbed by atmosphere and clouds. In term of reflection, 4 % of the radiation is from surface of Earth, 6 % is reflected by atmosphere and the rest is reflected by clouds. Find the percentage of radiation absorbed by and reflected by biological beings on Earth, with reason for your response.

Answer

QUANTUM BIOLOGY - ANSWER 33.8 : (a) Number of all of the grains of sand on the stars in the visible sky = D = B x C = 7 x (10 ^ 20) x 8 x (10 ^ 3) = 56 x (10 ^ 23) = [ 5.6 x (10 ^ 24) ] < (10 ^ 28), or D < A (proven). (b) Not sure. Reason : No data related to biological beings is provided in the question. 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 { 1278 }

QUANTUM BIOLOGY - EXAMPLE 33.9 : (a) Let ^ be the symbol of power where 10 ^ 1 = 10, 10 ^ 2 = 100, 10 ^ 3 = 1000 etc. Total energy consumption of the brain is about 25 Watts, whereas a Blue Gene computer requires 1.5 Mega Watts. Blue Gene computer performs at 1 petaflop. In a human body, there are approximately 10 ^ 16 synapse operations per second i. e. at least 10 petaflops. Prove by calculations that a human brain is more energy efficient than a Blue Gene computer. (b) Quantum effects and quantum entanglement in the brain are identical to quantum gravity and string theory. If one is true, the other is true. What conclusion can be made if quantum effects in the brain and quantum gravity are true?

Answer

QUANTUM BIOLOGY - ANSWER 33.9 : (a) Comparison of energy consumption per rate of operation is made : - Human brain : 25 Watts / (10 petaflops) = 2.5 Watts / petaflop = A. Blue Gene computer : 1.5 x (10 ^ 6) Watts / petaflop = B. Then A < B (Proven). (b) Quantum entanglement in the brain and string theory are true. 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 { 1370 }

QUANTUM BIOLOGY - EXAMPLE 33.10 : The wavefunction starts out in a superposition of possible states in a closed black box like this : Ψ (kitty) = 0.7071 Ψ (alive) + 0.7071 Ψ (dead) where Ψ (kitty) = wavefunction of a kitten, Ψ (alive) = wavefunction of a living kitten, Ψ (dead) = wavefunction of a dead kitten. By prediction and calculation, find the probability of : (a) a living kitten inside the black box; (b) a dead kitten inside the black box.

Answer

QUANTUM BIOLOGY - ANSWER 33.10 : By prediction, probability of a living kitten = probability of a dead kitten, inside a black box = 0.5 = P. By calculation, P = (0.7071) (0.7071) = 0.5. 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.