Hand in 10 Solutions(07)

Hand in 10 Solutions(07) - Solutions for Hand-In Set #10...

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Unformatted text preview: Solutions for Hand-In Set #10 PHYS 212 Spring 2007 Supplementary Reading 1.4 Supplementary Reading 1.10 When you plot the square of the wavefunction for the ground state you see a broad "bump" between 0 and L. From this graph you see that the particle is more likely to be found near the middle of the "box" but there is a large probability density throughout the box; the particle is said to be delocalized. When you add the wavefunctions for the five lowest-energy states together and square the result to get a new probability density, you get something quite different. The probability density is large in the left quarter of the "box" but almost zero everywhere else, i.e., the particle has been localized in the left quarter of the box. Our knowledge of the location of the particle has improved, but but there is a cost to this knowledge: we now no longer can say exactly what the energy of the particle is, nor can we say as much about the magnitude of its momentum. This is the uncertainty principle in action, and this is a direct consequence of the wave nature of particles. With the addition of more modes, the localization improves even more, but it takes an even bigger spread of energies. (If you study quantum mechanics more you will learn how to localize particles in other regions of the box.) (a) To show that the state is normalized, we calculate the quantity | : | = (0.7 1| - 0.5 2| + 0.4 3| + 0.3 4| + 0.1 5|) (0.7|1 + 0.5|2 + 0.4|3 + 0.3|4 + 0.1|5 ) = 0.72 1|1 + 0.52 2|2 + 0.42 3|3 + 0.32 4|4 + 0.12 5|5 + (0.7 0.5) 1|2 + (0.7 0.4) 1|3 + other "cross terms" = 0.72 + 0.52 + 0.42 + 0.32 + 0.12 + 0 + 0 + . . . = 1. Because | = 1, the state is normalized. (b) Because the states |1 , |2 , |3 , etc., are orthogonal and normalized, the probabilities come from the squares of the coefficients in the state | : Prob(E1 ) = |0.7|2 Prob(E2 ) = |0.5|2 Prob(E3 ) = |0.4|2 Prob(E4 ) = |0.3|2 Prob(E5 ) = |0.1|2 = 0.49 - 4,900 electrons = 0.25 - 2,500 electrons = 0.16 - 1,600 electrons = 0.09 - 900 electrons = 0.01 - 100 electrons Supplementary Reading 1.7 (c) The expectation value E is just the weighted average of the measurements: E = Prob(E1 ) E1 + Prob(E2 ) E2 + Prob(E3 ) E3 + Prob(E4 ) E4 + Prob(E5 ) E5 -8.0 eV -8.0 eV = 0.49 + 0.25 1 22 -8.0 eV -8.0 eV + 0.09 + 0.16 32 42 -8.0 eV + 0.01 52 = -4.610 eV. For a photon in the state | = a|R + b|T , the probability of being reflected is |a|2 , which we are told is 0.3 (30%). Here are five choices for a, all of which satisfy the condition |a|2 = 0.3: a = 0.3 0.5477 a = - 0.3 -0.5477 a = i 0.3 a= 0.5477i a = -i 0.3 -0.5477i 0.15 + i 0.15 0.3873 + 0.3873i The probablity of being transmitted is |b|2 , which we are told is 0.7. Here are five choices for b, all of which satisfy the condition |b|2 = 0.7: b = 0.7 0.8367 b = - 0.7 -0.8367 b = i 0.7 b= 0.8367i b = -i 0.7 -0.8367i 0.35 - i 0.35 0.5916 - 0.5916 Supplementary Reading 1.13 A76 -- Population Inversion (a) The photon is in state |-pol , so the probability that it is in this state is just 1. This simple idea is expressed by the normalization condition: -pol|-pol = 1. Using the expression for |-pol from the problem statement gives -pol|-pol = (cos x-pol| + sin y-pol|) (cos |x-pol + sin |y-pol ) = cos2 x-pol|x-pol + sin2 y-pol|y-pol + cos sin x-pol|y-pol + cos sin y-pol|x-pol = cos2 + sin2 + 0 + 0 = 1. (b) and (c) After passing through the first polarizer the photon is, by definition, in the state |x-pol . After passing through the second polarizer the photon is in the state |-pol = cos |x-pol + sin |y-pol . The probability that photons "prepared" to be in state |x-pol (i.e., passed through the first polarizer) are found after a measurement to be in state |-pol , (i.e., passed through the second polarizer) is | -pol|x-pol |2 = | (cos x-pol| + sin y-pol|) |x-pol |2 2 When a photon interacts with atoms, the probability that the photon will be absorbed by an atom in its ground state is identical to the probability that the photon will induce stimulated emission from an atom initially in its excited state. If there are more atoms in the ground state than the excited state, the probability that the photon will interact with a ground state atom and be absorbed is greater than the probability that the photon will interact with an excited state atom and induce emission of a "copy" of itself. This is bad for a laser. Conversely, if there are more atoms in the excited state this is called a population inversion; in this case the probability that the photon will interact with an excited atom is greater than the probability that it will be absorbed by a ground state atom. The net result is then amplification, which is necessary for laser action. Tipler 36-8 For the an electron with principal quantum number n, the orbital quantum number l can take on the values 0, 1, 2, or 3. The possble values of the magnetic quantum number for each of these values of l are given below: l = 0 ml = 0 only. l = 1 ml = -1, 0, or 1. l = 2 ml = -2, -1, 0, 1, or 2 l = -3 ml = -3, -2, -1, 0, 1, 2, or 3 Thus there are seven different values of ml : -3, -2, -1, 0, 1, 2, and 3. = |cos x-pol|x-pol + sin y-pol|x-pol | = | cos + 0|2 = cos2 . (d) The probability for a photon in state |-pol (passed by second polarizer) to be found after a measurement to be in state |y-pol (passed by the third polarizer) is | y-pol|-pol |2 = = = = y-pol| (cos |x-pol + |y-pol ) |2 | cos y-pol|x-pol + sin y-pol|y-pol |2 |0 + sin |2 sin2 . Tipler 36-34 (a) The possible states are n = 4, l = 0 ml = 0, and ms = 1/2 for each possible ml ; two states. n = 4, l = 1 ml = -1, 0,or 1, and ms = 1/2 for each possible ml ; six states. n = 4, l = 2 ml = -2, -1, 0, 1,or 2, and ms = 1/2 for each possible ml ; 10 states. n = 4, l = 3 ml = -3, -2, -1, 0, 1, 2, or 3, and ms = 1/2 for each possible ml ; 14 states. There are a total of 2 + 6 + 10 + 14 = 32 states. (b) The possible states are (e) The probability for the photon to make it through the second polarizer and the third polarizer is the product of the two probabilities calculated above: Probability = sin cos . This should be consistent with what you observed in lab. Note: This probability can be rewritten as (cos sin ) = 2 2 2 1 sin 2 2 2 1 = sin2 2. 4 n = 2, l = 0 ml = 0, and ms = 1/2 for each possible ml ; two states. n = 2, l = 1 ml = -1, 0,or 1, and ms = 1/2 for each possible ml ; six states. There are a total of 2 + 6 = 8 states. NOTE: It isn't too hard to derive a general formula for the number of possible states with principal quantum number n. We leave it as an exercise to show that this is equal to 2n2 . but to complete the problem we must know the values of -x|+z and -x|-z . These we can get by using Eq. (2.2) to write -x| in terms of +z| and -z|: -x|+z = = 1 +z| - 2 1 , 2 1 +z| - 2 1 . 2 1 -z| |-z 2 1 -z| |+z 2 -x|-z Supplementary Reading 2.3 = = - (a) The probability that the electron is in the state |+x is given by P+x = | +x| |2 = = +x| 4 3 |+z + |-z 5 5 2 Using these expressions gives 2 P-x = 3 5 1 4 - 2 5 1 2 = 0.02. (b) The probability that the electron is in the state |+y is given by P+y = | +y| |2 = = +y| 4 3 |+z + |-z 5 5 2 4 3 +x|+z + +x|-z 5 5 2 , but to complete the problem we must know the values of +x|+z and +x|-z . These we can get by using Eq. (2.2) to write +x| in terms of +z| and -z|: +x|+z = = 1 +z| + 2 1 , 2 1 +z| + 2 1 . 2 1 -z| |-z 2 1 -z| |+z 2 4 3 +y|+z + +y|-z 5 5 2 , but to complete the problem we must know the values of +y|+z and +y|-z . These we can get by using Eq. (2.2) to write +y| in terms of +z| and -z|: +y|+z = = 1 +z| + i 2 1 , 2 1 +z| + i 2 1 . 2 1 -z| |-z 2 1 -z| |+z 2 +x|-z = = Using these expressions gives 2 +y|-z 1 4 + 2 5 1 2 = = i P+x = 3 5 = 0.98. The probability that the electron is in the state |-x is given by P-x = | -x| | = = -x| 2 2 Using these expressions gives 2 P+y = 3 5 4 1 +i 2 5 1 2 4 3 |+z + |-z 5 5 = 0.5. The probability that the electron is in the state |-y is given by , P-y = | -y| |2 4 3 -x|+z + -x|-z 5 5 2 = = -y| 3 4 |+z + |-z 5 5 2 Supplementary Reading 2.6 4 3 -y|+z + -y|-z 5 5 2 , but to complete the problem we must know the values of -y|+z and -y|-z . These we can get by using Eq. (2.2) to write -y| in terms of +z| and -z|: -y|+z = = 1 +z| - i 2 1 , 2 1 +z| - i 2 1 . 2 1 -z| |-z 2 1 -z| |+z 2 -y|-z = = -i The field of chemistry would be so different if electrons were bosons that it is hard to imagine. Take, for instance, a carbon atom. Carbon has 6 protons 6 electrons. Because the electrons are fermions they obey the Pauli exclusion principle, and no two electrons can occupy the same quantum state. This leads to various distinct "shells" and orbitals: there are 2 electrons in the `closed' n = 1 shell, and four electrons in the unfilled n = 2 shell that are free to form bonds, leading to molecules like methane (CH4 ) and the rest of the rich field of carbon chemistry (on which life is based). If electrons were bosons, all electrons would go into the ground state, and the electron configuration would be 1s6 (instead of 1s2 2p4 ). All of chemistry would be different, and there certainly would be no periodic table as we know it. Using these expressions gives 2 P-y = 3 5 4 1 -i 2 5 1 2 = 0.5 ...
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This homework help was uploaded on 04/07/2008 for the course PHYS 212 taught by Professor Ladd during the Spring '08 term at Bucknell.

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