handin10

handin10 - Physics 212 Classical and Modern Physics Spring...

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Unformatted text preview: Physics 212 Classical and Modern Physics Spring 2008 Solutions for Hand-In Set #10 A96 More practice with complex numbers (a) z * 1 = 0 . 25 * = . 25. (b) z * 2 = (0 . 25 i ) * =- . 25 i . (c) | z 3 | 2 = z 3 z * 3 = (0 . 5 + 0 . 3 i )(0 . 5- . 3 i ) = 0 . 5 2 + (0 . 3)(0 . 5) i- (0 . 5)(0 . 3) i- . 3 2 i 2 = 0 . 25 + 0 . 09 = . 34. Supplementary Reading 1.4 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 den- sity 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 lo- cation of the particle has improved, but but there is a cost to this knowledge: we now no longer can say ex- actly 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.) Supplementary Reading 1.7 For a photon in the state | i = a | R i + b | T i , 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 = . 3 ' . 5477 a =- . 3 ' - . 5477 a = i . 3 ' . 5477 i a =- i . 3 ' - . 5477 i a = . 15 + i . 15 ' . 3873 + 0 . 3873 i 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 = . 7 ' . 8367 b =- . 7 ' - . 8367 b = i . 7 ' . 8367 i b =- i . 7 ' - . 8367 i b = . 35- i . 35 ' . 5916- . 5916 i Supplementary Reading 1.10 (a) To show that the state is normalized, we calculate the quantity h | i : h | i = ( . 7 h 1 | - . 5 h 2 | + 0...
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handin10 - Physics 212 Classical and Modern Physics Spring...

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