Lect18 - Lecture 18: 3D Review, MRI, Examples A real (2D)...

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Lecture 18, p 1 Lecture 18: 3D Review, MRI, Examples A real (2D) “quantum dot” http://pages.unibas.ch/phys- meso/Pictures/pictures.html
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Lecture 18, p 2 D=1 6E o (2,1,1) (1,2,1) (1,1,2) D=3 (1,1,1) 3E o (n x ,n y n z ) Cubical Box Exercise Consider a 3D cubic box: Show energies and label (n x ,n y ,n z ) for the first 11 states of the particle in the 3D box, and write the degeneracy, D, for each allowed energy. Define E o = h 2 /8mL 2 . E y z x L L L Degeneracy
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Lecture 18, p 3 D=1 6E o (2,1,1) (1,2,1) (1,1,2) D=3 (1,1,1) 3E o (n x ,n y n z ) Solution E y z x L L L 9E o D=3 (2,2,1) (2,1,2) (1,2,2) 11E o D=3 (3,1,1) (1,3,1) (1,1,3) 12E o D=1 (2,2,2)   2 2 2 2 2 8 x y z n n n x y z h E n n n mL n x ,n y ,n z = 1,2,3,. .. Degeneracy Consider a 3D cubic box: Show energies and label (n x ,n y ,n z ) for the first 11 states of the particle in the 3D box, and write the degeneracy, D, for each allowed energy. Define E o = h 2 /8mL 2 .
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Lecture 18, p 4 For a cubical box, we just saw that the 5 th energy level is at 12 E 0 , with a degeneracy of 1 and quantum numbers (2,2,2). 1. What is the energy of the next energy level? a. 13E 0 b. 14E 0 c. 15E 0 2. What is the degeneracy of this energy level? a. 2 b. 4 c. 6 Act 1
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Lecture 18, p 5 For a cubical box, we just saw that the 5 th energy level is at 12 E 0 , with a degeneracy of 1 and quantum numbers (2,2,2). 1. What is the energy of the next energy level? a. 13E 0 b. 14E 0 c. 15E 0 2. What is the degeneracy of this energy level? a. 2 b. 4 c. 6 Solution E 1,2,3 = E 0 (1 2 + 2 2 + 3 2 ) = 14 E 0
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Lecture 18, p 6 For a cubical box, we just saw that the 5 th energy level is at 12 E 0 , with a degeneracy of 1 and quantum numbers (2,2,2). 1. What is the energy of the next energy level? a. 13E 0 b. 14E 0 c. 15E 0 2. What is the degeneracy of this energy level? a. 2 b. 4 c. 6 Solution E 1,2,3 = E 0 (1 2 + 2 2 + 3 2 ) = 14 E 0 Any ordering of the three numbers will give the same energy. Because they are all different (distinguishable), the answer is 3! = 6. Question: Is it possible to have D > 6? Hint: Consider E = 62E 0 .
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Lecture 18, p 7 E 3E o 6E o 9E o 11E o Consider a non-cubic box: The box is stretched along the y-direction. What will happen to the energy levels? Define E o = h 2 /8mL 1 2 z x y L 1 L 2 > L 1 L 1 Non-cubic Box
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Lecture 18, p 8 1: The symmetry is “broken” for y, so the 3-fold degeneracy is lowered. A 2-fold degeneracy remains, because x and z are still symmetric. 2: There is an overall lowering of energies due to decreased confinement along y. (1,1,1) D=1 (1,2,1) D=1 D=2 (2,1,1) (1,1,2) E 3E o 6E o 9E o 11E o Consider a non-cubic box: The box is stretched along the y-direction. What will happen to the energy levels? Define E o = h 2 /8mL 1 2     2 2 2 2 0 2 2 8 x y z n n n x z y h E E n n n mL Solution z x y L 1 L 2 > L 1 L 1 The others are left for you.
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Lect18 - Lecture 18: 3D Review, MRI, Examples A real (2D)...

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