237solutions4

# 237solutions4 - 1 2 \$& ’ sin 2 L x \$& ’ = 2 2 5 L...

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Solutions/Hints – Module #4 & GRE Practice #4 Module #4: Getting Started 1. a. " b. 0 c. " 2 2. n ' n = " n',n 3. See page 226. Discussion Questions 1. In class we examined how the Dromedary ground state camel can be made as flat (uncurvy) as we like so we can give it as low an energy as we like. 2. They resemble each other because in each case, the energies of the eigenfunctions increase with curviness. 3. An odd parity Bactrian camel is curvier (more energetic) than a Dromedary camel. Enrichment Problem 1. First, the problem should be modified to say, “… at t=0, the wave function is equal to " ( x , t = 0) = 8 2 5 L sin 3 # 2 L x \$ % & ( ) cos # 2 L x \$ % & ( ) So I am giving you the normalization. Next use trigonometry identities: " ( x , t = 0) = 8 2 5 L sin 3 # 2 L x \$ % & ( ) cos # 2 L x \$ % & ( ) = 8 2 5 L sin 2 # 2 L x \$ % & ( ) * + , - . / sin # 2 L x \$ % & ( ) cos # 2 L x \$ % & ( ) * + , - . / = 8 2 5 L 1 2 \$ % & ( ) 1 0 cos # L x \$ % & ( ) 1 2 \$ % & ( ) sin # L x \$ % & ( ) * + , - . / = 2 2 5 L sin # L x \$ % & ( ) 0 sin # L x \$ % & ( ) cos # L x \$ % & ( ) * + , - . / = 2 2 5

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Unformatted text preview: ) 1 2 \$ % & ’ ( ) sin 2 L x \$ % & ’ ( ) * + ,-. / = 2 2 5 L sin L x \$ % & ’ ( ) 2 5 L sin 2 L x \$ % & ’ ( ) = 2 2 5 L L 2 2 L sin L x \$ % & ’ ( ) 2 5 L L 2 2 L sin 2 L x \$ % & ’ ( ) = 2 2 5 L L 2 " 1 ( x , t = 0) 2 5 L L 2 " 2 ( x , t = 0) = 2 1 5 " 1 ( x , t = 0) 1 5 " 2 ( x , t = 0) = 4 5 " 1 ( x , t = 0) 1 5 " 2 ( x , t = 0) Yikes! Notice that c 1 = 4 5 and c 2 = 1 5 and c 1 2 + c 2 2 = 1. Finally, we can answer the question: " ( x , t ) = 4 5 " 1 ( x , t ) # 1 5 " 2 ( x , t ) = 4 5 \$ 1 ( x ) % 1 ( t ) # 1 5 2 ( x ) 2 ( t ) . 2. This takes a while and will be done later in the course. GRE Practice #4: 42-B, 60-B, 51-B, 98-D....
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