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Unformatted text preview: CHAPTER 13 1. (a) electronproton system m r = m e 1 + m e / M p = (1 5.45 × 10 4 ) m e (b) electrondeuteron system m r = m e 1 + m e / M d = (1 2.722 × 10 4 ) m e (c) For two identical particles of mass m , we have m r = m 2 2. One way to see that P 12 is hermitian, is to note that the eigenvalues ±1 are both real. Another way is to consider i , j ∑ dx 1 ∫ dx 2 ψ ij * ( x 1 , x 2 ) P 12 ψ ij ( x 1 , x 2 ) = i , j ∑ dx 1 dx 2 ∫ ψ ij * ( x 1 , x 2 ) ψ ji ( x 2 , x 1 ) = j , i ∑ dy 1 ∫ dy 2 ψ ji * ( y 2 , y 1 ) ψ ij ( y 1 , y 2 ) = dy 1 ∫ dy 2 ( P 12 ψ ij ( y 1 , y 2 )) * ψ ij ( y 1 , y 2 ) j , i ∑ 3. If the two electrons are in the same spin state, then the spatial wave function must be antisymmetric. One of the electrons can be in the ground state, corresponding to n = 1, but the other must be in the next lowest energy state, corresponding to n = 2. The wave function will be ψ ground ( x 1 , x 2 ) = 1 2 u 1 ( x 1 ) u 2 ( x 2 ) u 2 ( x 1 ) u 1 ( x 2 ) ( 29 4. The energy for the nth level is E n = h 2 π 2 2 ma 2 n 2 ≡ ε n 2 Only two electrons can go into a particular level, so that with N electrons, the lowest N /2 levels must be filled. The energy thus is E tot = 2 ε n 2 n = 1 N /2 ∑ ≈ 2 ε 1 3 N 2 3 = ε N 3 12 If N is odd, then the above is uncertain by a factor of ε N 2 which differs from the above by (12/ N ) ε , a small number if N is very large. 5. The problem is one of two electrons interacting with each other. The form of the interaction is a square well potential. The reduction of the twobody problem to a oneparticle system is straightforward. With the notation 1 x = x 1 x 2 ; X = x 1 + x 2 2 ; P = p 1 + p 2 , the wave function has the form ψ ( x 1 , x 2 ) = e iPX u ( x ) , where u ( x ) is a solution of  h 2 m d 2 u ( x ) dx 2 + V ( x ) u ( x ) = Eu ( x ) Note that we have taken into account the fact that the reduced mass is m /2. The spatial interchange of the two electrons corresponds to the exchange x...
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 Spring '05
 mokhtari
 Mass, wave function

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