Problem 1, solution
(a) In the {|1>, |2>} basis the matrix of H is
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The eigenvalues of H are found from
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For the eigenvectors we find
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(b) The matrix of U has the eigenvectors as its columns.
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Problem 2, solution
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(a) [Q,P]=ih, [Q,P2]=P[Q,P]+[Q,P]P=2ihP. |
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(b) [Q,Pn+1]=[Q,PPn]=P[Q,Pn]+[Q,P]Pn. If [Q,Pn]=ihnPn-1, then [Q,Pn+1]=PihnPn-1+ihPn=ih(n+1)Pn. [Q,Pn]=ihnPn-1 holds for n=1 and n=2. Therefore it holds for all n. |
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(c) [P,Q]=-ih, [P,Q2]=Q[P,Q]+[P,Q]Q=-2ihQ. [P,Qn+1]=[P,QQn]=Q[P,Qn]+[P,Q]Qn. If [P,Qn]=-ihnQn-1, then [P,Qn+1]=-QihnQn-1-ihQn=-ih(n+1)Qn. [P,Qn]=-ihnQn-1 holds for n=1 and n=2. Therefore it holds for all n. |
Problem
3, solution
|i> and |j> are eigenkets of A.
A|i> = ai|i>, A|j> = aj|j>.
A(|i>+|j>) = ai|i>+aj|j> ¹ aij(|i>+|j>) unless ai=aj, i.e. unless |i> and |j> have the same eigenvalue.
Problem 4, solution
| (a) Let {|i>} denote a basis for the vector space.
(<i|X|j> and <j|Y|i> are numbers and we can reverse their order when multiplying.)
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| (b) <i|(XY)T|j>=<XYi|j>=<j|XY|i>*=<XTj|Y|i>*=<Yi|XT|j>=<i|YTXT|j>
for any |i> and |j>. |
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if |n> is an eigenvector of A with eigenvalue an. For an arbitrary vector |y> we have
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(d) if {|a’>} is a basis.
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Problem 5, solution
| (a) W is Hermitian, Wij=Wji*. |
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(b) w1=1: -c1+c3=0, c2=0 w2=0: c1=c3=0, c2=arbitrary w1=-1: c1+c3=0, c2=0
are normalized eigenvectors. |
| (c) Uij=<i|wj>
The matrix has the eigenvectors as its columns. |
| (d) UTij=Uij* |
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(e)  .
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(f)  .
UTWU is a diagonal matrix with the eigenvalues along its diagonal. |
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