Quiz2 - PHY4221 Quantum Mechanics I (Fall 2005) Quiz 2...

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PHY4221 Quantum Mechanics I (Fall 2005) Quiz 2 (Suggested solution) For the words ‘ eigenvalue ’ and ‘ eigenvector ’ of an operator, we mean the following relation: ˆ || O α αα = So it is obvious that if we have ˆ |1 1 |1 O = × , then 1 is an eigenvalue of the operator with |1 the corresponding eigenvector. Since the Hilbert space is spanned by three independent basis vectors, we say that the Hilbert space has 3 dimensions and as a result operator O in such a Hilbert space should have 3 eigenvectors (but not necessary 3 eigenvalues because of the possible degeneration). Now that we have: ˆ |2 1 |3 ˆ |3 1 |2 O O =−× We can consider an arbitrary combination of these two basis vectors ab + , where a and b are two complex c numbers. Assuming this new-constructed vector is an eigenvector of operator O , we have the following relation: () ˆ ˆˆ ˆ Oa b a b b Ob a b λ += + + = The above two equations give: ( ) 22 0 ,0 ai bi a o r b i a a b ba + =− ⇒⇒
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This note was uploaded on 12/10/2011 for the course PHYS 4221 taught by Professor Cflo during the Spring '11 term at CUHK.

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