Answer_To_The_3rd_Quiz - a b b a a b a i = = ⇒ ⇒ =...

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PHY4221 Quantum Mechanics I (Fall 2004) Quiz 3 (Answer) For the words ‘ eigenvalue ’ and ‘ eigenvector ’ of an operator, we mean the following relation: ˆ | | O α α α = 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 | 2 | 3 a b + , 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: ( 29 ( 29 ( 29 ( 29 ( 29 ˆ | 2 | 3 | 2 | 3 ˆˆˆ | 2 | 3 | 2 | 3 | 3 | 2 O a b a b O a b O a O b a b λ + = + + = + = - The above two equations give: ( 29 ( 29 2 2 | 2 | 3 0 | 2 | 3 , 0 | 2 | 3 a i a b a b b ia or b ia

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Unformatted text preview: a b b a a b a i + = - + = ⇒ ⇒ = = -⇒ + = = ≠- The normalization will show you that 1 2 a = . Furthermore, from above relation we have / a b i or i = =-(λ here is in fact the eigenvalue). You see that: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 1 1 ˆ | 2 | 3 | 3 | 2 | 2 | 3 2 2 2 ˆ | 2 | 3 1 1 ˆ | 3 | 2 | 2 | 3 | 2 | 3 2 2 2 i O i i i O a b i i i O i - +-+ + = = = +-- So you see that ( 29 1 | 2 | 3 2 i ± are the eigenvectors corresponding to eigenvalues – i and i . Thus we obtained the three eigenvectors and their corresponding eigenvalues. Note, however, that the operator O here is not Hermitian because its 3 eigenvalues are not all real numbers....
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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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Answer_To_The_3rd_Quiz - a b b a a b a i = = ⇒ ⇒ =...

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