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QM: Sum of projection operators = identity operator?

  • Thread starter Simfish
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  • #1
Simfish
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Homework Statement



So we have an observable K = [tex] \begin{bmatrix} 0 & -i \\ -i & 0 \end{bmatrix}[/tex]

and its eigenvectors are v1 = (-i, 1)T and v2 = (i, 1)T corresponding to eigenvalues 1 and -1, respectively.

Now if we take the outer products, we get these...

|1><1| = (-i, 1)T*(i, 1) = [tex] \begin{bmatrix} 1 & -i \\ i & 1 \end{bmatrix}[/tex]

|-1><-1| = (i, 1)T*(-i, 1) = [tex] \begin{bmatrix} 1 & i \\ -i & 1 \end{bmatrix}[/tex]

Then we add them and they sum up to form 2*(identity operator).

But isn't it supposed to sum up to the identity operator? What's wrong, or what happened?
 
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Answers and Replies

  • #2
tiny-tim
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Hi Simfish! :wink:
… and its eigenvectors are v1 = (-i, 1)T and v2 = (i, 1)T
Don't you need to normalise them (to unit vectors), by multiplying by 1/√2 ? :smile:
 
  • #3
Simfish
Gold Member
818
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Hi. :) Okay, good idea. Thanks!
 

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