QM: Sum of projection operators = identity operator?

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The discussion revolves around the calculation of the sum of projection operators for an observable represented by the matrix K. The eigenvectors corresponding to the eigenvalues 1 and -1 are identified, and their outer products are computed, resulting in matrices that sum to 2 times the identity operator. A key point raised is the necessity of normalizing the eigenvectors to unit vectors, which would correct the sum to equal the identity operator. The normalization involves multiplying the eigenvectors by 1/√2. Proper normalization is essential for ensuring the projection operators correctly represent the identity operator in quantum mechanics.
Simfish
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Homework Statement



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

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) = \begin{bmatrix} 1 &amp; -i \\ i &amp; 1 \end{bmatrix}

|-1><-1| = (i, 1)T*(-i, 1) = \begin{bmatrix} 1 &amp; i \\ -i &amp; 1 \end{bmatrix}

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?
 
Last edited:
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Hi Simfish! :wink:
Simfish said:
… 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:
 
Hi. :) Okay, good idea. Thanks!
 
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