Ket Notation - Effects of the Projection Operator

AI Thread Summary
The discussion centers on the effects of the projection operator in ket notation, specifically in the context of deriving an equation from Sakurai's Modern Quantum Mechanics. It explores whether the inner product of a bra and a ket, specifically <α|a'>, can be simplified to a summation over eigenkets of a Hermitian operator A. The participants clarify that while the inner product does not assume real values, it does relate to the complex conjugate of the reverse inner product. The conversation emphasizes the importance of understanding the properties of inner products in different bases and the implications of using eigenkets of Hermitian operators. Overall, the discussion seeks to clarify the mathematical relationships and properties inherent in quantum mechanics notation.
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Ket Notation -- Effects of the Projection Operator

Homework Statement


From Sakurai's Modern Quantum Mechanics (Revised Edition), it is just deriving equation 1.3.12.


Homework Equations


\begin{eqnarray*}\langle \alpha |\cdot (\sum_{a&#039;}^N |a&#039;\rangle \langle a&#039;|) \cdot|\alpha \rangle \end{eqnarray*}

The Attempt at a Solution


The summation can be moved to the left, so everything is being summed from a' to N, but does an alpha bra inner product with a' (or <α|a'>) does the sum of this from all a' to N equal Ʃ<a'|α>? maybe this is simple and I just can't see it?
 
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Check (1.2.12), which is a fundamental property of the inner product. That property holds even when the bras and kets correspond to different bases.
 


fzero said:
Check (1.2.12), which is a fundamental property of the inner product. That property holds even when the bras and kets correspond to different bases.

So, is it in this particular case that they are equal because we are considering the eigenkets of A, a hermitian operator? Because these are the eigenkets of A, does that mean that all operators on it are real? Even though it is the operator <α| acting on it and not A?
 


Questioneer said:
So, is it in this particular case that they are equal because we are considering the eigenkets of A, a hermitian operator? Because these are the eigenkets of A, does that mean that all operators on it are real? Even though it is the operator <α| acting on it and not A?

There's no assumption here that \langle \alpha | a&#039;\rangle is real, just that \langle \alpha | a&#039;\rangle = \langle a&#039;| \alpha\rangle^*. That is why the absolute value appears in the formula.
 
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