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I am reading Modern Quantum Mechanics by Sakuria and Napolitano.


Background Information from the Textbook

a' and a'' are eigenvalues of A.

A is a Hermitian operator.

The symbol, * , implies complex conjugation.

|a'> is an eigenket of A.


The reality condition is proved by assuming that a' = a''. In other words, the two eigenvalues are the same.


Then the authors start proving the orthogonality property by assuming that a' is no longer equal to a''. The eignevalues are assumed to be different now.


More Background Information from the Textbook (bottom of page 17)

Consider the following Equations.

A |a'> = a'|a'> (Equation 1)

< a''|A = a''* < a''| (Equation 2)

Multiply both sides of Equation 1 by < a''| on the left.

Multiply both sides of Equation 2 by |a'> on the right.

Subtract the Equation 2 from Equation 1.

The following expression is obtained.

(a' - a''*) < a'' | a'> = 0 (Equation 3)


After deriving Equation 3, the authors proceed to choose whether a' or a'' are the same or not.

The eigenvalues are the same when proving the reality condition.

The eigenvalues are different when proving the orthogonality property.

Is there a contradiction? On page 18, the authors prove the orthogonality property "because of the just-proved reality condition."

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    $\begingroup$ Questions should be self contained; please show the definitions of all symbols. $\endgroup$ – knzhou May 5 '17 at 23:35
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There is no contradiction. The equation

$$ (a' - a^{*''}) \langle a' | a'' \rangle =0 $$

should be used in the following way. Take $| a' \rangle = | a'' \rangle$, thus you prove reality $a' = a^{*''} = a^{'*}$. It follows because same eigenstate implies same eigenvalue.

This step assumes only that the norm of $| a' \rangle$ is non-vanishing.

Now take $|a' \rangle \neq | a^{''} \rangle$. This implies -- for non-degenerate spectrum -- that $a' \neq a''$. Nonetheless, you now know that $a^{*''} = a^{''}$ by the first argument. This only tells you that the difference $(a' - a^{''})$ is non-vanshing. Therefore, $\langle a^{'} | a^{''} \rangle = 0$ and the states are orthogonal.

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  • $\begingroup$ Thank you! That answers my question. I had previously misunderstood the implication of the phrase "Because of the just-proved reality condition" on page 18. $\endgroup$ – Math12345 May 6 '17 at 1:51
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Now I realize: The reason why the reality condition is used to prove the orthogonality property is that one needs to consider the case for which the eigenvalues are the same before considering the case when the eigenvalues are different, in order to eliminate the possibility of (a' - a''*) being 0. This leads to the necessity of the inner product being zero.

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