Is the Expectation Value Always 1 for Normalized State Vectors?

[tetris11]

Homework Statement

|O> = k |R1> + ¹/₉ |R2>

a) Find k if |O> has already been normalized, and b) then the expectation value.

The Attempt at a Solution

a)
To Normalise:
|(|O>)|² = (¹/₉ |R2> + k |R1>).(¹/₉ |R2> - k|R1>) = ¹/₈₁|R2>² - k²|R1>² = 1

I just assumed that |k| = (1-(¹/₈₁))^0.5, but I can't justify why. Is this answer correct?

b)
Expectation value:
Operator A of the system D

<D|A|D> = ∫D*AD dt = is the expectation value, where:

A|n> = λ_n|n>
|D> = Σ C_n |Q_n>
<D| = Σ C_n^* <Q_n|

So: <D|A|D> = Σ |C_n|² λ_n

So in the case of O:
<O|A|O> = ΣΣ [(¹/₉<R2| - k<R1|).A.(¹/₉|R2> + k|R1>)]
= ΣΣ[(¹/₈₁<R2|R2> + ^k/₉<R2|R1> - ^k/₉<R1|R2> - k²<R1|R1>)].A
= ? Help

 

[arkajad]

You need to read somewhere how to calculate the norm of a state. Read and try to understand. If you have problems at this point - you will never be able to understand anything that follows. Find it on the net, in textbooks, in your notes.

 

[tetris11]

Ok, so I'm fairly sure that the value of k = \sqrt{1 - (1/9)^2} = \sqrt{80/81}
Tricky part now is the expectation value:<A> = \sum |C_n|² a_n = |C₁|² a1 + |C₂|² a2

I'm told that the corresponding eigenvalues are +1 and -1.

So <A>= |1/9|² (+1) - |(80/81|(-1) = 81/81 = 1 ?

So for a normalised state vector, the expectation value will always be 1 for these eigenvalues? Or have I done something wrong?