Quantum Mechanics - quick question about probability

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SUMMARY

The discussion centers on calculating the probability of measuring an eigenvalue (d) in a quantum state (|k>). The correct equation for determining this probability is given by the expression ||^2, where |d> represents the normalized eigenstate corresponding to the eigenvalue d. Additionally, the average value of the observable D when measuring many identically prepared systems is expressed as . The inner product serves as the expansion coefficient, indicating the presence of the eigenstate |d> within the state |k>.

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Quantum - probability in a state
I have an eigenvalue (d) and i need to find the probability of it in a state k.

What is the equation?
<k|d|k> ?

I have spent some thought on this and it seems to simple.

thanks
 
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Note that <k|d|k>=d<k|k>=d.

I'm not sure what the exact question is. I'm guessing that what you have in mind is that the state preparation procedure has put the system in state |k>, that you're going to measure an observable D, and that d is an eigenvalue of D with a 1-dimensional eigenspace. Then the probability that the result of the D measurement will be d is ##|\langle d|k\rangle|^2##, where |d> is a normalized eigenstate of D with eigenvalue d.

<k|D|k> is the average value you will get if you do this measurement on many identically prepared systems.
 
If you have a state |k> then you want to take the inner product with the eigenbra <d|

If you expand |k> in d eigens then
|k> = <d|k>|d>
(Sum over d)

So <d|k> is the expansion coefficient for |d> or how much |d> is in |k> eg, it is the 'probability' of |k> as being in the eigenstate |d>

Taking the inner product with your chosen <d'| gives

<d'|k> = <d|k><d'|d>
<d'|k> = <d'|k>
(Sum over d)

This happens since the eigenstates are orthogonal (I'm assuming they're orthogonal)
The probability is then |<d'|k>|^2

Do you understand what I have done here?
 

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