Is I an observable?

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So, hermitian linear operators represent observables in QM. I (a matrix whose elements are all 1) is certainly a hermitian linear operator. Does this mean that I represent a measurable property? If so, what do we call that property? Identity? Moreover, for any state-vector A, A would be an eigenvector of I with the eigenvalue of 1. What does this all mean? What are the physical meaning of I as an observable (if it is) and its eigenvectors and the eigenvalue? How can we 'measure' I to get the value 1?
 

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  • #2
PeterDonis
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I (a matrix whose elements are all 1)
The symbol ##I## is usually used to denote the identity matrix (which has 1's all along the diagonal and 0's elsewhere). Is that what you meant? Or did you actually mean a matrix with every single element (off diagonal as well as on) 1?

I suspect you mean the identity matrix, since you say this:

for any state-vector A, A would be an eigenvector of I with the eigenvalue of 1.
Which is true for the identity matrix, but false for a matrix with all elements 1.
 
  • #3
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Interesting question... so the eigenvalues are 1. That means the result of any measurement is 1. And 1 commutes with any Hamiltonian, so it is conserved in any system. The expectation value is also 1, so the average value of this observable is one.

So if you have a black box that, whenever you apply it to any system, it gives you back a 1. That is the measurement. Not a very interesting black box, I might add, since no matter what you do with it it gives you back the same result.
 
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PeterDonis
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Not a very interesting black box, I might add, since no matter what you do with it it gives you back the same result.
Yes. The identity matrix is the mathematical description of the physical operation "do nothing at all". Which just gives you back whatever state you hand it, multiplied by the eigenvalue ##1##, i.e., the same state.
 

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