Solving Projection Operator Questions - QM Basics

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SUMMARY

The discussion centers on the properties of projection operators in quantum mechanics (QM), specifically addressing the invertibility of the operator I + P, where P is a projection operator. It is established that I + P is invertible because it is diagonalizable, and its inverse can be computed using the diagonal matrix representation. Additionally, the physical meaning of a projection operator is clarified, indicating that it provides the expected value of observing a quantum state, as defined by the Born rule.

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  • Understanding of quantum mechanics fundamentals
  • Familiarity with linear algebra concepts, particularly diagonalization
  • Knowledge of the Born rule in quantum mechanics
  • Basic grasp of operators in quantum theory
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  • Study the properties of diagonalizable matrices in linear algebra
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Students and researchers in quantum mechanics, physicists interested in operator theory, and anyone seeking to understand the mathematical foundations of projection operators.

White_M
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Hello,

Suppose P is a projection operator.
How can I show that I+P is inertible and find (I+P)^-1?
And is there a phisical meaning to a projection operator?

(Please be patient I have just started with QM).

Thanks.
Y.
 
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Well I + |u><u| is diagonalizeable, hence invertable. If A is diagonalizeable it is of the form P(-1)DP where D is a diagonal matrix and hence easily and obviously invertable (eigenvalues non zero). Simply take the inverse to get P(-1)D(-1)P.

|u><u| is a projection operator and represents an operator that as an observable gives the expected value of observing a state to determine if its in state |u> - with 1 if it is in that state and zero otherwise ie the expected number of times it gives a true result.

This follows directly from the Born rule, which is the expected value of observing a system in state P with observable O is trace(OP).

Actually a projection operator in general form is Ʃ |ui><ui|, but I will leave you to do the grunt work of generalizing it - it's a good exercise.

Thanks
Bill
 
Last edited:

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