Assume that you have a quantum mechanical observable N with eigenvalues W; and corresponding eigenvectors |w;). Show that Vn?) – (N)2 = E P(w,) (a) – (1))* where P(w;) is the probability of measuring the eigenvalue w¿ for some quantum state.
Assume that you have a quantum mechanical observable N with eigenvalues W; and corresponding eigenvectors |w;). Show that Vn?) – (N)2 = E P(w,) (a) – (1))* where P(w;) is the probability of measuring the eigenvalue w¿ for some quantum state.
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![Assume that you have a quantum mechanical observable \( \Omega \) with eigenvalues \( \omega_i \) and corresponding eigenvectors \( |\omega_i\rangle \). Show that
\[
\sqrt{\langle \Omega^2 \rangle - \langle \Omega \rangle^2} = \sqrt{\sum_j P(\omega_j) (\omega_j - \langle \Omega \rangle)^2}
\]
where \( P(\omega_j) \) is the probability of measuring the eigenvalue \( \omega_i \) for some quantum state.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbbeeb476-a64f-4459-b8af-acf5dbdffec4%2F817fd2cf-d1d0-4bc9-a1d7-67b6324bf8a9%2Fboz2a1e_processed.png&w=3840&q=75)
Transcribed Image Text:Assume that you have a quantum mechanical observable \( \Omega \) with eigenvalues \( \omega_i \) and corresponding eigenvectors \( |\omega_i\rangle \). Show that
\[
\sqrt{\langle \Omega^2 \rangle - \langle \Omega \rangle^2} = \sqrt{\sum_j P(\omega_j) (\omega_j - \langle \Omega \rangle)^2}
\]
where \( P(\omega_j) \) is the probability of measuring the eigenvalue \( \omega_i \) for some quantum state.
Expert Solution
Step 1
The average value of a function (f) of a variable (j) is the sum of all the products of the individual values (f (j)) of the function and their probabilities (P (j)).
It may be expressed as follows:
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