Expectation value

The expectation value of a quantum-mechanical operator \hat{O} is the weighted average value of its observable, and is defined as:

\langle O\rangle=\langle\psi\vert\hat{O}\vert\psi\rangle\; \; \; \; \; \; \; \; \; \; \; 34

The above equation has roots in probability theory, where the expectation value or expected value of an observable O is \langle O\rangle=\sum_{i=1}^{N}p_io_i, with p_i being the probability of observing the outcome o_i.

\langle O\rangle=\sum_{i=1}^{N}p_io_i=\sum_{i=1}^{N}\left | \langle\phi_i\vert\psi \rangle\right |^{2}o_i=\sum_{i=1}^{N} \langle\phi_i\vert\psi \rangle^{*}\langle\phi_i\vert\psi \rangle o_i

=\sum_{i=1}^{N} \langle\psi\vert\phi_i \rangle\langle\phi_i\vert\psi \rangle o_i=\langle\psi\vert(\sum_{i=1}^{N}o_i\vert\phi_i \rangle\langle\phi_i)\vert\psi \rangle

From eq30, \hat{O}=\sum_{i=1}^{N}o_i\vert\phi_i \rangle\langle\phi_i\vert, and so

\langle O\rangle=\langle\psi\vert\hat{O}\vert\psi \rangle

We further postulate that eq34 is valid for an infinite dimensional Hilbert space.



Why is p_i=\vert\langle\phi_i\vert\psi\rangle\vert^{2} ?


Consider an operator with a complete set of orthonormal basis eigenfunctions \left \{ \phi_i \right \}. So, any eigenfunction \psi can be written as a linear combination of these basis eigenfunctions, i.e. \psi=\sum_{i=1}^{N}c_i\phi_i. According to the Born rule, the probability that a measurement will yield a given result is \vert\psi\vert^{2}=\psi^{*}\psi, where \int\psi^{*}\psi\: d \tau=1. So,

\int\psi^{*}\psi\: d \tau=\int \sum_{i=1}^{N}c_{i}^{*}\phi_{i}^{*}\sum_{i=1}^{N}c_{i}\phi_{i}d\tau=\sum_{i=1}^{N}\vert c_{i}\vert^{2}\int \phi_{i}^{*}\phi_i d\tau=\sum_{i=1}^{N}\vert c_{i}\vert^{2}=1

We have used the orthonormal property of  in the 2nd and 3rd equalities. \vert c_i\vert^2 is interpreted as the probability that a measurement of a system will yield an eigenvalue associated with the eigenfunction . Therefore,

\vert\langle\phi_i\vert\psi\rangle\vert^{2}=\vert\langle\phi_i\vert\sum_j c_j\phi_j\rangle\vert^{2}=\vert\langle\phi_i\vert\ c_i\phi_i\rangle\vert^{2}=\vert c_i\vert^{2}



Using the Schrodinger equation, show that the expectation value of the Hamiltonian is E=\int \psi_i^{*}\hat{H}\psi_i\: d\tau.


Multiplying both sides of eq40 on the left by \psi_i^{*} and integrating over all space, we have

E=\frac{\int \psi_i^{*}\hat{H}\psi_i\: d\tau}{\int \psi_i^{*}\psi_i\: d\tau}

If the wavefunction is normalized, the above equation becomes E=\int \psi_i^{*}\hat{H}\psi_i\: d\tau.



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