Expectation value (quantum mechanics)

The expectation value of a quantum-mechanical operator $\hat{O}$ is the weighted average value of its observable. It 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.

Question

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

Answer

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 2^ndand 3^rd 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,

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

Question

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

Answer

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

Expectation value (quantum mechanics)

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