Expectation Values

Operators allow us to compute the expectation value of some physics quantity given the wavefunction. If a particle is in the state $\psi(x,t)$, the normal way to compute the expectation value of $f(x)$ is

$$\bgroup\color{black}\langle f(x)\rangle_\psi = \int\limits_{-... ...= \int\limits_{-\infty}^\infty \psi^*(x)\psi(x) f(x) dx.\egroup$$

We can move the $f(x)$ between just before $\psi$ anticipating the use of linear operators.

$$\bgroup\color{black}\langle f(x)\rangle_\psi=\int\limits_{-\infty}^\infty \psi^*(x) f(x) \psi(x) dx\egroup$$

If the variable we wish to compute the expectation value of (like $p$) is not a simple function of $x$, let its operator act on $\psi(x)$. The expectation value of $p$ in the state $\psi$ is

$$\langle p\rangle_\psi =\langle\psi\vert p\vert\psi\rangle= \int\limits_{-\infty}^\infty\psi^*(x)p^{(op)}\psi(x) dx$$

The Dirac Bra-ket notation shown above is a convenient way to represent the expectation value of a variable given some state.

* Example: A particle is in the state $\psi(x)=\left({1\over 2\pi\alpha}\right)^{1/4} e^{ik_0x} e^{-{x^2\over 4\alpha}}$. What is the expectation value of $p$?*

For any physical quantity $v$, the expectation value of $v$ in an arbitrary state $\psi$ is

$$\bgroup\color{black}\langle\psi\vert v\vert\psi\rangle=\int\limits_{-\infty}^\infty\psi^*(x)v^{(op)}\psi(x) dx\egroup$$

The expectation values of physical quantities should be real.

Gasiorowicz Chapter 3

Griffiths Chapter 1

Cohen-Tannoudji et al. Chapter