Expectation Values

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

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

We can move the \bgroup\color{black}$f(x)$\egroup between just before \bgroup\color{black}$\psi$\egroup anticipating the use of linear operators.

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

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

\bgroup\color{black}$\displaystyle \langle p\rangle_\psi =\langle\psi\vert p\vert\psi\rangle= \int\limits_{-\infty}^\infty\psi^*(x)p^{(op)}\psi(x) dx$\egroup
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 \bgroup\color{black}$v$\egroup, the expectation value of \bgroup\color{black}$v$\egroup in an arbitrary state \bgroup\color{black}$\psi$\egroup is

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

The expectation values of physical quantities should be real.

Gasiorowicz Chapter 3

Griffiths Chapter 1

Cohen-Tannoudji et al. Chapter

Jim Branson 2013-04-22