The Momentum Operator

We determine the momentum operator by requiring that, when we operate with \bgroup\color{black}$p_x^{(op)}$\egroup on \bgroup\color{black}$u_{p0}(x,t)$\egroup, we get \bgroup\color{black}$p_0$\egroup times the same wave function.


This means that for these definite momentum states, multiplying by \bgroup\color{black}$p_x^{(op)}$\egroup is the same as multiplying by the variable \bgroup\color{black}$p$\egroup. We find that this is true for the following momentum operator.
\bgroup\color{black}$\displaystyle p^{(op)}={\hbar\over i}{\partial\over\partial x}$\egroup
We can verify that this works by explicit calculation.

If we take our momentum operator and act on a arbitrary state,

=\int\limits_{-\infty}^{\infty}\phi(p) p u_p(x,t) dp\egroup\end{displaymath}

it gives us the right \bgroup\color{black}$p$\egroup for each term in the integral. This will allow us to compute expectation values for any variable we can represent by an operator.

Jim Branson 2013-04-22