The Momentum Operator

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

$$\bgroup\color{black}p^{(op)}u_{p0}(x,t)=p_0u_{p0}(x,t)\egroup$$

This means that for these definite momentum states, multiplying by $p_x^{(op)}$ is the same as multiplying by the variable $p$. We find that this is true for the following momentum operator.

$$p^{(op)}={\hbar\over i}{\partial\over\partial x}$$

We can verify that this works by explicit calculation.

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

$$\bgroup\color{black}p^{(op)}\psi(x,t)=p^{(op)}\int\limits_{-\... ...p =\int\limits_{-\infty}^{\infty}\phi(p) p u_p(x,t) dp\egroup$$

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