Momentum Operator

Quantum mechanical wave functions are analogous to electromagnetic radiation waves in many aspects. Much like how waves define the properties of electromagnetic radiation, wave functions can also define a system's properties. To find the momentum of a system, we must use the momentum operator, a Hermitian operator which returns the momentum of the system like so: $\hat{p} \psi= \mathbf{p} \psi$ where $\psi$ is the wave function of the system, $\hat{p}$ is the momentum operator, and $\mathbf{p}$ is the momentum. When deriving the momentum operator, it is helpful to recall that the momentum of a photon is related directly to its wavenumber, $k$, by the coefficient $\hbar$, the reduced Planck constant:

\begin{equation}
\mathbf{p}= \hbar\mathbf{k}
\end{equation}

The momentum of a quantum mechanical system is actually given by the same formula, so it can be said of the momentum operator that $\hat{p}\phi \left(\mathbf{k}\right) = \hbar{\mathbf{k}} \phi\left(\mathbf{k} \right)$ if $\phi\left( \mathbf{k} \right)$ is the system's wave function in momentum space. But say we had the position space wave function, $\psi(x)$ (let's work only in the $x$ dimension for now), for which we wanted to find the momentum. To do this, we need to derive the momentum operator in position space whose $x$-component is defined as $\hat{p}_x\psi(x) = p_x\psi(x)$. Recall the Fourier transform equation for transforming from position to momentum space:

\begin{equation}
\phi(k_x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \psi(x)e^{-ik_xx} dx
\end{equation}

We know that applying the $x$-component momentum operator to both sides will give this:

\begin{equation}
\hat{p}_x\phi(k_x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \hbar{k_x}\psi(x) e^{-ik_xx} dx
\end{equation}

Integrating by parts,

\begin{equation}
\hat{p}_x\phi(k_x) = \frac{1}{\sqrt{2\pi}} \left[\frac{\hbar k_x}{-ik_x} \psi(x) e^{-ik_xx}- \int \frac{\hbar k_x}{-ik_x} \frac{\partial \psi(x)}{\partial x} e^{-ik_xx} dx\right]_{x =-\infty}^{x =\infty}
\end{equation}

Because $\lim_{x \to \pm\infty} \psi(x)=0$,

\begin{equation}
\hat{p}_x\phi(k_x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \frac{\hbar}{i} \frac{\partial \psi(x)}{\partial x}e^{-ik_xx} dx
\end{equation}

Comparing this to equation $(3)$, we can complete the derivation of the $x$-component momentum operator in position space:

\begin{equation}
\hat{p}_x = -i\hbar \frac{\partial}{\partial x}
\end{equation}

(If you didn't know before, $\frac{1}{i} = \frac{i}{i^2} = -i$.) Similarly, for the $z$- and $y$-components, we can prove:

\begin{equation}
\hat{p}_y = -i\hbar \frac{\partial}{\partial y}
\end{equation}
\begin{equation}
\hat{p}_z = -i\hbar \frac{\partial}{\partial z}
\end{equation}

Furthermore, because $\hat{p} = \mathbf{x}\hat{p}_x + \mathbf{y}\hat{p}_y + \mathbf{z}\hat{p}_z$ where $\mathbf{x}$, $\mathbf{y}$, and $\mathbf{z}$ are unit vectors in the $x$, $y$, and $z$ directions respectively, we can write $\hat{p}$ in terms of the del operator, $\nabla = \mathbf{x} \frac{\partial}{\partial x} + \mathbf{y} \frac{\partial}{\partial y} + \mathbf{z} \frac{\partial}{\partial z}$:

\begin{equation}
\hat{p} = -i\hbar\nabla
\end{equation}