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+---
+title: "Propagator"
+date: 2021-07-04
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+In quantum mechanics, the **propagator** $K(x_f, t_f; x_i, t_i)$
+gives the probability amplitude that a particle
+starting at $x_i$ at $t_i$ ends up at position $x_f$ at $t_f$.
+It is defined as follows:
+
+$$\begin{aligned}
+    \boxed{
+        K(x_f, t_f; x_i, t_i)
+        \equiv \matrixel{x_f}{\hat{U}(t_f, t_i)}{x_i}
+    }
+\end{aligned}$$
+
+Where $\hat{U} \equiv \exp(- i t \hat{H} / \hbar)$ is the time-evolution operator.
+The probability that a particle travels
+from $(x_i, t_i)$ to $(x_f, t_f)$ is then given by:
+
+$$\begin{aligned}
+    P
+    &= \big| K(x_f, t_f; x_i, t_i) \big|^2
+\end{aligned}$$
+
+Given a general (i.e. non-collapsed) initial state $\psi_i(x) \equiv \psi(x, t_i)$,
+we must integrate over $x_i$:
+
+$$\begin{aligned}
+    P
+    &= \bigg| \int_{-\infty}^\infty K(x_f, t_f; x_i, t_i) \: \psi_i(x_i) \dd{x_i} \bigg|^2
+\end{aligned}$$
+
+And if the final state $\psi_f(x) \equiv \psi(x, t_f)$
+is not a basis vector either, then we integrate twice:
+
+$$\begin{aligned}
+    P
+    &= \bigg| \iint_{-\infty}^\infty \psi_f^*(x_f) \: K(x_f, t_f; x_i, t_i) \: \psi_i(x_i) \dd{x_i} \dd{x_f} \bigg|^2
+\end{aligned}$$
+
+Given a $\psi_i(x)$, the propagator can also be used
+to find the full final wave function:
+
+$$\begin{aligned}
+    \boxed{
+        \psi(x_f, t_f)
+        = \int_{-\infty}^\infty \psi_i(x_i) K(x_f, t_f; x_i, t_i) \:dx_i
+    }
+\end{aligned}$$
+
+Sometimes the name "propagator" is also used to refer to
+the [fundamental solution](/know/concept/fundamental-solution/) $G$
+of the time-dependent Schrödinger equation,
+which is related to $K$ by:
+
+$$\begin{aligned}
+    \boxed{
+        G(x_f, t_f; x_i, t_i)
+        = - \frac{i}{\hbar} \: \Theta(t_f - t_i) \: K(x_f, t_f; x_i, t_i)
+    }
+\end{aligned}$$
+
+Where $\Theta(t)$ is the [Heaviside step function](/know/concept/heaviside-step-function/).