---
tag: 9DYE6V628H
kind: definition
name: Whittaker function
msc: [33C15]
defines: 7EFDMJSVDP
uses: [6AZ4ZBN1M8, 234F5DNAJH]
sources:
  - url: https://en.wikipedia.org/wiki/Whittaker_function
---

## Definition

Let $\kappa$ and $\mu$ be complex numbers, let $z \in \C \setminus (-\infty, 0]$, and use principal branches of powers of $z$.
Let $M(a, b, z)$ be [[Kummer's function|6AZ4ZBN1M8]] and $U(a, b, z)$ the [[Tricomi function|234F5DNAJH]].
If $2\mu \notin \set{-1, -2, -3, \dots}$, the *Whittaker function* $M_{\kappa, \mu}$ is
$$M_{\kappa, \mu}(z) = e^{-z/2} z^{\mu + \frac12}\, M\left( \mu - \kappa + \tfrac12,\ 1 + 2\mu,\ z \right).$$
For all $\kappa, \mu$, the *Whittaker function* $W_{\kappa, \mu}$ is
$$W_{\kappa, \mu}(z) = e^{-z/2} z^{\mu + \frac12}\, U\left( \mu - \kappa + \tfrac12,\ 1 + 2\mu,\ z \right).$$

## Remarks

Both functions solve *Whittaker's equation*
$$\frac{d^2 w}{dz^2} + \left( -\frac14 + \frac{\kappa}{z} + \frac{\frac14 - \mu^2}{z^2} \right) w = 0.$$
This follows from Kummer's equation for $M$ and $U$ by the substitution $w = e^{-z/2} z^{b/2} y$ with $b = 1 + 2\mu$.
Example: for $\kappa = 0$ and $\mu = \frac12$, one has $U(1, 2, z) = z^{-1}$, so $W_{0, 1/2}(z) = e^{-z/2}$.
Check: $w = e^{-z/2}$ gives $w'' = \frac14 w$, and the equation becomes $\left( \frac14 - \frac14 + 0 + 0 \right) w = 0$.
Non-example: for $\kappa = 1$ and $\mu = \frac12$, the function $e^{-z/2}$ is not a solution, since the equation leaves the term $\frac{1}{z} e^{-z/2} \ne 0$.
As $z \to \infty$ with $\abs{\arg z} < \frac{3\pi}{2} - \delta$, one has $W_{\kappa, \mu}(z) \sim e^{-z/2} z^{\kappa}$.
The equation depends on $\mu$ only through $\mu^2$, and indeed $W_{\kappa, -\mu} = W_{\kappa, \mu}$.