In triangle $ABC,$ $\angle A,$ $\angle B,$ and $\angle C$ form an arithmetic sequence.  The length of the altitude from $B$ to $\overline{AC}$ is equal to $AB - BC.$  Find all possible values of
\[\sin \frac{C - A}{2}.\]Enter all possible values, separated by commas.
Since $\angle A,$ $\angle B,$ $\angle C$ form an arithmetic sequence, $2 \angle B = \angle A + \angle C.$  Then
\[3 \angle B = \angle A + \angle B + \angle C = 180^\circ,\]which means $\angle B = 60^\circ,$ and $\angle A + \angle C = 120^\circ.$

Let $h$ be the altitude from $B.$

[asy]
unitsize (1 cm);

pair A, B, C, D;

A = (0,0);
B = 5*dir(40);
C = (5,0);
D = (B.x,0);

draw(A--B--C--cycle);
draw(B--D);

label("$A$", A, SW);
label("$B$", B, N);
label("$C$", C, SE);
label("$h$", (B + D)/2, E);
[/asy]

Then
\[h = AB - BC = \frac{h}{\sin A} - \frac{h}{\sin C},\]so $1 = \frac{1}{\sin A} - \frac{1}{\sin C}.$  Hence,
\[\sin C - \sin A = \sin A \sin C.\]We can write this as
\[2 \sin \frac{C - A}{2} \cos \frac{A + C}{2} = \frac{\cos (A - C) - \cos (A + C)}{2}.\]Since $A + C = 120^\circ,$
\[\sin \frac{C - A}{2} = \frac{\cos (A - C) + \frac{1}{2}}{2} = \frac{\cos (C - A) + \frac{1}{2}}{2}.\]Then
\begin{align*}
4 \sin \frac{C - A}{2} &= 2 \cos (C - A) + 1 \\
&= 2 \left( 1 - 2 \sin^2 \frac{C - A}{2} \right) + 1 \\
&= 3 - 4 \sin^2 \frac{C - A}{2},
\end{align*}so
\[4 \sin^2 \frac{C - A}{2} + 4 \sin \frac{C - A}{2} - 3 = 0.\]This factors as
\[\left( 2 \sin \frac{C - A}{2} - 1 \right) \left( 2 \sin \frac{C - A}{2} + 3 \right) = 0.\]Thus, the only possible value of $\sin \frac{C - A}{2}$ is $\boxed{\frac{1}{2}}.$