# Why $2j+1$ number of states?

In this statement from Modern Quantum Mechanics by J.J. Sakurai:

If $j$ is an integer, all $m$ values are integers; if $j$ is a half-integer, all $m$ values are half-integers. The allowed $m$-values for a given $j$ are $$m = \underbrace{-j,-j+1,\ldots,j-1,j}_{2j+1 {~\rm{states}}}$$

It says that $m$ will have a total of $2j+1$ states. I do not see this, however. Perhaps it is obvious, but could someone explain or show me why if $m$ goes from $-j,\ldots,j$ it will give $2j+1$ number of states?

The $2j$ is from the positive and negative j values, and the additional $+1$ accounts for $j=0$.
• It is not correct for half integers, which has no $m=0$ Dec 27, 2014 at 5:33
• @hwlau the reasoning doesn't work for half-integers but the expression $2j+1$ is still valid. Dec 27, 2014 at 6:36
• @DavidZ Is it just coincidence that $\frac{d}{dj}(j(j+1)) = 2j+1$. I mean is there any connection between $2j+1$ and the eigenvalue of $J^2$ Dec 27, 2014 at 9:11