0
$\begingroup$

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?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
5
$\begingroup$

The $2j$ is from the positive and negative j values, and the additional $+1$ accounts for $j=0$.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ It is not correct for half integers, which has no $m=0$ $\endgroup$
    – unsym
    Commented Dec 27, 2014 at 5:33
  • 1
    $\begingroup$ @hwlau the reasoning doesn't work for half-integers but the expression $2j+1$ is still valid. $\endgroup$
    – David Z
    Commented Dec 27, 2014 at 6:36
  • $\begingroup$ True. I was trying to be instructive by considering the simple case, but my terseness could have been more precise I suppose. $\endgroup$
    – kbh
    Commented Dec 27, 2014 at 6:49
  • $\begingroup$ @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 $J^2$ $\endgroup$
    – Hubble07
    Commented Dec 27, 2014 at 9:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.