12.5 A warning about angular momentum

Normally, eigen­states are indeterminate by a complex number of magnitude one. If you so desire, you can multiply any normalized eigenstate by a number of unit magnitude of your own choosing, and it is still a normalized eigenstate. It is important to remember that in analytical expressions involving angular momentum, you are not allowed to do this.

As an example, consider a pair of spin 1/2 particles, call them $a$ and $b$, in the “singlet state”, in which their spins cancel and there is no net angular momentum. It was noted in chapter 5.5.6 that this state takes the form

$${\big\vert\:0\big\rangle }_{ab} = \frac{{\big\vert\leave... ...x\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_b}{\sqrt 2}$$

(This section will use kets rather than arrows for spin states.) But if you were allowed to arbitrarily change the definition of say the spin state ${\big\vert\leavevmode\kern.03em \raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /... ...rn-.2em /\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\big\rangle }_a$ by a minus sign, then the minus sign in the singlet state above would turn in a plus sign. The given expression for the singlet state, with its minus sign, is only correct if you use the right normal­ization factors for the individual states.

It all has to do with the ladder operators ${\widehat J}^+$ and ${\widehat J}^-$. They are very convenient for analysis, but to make that easiest, you would like to know exactly what they do to the angular momentum states $\big\vert j\:m\big\rangle$. What you have seen so far is that ${\widehat J}^+\big\vert j\:m\big\rangle$ produces a state with the same square angular momentum, and with angular momentum in the $z$-direction equal to $(m+1)\hbar$. In other words, ${\widehat J}^+\big\vert j\:m\big\rangle$ is some multiple of a suitably normalized eigenstate $\big\vert j\:m{+}1\big\rangle$;

$${\widehat J}^+ \big\vert j\:m\big\rangle = C \big\vert j\:m{+}1\big\rangle$$

where the number $C$ is the multiple. What is that multiple? Well, from the magnitude of ${\widehat J}^+\big\vert j\:m\big\rangle$, derived earlier in (12.6) you know that its square magnitude is

$$\vert C\vert^2 = j(j+1)\hbar^2 - m^2 \hbar^2 - m\hbar^2.$$

But that still leaves $C$ indeterminate by a factor of unit magnitude. Which would be very inconvenient in the analysis of angular momentum.

To resolve this conundrum, restrictions are put on the normal­ization factors of the angular momentum states $\big\vert j\:m\big\rangle$ in ladders. It is required that the normal­ization factors are chosen such that the ladder operator constants are positive real numbers. That really leaves only one normal­ization factor in an entire ladder freely selectable, say the one of the top rung.

Most of the time, this is not a big deal. Only when you start trying to get too clever with angular momentum normal­ization factors, then you want to remember that you cannot really choose them to your own liking.

The good news is that in this convention, you know precisely what the ladder operators do {D.65}:

$${\widehat J}^+ \big\vert j\:m\big\rangle = \hbar \sqrt{j... ...\big) - m\big(1+m\big)} \; \big\vert j\:m{+}1\big\rangle %$$ (12.9)

$${\widehat J}^- \big\vert j\:m\big\rangle = \hbar \sqrt{j... ...\big) + m\big(1-m\big)} \; \big\vert j\:m{-}1\big\rangle %$$ (12.10)