# Half integer orbital momentum, following Griffiths

1. Oct 7, 2011

### Tomer

Hello everyone, and thanks for reading.

I'm having a difficult time understanding something.
On yet another attempt to deepen my quantum mechanics understanding I referred to the widely recommended book of Griffiths.
I find the book indeed very good and pretty thorough in it's scope. There is one thing however I was unable to get from the book.
Griffiths develops the quantum number "l" in two different ways: In the first way - the "ugly" way, "l"'s origin is being a separation variable (more accurately, l(l+1) ) between the radial and angular expressions retrieved from the spherical Hamiltonian while looking for separable solutions. In this development however he concludes that l has to be a whole integer because of the cyclic nature $\phi$ imposes. ($e^{i\phi} = e^{i\phi + 2\pi})$.

Later, he develops the angular momentum operators which carry their "l"'s as their eigenvalues, and with help of ladder operators shows eventually the equivalence of this l with the last l.
However, in this way, he concludes that l isn't nessecarily whole, and that also half-integers are allowed.

What he doesn't bother to explain, is how come we didn't get these values from the first technique? How come we get a condition that forces that l's to be whole? And how do the half integers not contradict the cyclic nature of $\phi$?

Thanks a lot!

Tomer.

2. Oct 7, 2011

### dextercioby

Well, essentially it boils down to what the small <l> stands for. In technical terms, it's the related to the spectral value of a linear operator which represents the square of the ORBITAL angular momentum operator, denoted by $\displaystyle{\hat{\vec{L}}}$ . Since we know that

$$\hat{\vec{L}} = \hat{\vec{r}} \times \hat{\vec{p}}$$

and we're asking for L's components and for L^2 to be self-adjoint, solving the spectral problem for L^2 (= solving 2 Sturm-Liouville ODE's derived from a linear PDE) we're forced to conclude that <l> must be a positive integer and <m> be linked to <l> in a specific way (<m> is the spectral value of $\displaystyle{\hat{L}}_z$).

When people talk in a general manner about angular momentum, not necessarily orbital one, they should use a different notation, namely J and <j>, respectively.

The general theory of J makes <j> to be either positive and integer, or positive and semi-integer. In a way, the set of all <l>'s in included in the set of all <j>'s.

3. Oct 8, 2011

### Tomer

Alright, so orbital angular momentum has to be a whole integer, and the parallel, algebraic development shows the possible values of a somewhat "generalized" angular momentum which includes the spin?

4. Oct 8, 2011

### strangerep

The "algebraic" development (i.e., representing the generic rotation group generators as Hermitian operators on a Hilbert space and deriving their spectrum) should be regarded as more fundamental. From this comes the result that angular momentum is quantized to integer and half-integer values.

Adding the extra restriction on the form of the rotation generators, i.e., insisting that $L = r \times p$, has the consequence that only integral values of quantized angular momentum are allowed in that case.

Ballentine explains all this pretty well in his chapter on angular momentum.

5. Oct 9, 2011

### Tomer

Thanks very much for the replies, I think I got it, more or less :-)