Half integer orbital momentum, following Griffiths

In summary, the author argues that the angular momentum value <l> must be a whole integer because of the cyclic nature of \phi. Half integers are allowed because the cyclic nature of \phi imposes a requirement that <l> be a whole integer.
  • #1
Tomer
202
0
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 [itex]\phi[/itex] imposes. ([itex] e^{i\phi} = e^{i\phi + 2\pi})[/itex].

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 [itex]\phi[/itex]?

Thanks a lot!

Tomer.
 
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  • #2
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 [itex] \displaystyle{\hat{\vec{L}}}[/itex] . Since we know that

[tex] \hat{\vec{L}} = \hat{\vec{r}} \times \hat{\vec{p}} [/tex]

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 [itex] \displaystyle{\hat{L}}_z [/itex]).

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
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
Tomer said:
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?

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 [itex]L = r \times p[/itex], 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
Thanks very much for the replies, I think I got it, more or less :-)
 

FAQ: Half integer orbital momentum, following Griffiths

What is half integer orbital momentum?

Half integer orbital momentum refers to a property of subatomic particles, specifically electrons, which have a spin quantum number of 1/2. This means that they can have a spin of either +1/2 or -1/2, as opposed to integer spin particles which can only have a spin of +1 or -1.

How is half integer orbital momentum related to angular momentum?

Half integer orbital momentum is a form of angular momentum, specifically orbital angular momentum. This is because the spin of an electron contributes to its overall angular momentum, along with its orbital motion around the nucleus.

What are some examples of particles with half integer orbital momentum?

Electrons, as mentioned before, are the most well-known particles with half integer orbital momentum. However, other subatomic particles such as protons and neutrons also have half integer orbital momentum due to their composite structure and internal quark spin.

How is half integer orbital momentum represented mathematically?

Half integer orbital momentum is represented by the quantum number s, which takes on values of +1/2 or -1/2. It is also represented by the spin operator, which is a mathematical operator that describes the behavior of spin in quantum mechanics.

What are the practical applications of understanding half integer orbital momentum?

Understanding half integer orbital momentum is crucial in many areas of physics, such as quantum mechanics, atomic and molecular physics, and solid state physics. It also has practical applications in technology, such as in the development of magnetic materials and spintronics devices.

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