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I'm studying for my nuclear physics exam and the book we use is Introductory Nuclear Physics by K.S. Krane. In the chapter on Basic Nuclear Structure, we research the deuteron. However, when discussing the total angular momentum of the deuteron, something confuses me (p84 if someone has the book). I'll copy the paragraph:

Spin and Parity

The total angular momentum $I$ of the deuteron should have three components: the individual spins $s_n$ and $s_p$ of the neutron and proton (each equal to $\frac{1}{2}$), and the orbital angular momentum $l$ of the nucleons as they move about their common center of mass: $$I = s_n + s_p + l $$ [...]

The measured spin of the deuteron is $I = 1$ [...]. Since the neutron and proton spins can be either parallel (for a total of $1$) or antiparallel (for a total of zero), there are four ways to couple $s_n$, $s_p$, and $l$ to get a total of $1$:

(a) $s_n$ and $s_p$ parallel with $l = 0$

(b) $s_n$ and $s_p$ antiparallel with $l = 1$

(c) $s_n$ and $s_p$ parallel with $l = 1$

(d) $s_n$ and $s_p$ parallel with $l = 2$

Unfortunately, this is not explained deeper and options (b) and (c) are shown to be invalid because the measured parity is $1$, whereas those options have parity $-1$.

Now, (a), (b) and (d) seem okay with me to get total angular momentum $1$, but I don't understand how we can get to $1$ in (c): $I = 1$ if $l=1$ and $s_n + s_p = 1$. This gives either $0$ or $2$? Or can $I$ take all values between $l-s_n-s_p$ and $l+s_n+s_p$ rather than just those two extremes?

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Take a look here en.wikipedia.org/wiki/Deuterium at the "Approximated wavefunction of the deuteron" section. –  nijankowski Jun 22 '13 at 1:45

2 Answers 2

I think this is just standard angular momentum coupling, not anything specific to nuclear physics. In (c), you're coupling spin 1 to spin 1. There are 9 states, which can be broken down into multiplets with spins 0, 1, and 2.

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As I understand it, by experimental measurements, it is known that the deuteron has:

  • an "even" spatial parity
  • and total nuclear spin = 1 (as it's a boson)

And as parity is characterized by (-1) to the power l, any odd l value like l = 1 gives us an "odd" parity, which is inconsistent with our measurement, so options (b) and (c) must be eliminated.

Further, l = 0 and l = 2 options (which of course give "even" parity), when combined with spin s = 1 , give correct total spin = 1. So deuteron's ground state is a superposition of l = 0 and l = 2 states.

Note, that we have not talked about isospin yet, which ensures that the total wave function describing deuteron(which is a system of two fermions) is anti-symmetric, as required by Pauli-exclusion principle.

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Thanks! will try to use it. –  zixtor Aug 25 '13 at 12:57
@DImension10AbhimanyuPS To save work, you could just post these links in future: basics, reference –  Larry Harson Aug 25 '13 at 13:45
@LarryHarson: No, I used to use the autoreview extension, but as it doesn't work anymore, so I just saved all those commands in a notepad file and post them when needed. –  Dimensio1n0 Aug 25 '13 at 13:47

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