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Both $ \frac{1}{\sqrt{2}} (np - pn)$ and $\frac{1}{\sqrt{2}} (np + pn)$ has third component of isospin equal to 0. But the first combination has total isospin equal to 0 while the second combination has total isospin equal to 1. The third component equal to 0 is possible for both 0 and 1 total isospin. So, how do we decide which combination has total isospin equal to 1 and which has it equal to 0?

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To answer this question, you need to make decomposition of quantum states. Lets define protons has total isospin 1/2 and projection 1/2: $|1/2, 1/2>$, while neutron has the same isospin but projection -1/2: $|1/2, -1/2>$. Lets combine all four options. To help with it, we use Clebsch-Gordon coeffiicients for (1/2, 1/2) states: https://en.wikipedia.org/wiki/Table_of_Clebsch%E2%80%93Gordan_coefficients#.C2.A0j1.3D1.2F2.2C_j2.3D1.2F2

  1. $p+p \to |1/2, 1/2>|1/2, 1/2> = |1, 1>$
  2. $p+n \to |1/2, 1/2>|1/2, -1/2> = 1/2|1 0> + 1/2|0 0>$
  3. $n+p \to |1/2, -1/2>|1/2, 1/2> = 1/2|1 0> - 1/2|0 0>$
  4. $n+n \to |1/2, -1/2>|1/2, -1/2> = |1, -1>$

States, which have total isospin 1 form triplet, while those with 0 forms singlet. SO if you want to find composition of triplet state $|1 0>$, you add 2. and 3. together:

$(1/2|1 0> + 1/2|0 0>) + (1/2|1 0> - 1/2|0 0>) = 2 * 1/2 |1 0> = pn + np$

and therefore (please be aware that all 1/2 coefficients should be rooted like in wikipedia lin, but for simplicity root sign is often omitted, so it should look $2\sqrt{1/2}|1 0>$)

$|1 0> = \frac{1}{\sqrt2}(pn + np)$

If you subtract 3. from 2., you will find decomposition of singlet state being $|0 0> = \frac{1}{\sqrt2}(pn - np)$

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There are 4 combinations of 2 nucleons. These are pp, np, pn, nn. It is possible to see that pp, nn are both symmetrical whilst the other two have not got a defined symmetry.

It is possible to produce 3 symetric combinations, which will form an spin=1 triplet: pp, $\frac{1}{2}(np+pn)$, nn These have the Isospin values 1,0,-1.

A single antisymetric combination is also possible with spin=0: $\frac{1}{2}(pn-np)$ As this forms an isospin singlet it also has a isospin value of 0.

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