Elong-14-04-16

Deuteron Shape

From the video made by S.C. Pieper, et al., I extracted the tensor and vector polarization frames and made repeating videos of each. When we vector-polarize or tensor-polarize, the probability densities for the deuteron look like:

m_s=$pm;1	m_s=0

Tensor Polarization Relation to Vector Polarization

Tensor polarization is related to the vector polarization by <math>P_z=2-\sqrt{4-3P_{zz}^2}</math>

Deuteron States

From basic quantum mechanics, we know that the possible states for 2 nucleons are

the isospin singlet with S=0:

<math>|\uparrow \downarrow> - |\downarrow \uparrow></math> with <math>m_s=0</math>, <math>L=1</math>

and the isospin triplet with S=1:

<math>|\uparrow \uparrow></math> with <math>m_s=+1</math>, <math>L=0</math>

<math>|\uparrow \downarrow> + |\downarrow \uparrow></math> with <math>m_s=0</math>, <math>L=1</math>

<math>|\downarrow \downarrow></math> with <math>m_s=-1</math>, <math>L=2</math>

For the deuteron, <math>J=1</math> and <math>P=+1</math>. This kills both of the <math>m_s=0</math> states, since they cannot simultaneously have <math>J=1</math> and <math>P=+1</math> since <math>P=(-1)^L</math>.

This leaves only two possible states:

<math>|\uparrow \uparrow></math> with <math>m_s=+1</math>, <math>L=0</math>

<math>|\downarrow \downarrow></math> with <math>m_s=-1</math>, <math>L=2</math>

Angular Momentum Analysis

Now, let's look at each of these a bit more indepth according to their angular momentum components

J	<math>m_J</math>	L	<math>m_L</math>	S	<math>m_S</math>	<math>\pi = -1^L</math>	T	<math>^{2S+1}L_J</math>
0	0	0	0	1	0	+	1	<math>^1 S_0</math>
1	±1	0	0	1	±1	+	0	<math>^3 S_1</math>
1	1	2	1	1	0	-	0	<math>^1 P_1</math>
1	0	2	-1	1	1	-	1	<math>^3 P_0</math>
1	0	2	1	1	-1	-	1	<math>^3 P_0</math>
1	1	2	1	1	0	-	1	<math>^3 P_1</math>
1	2	2	1	1	1	-	1	<math>^3 P_2</math>
1	2	2	2	1	0	+	1	<math>^1 D_2</math>
1	1	2	2	1	-1	+	0	<math>^3 D_1</math>
1	-1	2	-2	1	+1	+	0	<math>^3 D_1</math>
3	3	2	2	1	1	+	0	<math>^3 D_2</math>
3	3	2	2	1	1	+	0	<math>^3 D_3</math>

When we remove all of the states that can't exist, we're left with two mentioned in the last section:

J	<math>m_J</math>	L	<math>m_L</math>	S	<math>m_S</math>	<math>\pi = -1^L</math>	T	<math>^{2S+1}L_J</math>
1	±1	0	0	1	±1	+	0	<math>^3 S_1</math>
1	1	2	2	1	-1	+	0	<math>^3 D_1</math>
1	-1	2	-2	1	+1	+	0	<math>^3 D_1</math>

S and D States

A variety of models have looked at the wavefunction from the S- and D-states.

File:2014-04-24-deuteron-wavefunction-s-d-states.pdf

--E. Long 19:53, 24 April 2014 (UTC)