Elong-14-04-16

Deuteron Polarization

Being a spin-1 particle, the deuteron has three spin sub-states that can be filled: <math>m_J=+1, m_J=0, m_J=-1</math>. For a given quantity of deuterons, the polarization is measured based on the population <math>p_m</math> in each substate. The population is defined as

<math>p_m=\frac{N_m}{N_1+N_0+N_{-1}}</math>

where <math>N_m</math> is the number of deuterons in a given substate. In a system of pure deuterons, the normalization of the populations is

<math>p_1+p_0+p_{-1}=1</math>.

Vector polarization depends only on the <math>m_J=\pm 1</math> populations, such that

<math>P_z = p_1 - p_{-1}~</math>.

It has a range of <math>-1<P_z<1~</math>.

If the <math>m_1</math> state is filled, then <math>p_1=1,~p_{-1}=0,~p_0=0</math> and <math>P_z=1~</math>.

If the <math>m_{-1}</math> state is filled, then <math>p_1=0,~p_{-1}=1,~p_0=0</math> and <math>P_z=-1~</math>.

If the <math>m_0</math> state is filled, then <math>p_1=0,~p_{-1}=0,~p_0=1</math> and <math>P_z=0~</math>.

Tensor polarization depends on the combined <math>m_J=\pm 1</math> populations, as well as the <math>m_J=0</math> population such that

<math>P_{zz} = (p_1 + p_{-1}) - 2p_0~</math>.

It has a range of <math>-2<P_{zz}<1~</math>.

If the <math>m_1</math> and <math>m_{-1}</math> states are filled, then <math>(p_1 + p_{-1})=1,~p_0=0</math> and <math>P_{zz}=1~</math>.

If the combined <math>m_1</math> and <math>m_{-1}</math> states equal the <math>m_0</math> state, then <math>(p_1+p_{-1})=0.5, p_0=0.5<math> and <math>P_{zz}=-0.5~</math>

If the <math>m_0</math> state is filled, then <math>(p_1 + p_{-1})=0,~p_0=1</math> and <math>P_{zz}=-2~</math>.

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=±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)