Difference between revisions of "Elong-14-04-16"

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(Deuteron Polarization)
(Deuteron Polarization)
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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>.
 
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>.
  
[[Image:2014-08-21-Pzz-vs-m0.png]]
+
[[Image:2014-08-21-Pzz-vs-m0.png|600px]]
  
 
Or, in terms of an overly-simplistic animation,
 
Or, in terms of an overly-simplistic animation,
  
[[Image:2014-08-22-Pzz-animation-food.gif]]
+
[[Image:2014-08-22-Pzz-animation-food.gif|300px]]
  
 
=Deuteron Shape=
 
=Deuteron Shape=

Revision as of 11:14, 22 August 2014

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>.

2014-08-21-Pzz-vs-m0.png

Or, in terms of an overly-simplistic animation,

2014-08-22-Pzz-animation-food.gif

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
2014-04-16-vector.gif 2014-04-16-tensor.gif


2014-06-23-deuteron-pol.gif

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>

2014-04-16-tensor-vector-plot.png

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.

2014-04-24-deuteron-wavefunction-s-d-states.png

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


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