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(New page: Problem (a) Find the total spin of a syatem of three spin <math>\frac{1}{2}</math> particles and derive the corresponding Clebsch-Gordan Coefficients. (b) Consider a system of three noni...)
 
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We are going to denote the joint eigenstates of <math>\hat{J_{1}}^{2}</math>,<math>\hat{J_{2}}^{2}</math>,<math>\hat{J_{3}}^{2}</math>,<math>\hat{J_{12}}^{2}</math>,<math>\hat{J}^{2}</math> and <math>J_{z}</math> by <math>|j_{12},j,m> </math> and the joint eigenstates of <math>\hat{J_{1}}^{2}</math>,<math>\hat{J_{2}}^{2}</math>,<math>\hat{J_{3}}^{2}</math>,<math>\hat{J_{1_{z}}}^{2}</math>,<math>\hat{J_{2_{z}}}^{2}</math> and <math>\hat{J_{3_{z}}}^{2}</math> by <math>|j_{1},J_{2},j_{3};m_{1},m_{2},m_{3}></math>; since <math>j_{1}=j_{2}=j_{3}=\frac{1}{2}</math> and <math>m_{1}=\pm \frac{1}{2}</math>,<math>m_{2}=\pm \frac{1}{2}</math>,<math>m_{3}=\pm \frac{1}{2}</math>,we will be using throughout this problem the lighter notation <math>|j_{1},j_{2},j_{3};\pm ,\pm,\pm> </math> to abbreviate <math>|\frac{1}{2},\frac{1}{2},\frac{1}{2};\pm \frac{1}{2},\pm \frac{1}{2},\pm \frac{1}{2}> </math>
We are going to denote the joint eigenstates of <math>\hat{J_{1}}^{2}</math>,<math>\hat{J_{2}}^{2}</math>,<math>\hat{J_{3}}^{2}</math>,<math>\hat{J_{12}}^{2}</math>,<math>\hat{J}^{2}</math> and <math>J_{z}</math> by <math>|j_{12},j,m> </math> and the joint eigenstates of <math>\hat{J_{1}}^{2}</math>,<math>\hat{J_{2}}^{2}</math>,<math>\hat{J_{3}}^{2}</math>,<math>\hat{J_{1_{z}}}^{2}</math>,<math>\hat{J_{2_{z}}}^{2}</math> and <math>\hat{J_{3_{z}}}^{2}</math> by <math>|j_{1},J_{2},j_{3};m_{1},m_{2},m_{3}></math>; since <math>j_{1}=j_{2}=j_{3}=\frac{1}{2}</math> and <math>m_{1}=\pm \frac{1}{2}</math>,<math>m_{2}=\pm \frac{1}{2}</math>,<math>m_{3}=\pm \frac{1}{2}</math>,we will be using throughout this problem the lighter notation <math>|j_{1},j_{2},j_{3};\pm ,\pm,\pm> </math> to abbreviate <math>|\frac{1}{2},\frac{1}{2},\frac{1}{2};\pm \frac{1}{2},\pm \frac{1}{2},\pm \frac{1}{2}> </math>


In total there are eight states <math>|j_{12},j,m></math> since <math>(2j_{1}+1)(2j_{2}+1)(2j_{3}+1)=8</math>. Four of these correspond to the subspace
In total there are eight states <math>|j_{12}, j, m></math> since <math>(2j_{1}+1)(2j_{2}+1)(2j_{3}+1)=8</math>. Four of these correspond to the subspace <math>j=\frac{3}{2}:|1,\frac{3}{2},\frac{3}{2}>,|1,\frac{3}{2},\frac{1}{2}>,|1,\frac{3}{2},-\frac{1}{2}>,|1,\frac{3}{2},-\frac{3}{2}></math>. The remaining four belong to the subspace <math>j=\frac{1}{2}:|0,\frac{1}{2},\frac{1}{2}>,|0,\frac{1}{2},-\frac{1}{2}>,|1,\frac{1}{2},\frac{1}{2}>,|1,\frac{1}{2},-\frac{1}{2}></math>. To construct the states <math>|j_{12},j,m> </math> in terms of <math>|j_{1},j_{2},j_{3};\pm,\pm,\pm</math>, we are going to consider the two subspaces <math>j=\frac{3}{2}</math> and <math>j=\frac{1}{2}</math> separately.
 
 
Subspace <math>j=\frac{3}{2}</math>
 
First, the states <math>|1,\frac{3}{2},\frac{3}{2}></math> and <math>|1,\frac{3}{2},-\frac{3}{2}></math> are clearly given by
 
<math>|1,\frac{3}{2},\frac{3}{2}>=|j_{1},j_{2},j_{3};+,+,+> </math>,              <math>|1,\frac{3}{2},-\frac{3}{2}>=|j_{1},j_{2},j_{3};-,-,-> </math>
 
To obtain <math>|1,\frac{3}{2},\frac{1}{2}></math>, we need to apply, on the one hand, <math>\hat{J_{-}}</math> on <math>|1,\frac{3}{2},\frac{3}{2}></math>
 
<math>\hat{J_{-}}|1,\frac{3}{2},\frac{3}{2}>=\hbar\sqrt{\frac{3}{2}\left ( \frac{3}{2}+1 \right )-\frac{3}{2}\left ( \frac{3}{2}-1 \right )}|1,\frac{3}{2},\frac{3}{2}>=\hbar\sqrt{3}|1,\frac{3}{2},\frac{1}{2}></math>
 
and, on the other hand, apply <math>(\hat{J_{1-}}+\hat{J_{2-}}+\hat{J_{3-}})</math> on <math>|j_{1},j_{2},j_{3};+,+,+></math>
 
This yields
<math>(\hat{J_{1-}}+\hat{J_{2-}}+\hat{J_{3-}})|j_{1},j_{2},j_{3};+,+,+>=\hbar\left (| j_{1},j_{2},j_{3};-,+,+>+|j_{1},j_{2},j_{3};+,-,+>+|j_{1},j_{2},j_{3};+,+,-> \right )</math>
 
Since <math>\sqrt{\frac{1}{2}\left ( \frac{1}{2} +1\right )-\frac{1}{2}\left ( \frac{1}{2} -1\right )}=1</math>. Equating above equations, we get
 
<math>|1,\frac{3}{2},\frac{1}{2}>=\frac{1}{\sqrt{3}}\left ( |j_{1},j_{2},j_{3};-,+,+>+|j_{1},j_{2},j_{3};+,-,+>+|j_{1},j_{2},j_{3};+,+,-> \right )</math>
 
Following the same method-applying <math>\hat{J_{-}}</math> on <math>(\hat{J_{1-}}+\hat{J_{2-}}+\hat{J_{3-}})</math> on the right hand side of above equation and then equating the two results-we find
 
<math>|1,\frac{3}{2},-\frac{1}{2}>=\frac{1}{\sqrt{3}}\left ( |j_{1},j_{2},j_{3};+,-,->+|j_{1},j_{2},j_{3};-,+,->+|j_{1},j_{2},j_{3};-,-,+> \right )</math>
 
 
Subspace--j=1/2
 
We can write <math>|0,\frac{1}{2},\frac{1}{2}></math> as a linear combination of <math>|j_{1},j_{2},j_{3};+,+,-> </math> and <math>|j_{1},j_{2},j_{3};-,+,+> </math>
 
<math>|0,\frac{1}{2},\frac{1}{2}>=\alpha |j_{1},j_{2},j_{3};+,+,-> +\beta |j_{1},j_{2},j_{3};-,+,+> </math>
 
Since <math>|0,\frac{1}{2},\frac{1}{2}></math> is normalized, while <math>(\hat{J_{1}}+\hat{J_{2}}+\hat{J_{3}})|j_{1},j_{2},j_{3};+,+,-></math> and <math>(\hat{J_{1}}+\hat{J_{2}}+\hat{J_{3}})|j_{1},j_{2},j_{3};-,+,+></math>, and since Clebsch-Gordan coefficients, such as <math>\alpha </math> and <math>\beta</math>, are real numbers, above equation yields
 
<math>\alpha ^{2}+\beta ^{2}=1</math>
 
On the other hand, since <math><1,\frac{3}{2},\frac{1}{2}|0,\frac{1}{2},\frac{1}{2}>=0</math>, hence we get,
 
<math>\frac{1}{\sqrt{3}}(\alpha +\beta )</math>          <math>\Rightarrow </math>          <math>\alpha =-\beta </math>
 
A substitution of <math>\alpha =-\beta </math> gives <math>\alpha =-\beta =\frac{1}{\sqrt{2}}</math>, and substituting this we get,
 
<math>|0,\frac{1}{2},\frac{1}{2}>= \frac{1}{\sqrt{2}}\left ( |j_{1},j_{2},j_{3};+,+,-> - |j_{1},j_{2},j_{3};-,+,+>  \right )</math>
 
Following the same procedure i.e. applying <math>\hat{J_{-}}</math> on the left of above equation and <math>(\hat{J_{1-}}+\hat{J_{2-}}+\hat{J_{3-}})</math> on the right hand side and then equating the two results-we find
 
<math>|0,\frac{1}{2},-\frac{1}{2}>=\frac{1}{\sqrt{2}}\left (- |j_{1},j_{2},j_{3};+,-,->+|j_{1},j_{2},j_{3};-,-,+> \right )</math>
 
Now to find <math>|1,\frac{1}{2},\frac{1}{2}></math>, we may write it as a linear combination of  <math>|j_{1},j_{2},j_{3};+,+,-></math>, <math>|j_{1},j_{2},j_{3};+,-,+></math>, and <math>|j_{1},j_{2},j_{3};-,+,+></math>
 
<math>|1,\frac{1}{2},\frac{1}{2}>=\alpha |j_{1},j_{2},j_{3};+,+,-> +\beta |j_{1},j_{2},j_{3};+,-,+> </math>+\gamma |j_{1},j_{2},j_{3};-,+,+>
 
This state is orthogonal to  <math>|0,\frac{1}{2},-\frac{1}{2}></math>, and hence <math>\alpha =\gamma </math> ; similarly, since this state is also orthogonal to <math>|1,\frac{3}{2},\frac{1}{2}></math>, we have <math>\alpha +\beta +\gamma =0</math>, and hence <math>2\alpha +\beta =0</math> or <math>\beta =-2\alpha =-2\gamma </math>. Now, since all the states of above equation are orthonormal, we have <math>\alpha ^{2}+\beta ^{2}+\gamma ^{2}=1</math>, which when combined with <math>\beta =-2\alpha =-2\gamma </math> leads to <math>\alpha =\gamma =-\frac{1}{\sqrt{6}} </math> and <math>\beta =\frac{2}{\sqrt{6}}</math>. We may thus write as
 
<math>|1,\frac{1}{2},\frac{1}{2}>=\frac{1}{\sqrt{6}}\left ( -|j_{1},j_{2},j_{3};+,+,-> +2|j_{1},j_{2},j_{3};+,-,+> -|j_{1},j_{2},j_{3};-,+,+> \right )</math>
 
Finally, applying <math>\hat{J_{-}}</math> on the left side of the above equation and <math>(\hat{J_{1-}}+\hat{J_{2-}}+\hat{J_{3-}})</math> on the right hand side and equating the two results, we get
 
<math>|1,\frac{1}{2},-\frac{1}{2}>=\frac{1}{\sqrt{6}}\left ( |j_{1},j_{2},j_{3};+,-,-> -2|j_{1},j_{2},j_{3};-,+,-> -|j_{1},j_{2},j_{3};-,-,+> \right )</math>
 
 
(b) Since we have three different (nonidentical) particles, their spin angular momentum mutually commute. We may thus write Hamiltonian as <math>\hat{H}=-\frac{\epsilon _{0}}{\hbar^{2}}\left ( \hat{S_{1}} + \hat{S_{2}} \right ).\hat{S_{3}}</math>. Due to this suggestive form of <math>hat{H}</math>, it is appropriate, as shown in (a), to start by coupling <math>\hat{S_{1}}</math> with <math>\hat{S_{2}}</math> to obtain <math>\hat{S_{12}}=\hat{S_{1}}+\hat{S_{2}}</math>, and then add <math>\hat{S_{12}}</math> to <math>\hat{S_{3}}</math> to generate the total spin; <math>\hat{S}=\hat{S_{12}}+\hat{S_{3}}</math>. We may thus write <math>\hat{H}</math> as
 
<math>\hat{H}=-\frac{\epsilon _{0}}{\hbar^{2}}\left ( \hat{S_{1}} + \hat{S_{2}} \right ).\hat{S_{3}}=-\frac{\epsilon _{0}}{\hbar^{2}}\hat{S_{12}}.\hat{S_{3}}=-\frac{\epsilon _{0}}{2\hbar^{2}}\left ( \hat{S}^{2}-\hat{S_{12}}^{2} -\hat{S_{3}}^{2}\right )</math>
 
since <math>\hat{S_{12}}.\hat{S_{3}}=\frac{1}{2}[(\hat{S_{12}}+\hat{S_{3}})^{2}-\hat{S_{12}}^{2}-\hat{S_{3}}^{2}]</math>. Since the operator <math>\hat{H}, \hat{S}^{2},\hat{S_{12}}^{2}</math>, and <math>\hat{S_{3}}^{2}</math> mutually commute , we may select as their joint eigenstates the kets <math>|{s_{12}},s,m></math>; we have seen in (a) how to construct these states. The eigenvalues of <math>\hat{H}</math> are thus given by
 
<math>\hat{H}|s_{12},s,m>=-\frac{\epsilon _{0}}{\hbar^{2}}(\hat{S}^{2}-\hat{S_{12}}^{2}-\hat{S_{3}}^{2})|s_{12},s,m>=-\frac{\epsilon _{0}}{\hbar^{2}}[s(s+1)-s_{12}(s_{12}+1-\frac{3}{4})]|s_{12},s,m></math>
 
since <math>s_{3}=\frac{1}{2} </math> and <math>\hat{S_{3}}^{2}|s_{12},s,m>=\hbar^{2}s_{3}(s_{3}+1)|s_{12},s,m>=(\frac{3\hbar^{2}}{4})|s_{12},s,m></math>
 
As shown in above, the energy levels of this system are degenerate with respect to m, since they depend on the quantum numbers <math>s</math> and <math>s_{12}</math> but not m:
 
<math>E_{12,s}=-\frac{\epsilon _{0}}{\hbar^{2}}[s(s+1)-s_{12}(s_{12}+1-\frac{3}{4})]</math>
 
For instance, the energy <math>E_{12,s}=E_{1,3/2}=-\frac{\epsilon _{0}}{2}</math> is fourfold degenerate, since it corresponds to four different states: <math>|s_{12},s,m>=|1,\frac{3}{2},\pm \frac{3}{2}></math> and <math>|1,\frac{3}{2},\pm \frac{1}{2}></math>. Similarly, the energy <math>E_{0,1/2}=0</math> is two fold degenerate; the corresponding states are <math>|0,\frac{1}{2},\pm \frac{1}{2}></math>. Finally, the energy <math>E_{1,1/2}=\epsilon _{0}</math> is also two fold degenerate since it corresponds to <math>|1,\frac{1}{2},\pm \frac{1}{2}></math>.
 
(Reference taken from Quantum mechanics Concepts and Applications by Nouredine Zetilli.)

Latest revision as of 22:35, 30 April 2010

Problem

(a) Find the total spin of a syatem of three spin Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2}} particles and derive the corresponding Clebsch-Gordan Coefficients.

(b) Consider a system of three nonidentical spin Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2}} particles whose Hamiltonian is given by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{H}=-\epsilon _{0}(\vec{S}_{1}.\vec{S}_{3}+\vec{S}_{2}.\vec{S}_{3})/\hbar^{2}} . Find the system's energy levels and their degeneracies.

Solution:-

(a) To add Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j_{1}=\frac{1}{2}} ,Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j_{2}=\frac{1}{2}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j_{3}=\frac{1}{2}} , we begin by coupling Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j_{1}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j_{2}} to form Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j_{12}=j_{1}+j_{2}} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_{1}-j_{2}|\leq j_{12}\leq |j_{1}+j_{2}|} , hence Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j_{12}=0,1} . Then we add jFailed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle _{12}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j_{3}} ; this leads to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_{12}-j_{3}|\leq j\leq |j_{12}+j_{3}|}

We are going to denote the joint eigenstates of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{J_{1}}^{2}} ,Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{J_{2}}^{2}} ,Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{J_{3}}^{2}} ,Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{J_{12}}^{2}} ,Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{J}^{2}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_{z}} by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_{12},j,m> } and the joint eigenstates of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{J_{1}}^{2}} ,Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{J_{2}}^{2}} ,Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{J_{3}}^{2}} ,Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{J_{1_{z}}}^{2}} ,Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{J_{2_{z}}}^{2}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{J_{3_{z}}}^{2}} by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_{1},J_{2},j_{3};m_{1},m_{2},m_{3}>} ; since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j_{1}=j_{2}=j_{3}=\frac{1}{2}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m_{1}=\pm \frac{1}{2}} ,Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m_{2}=\pm \frac{1}{2}} ,Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m_{3}=\pm \frac{1}{2}} ,we will be using throughout this problem the lighter notation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_{1},j_{2},j_{3};\pm ,\pm,\pm> } to abbreviate Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\frac{1}{2},\frac{1}{2},\frac{1}{2};\pm \frac{1}{2},\pm \frac{1}{2},\pm \frac{1}{2}> }

In total there are eight states Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_{12}, j, m>} since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (2j_{1}+1)(2j_{2}+1)(2j_{3}+1)=8} . Four of these correspond to the subspace Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j=\frac{3}{2}:|1,\frac{3}{2},\frac{3}{2}>,|1,\frac{3}{2},\frac{1}{2}>,|1,\frac{3}{2},-\frac{1}{2}>,|1,\frac{3}{2},-\frac{3}{2}>} . The remaining four belong to the subspace Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j=\frac{1}{2}:|0,\frac{1}{2},\frac{1}{2}>,|0,\frac{1}{2},-\frac{1}{2}>,|1,\frac{1}{2},\frac{1}{2}>,|1,\frac{1}{2},-\frac{1}{2}>} . To construct the states Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_{12},j,m> } in terms of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_{1},j_{2},j_{3};\pm,\pm,\pm} , we are going to consider the two subspaces Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j=\frac{3}{2}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j=\frac{1}{2}} separately.


Subspace Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j=\frac{3}{2}}

First, the states Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |1,\frac{3}{2},\frac{3}{2}>} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |1,\frac{3}{2},-\frac{3}{2}>} are clearly given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |1,\frac{3}{2},\frac{3}{2}>=|j_{1},j_{2},j_{3};+,+,+> } , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |1,\frac{3}{2},-\frac{3}{2}>=|j_{1},j_{2},j_{3};-,-,-> }

To obtain Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |1,\frac{3}{2},\frac{1}{2}>} , we need to apply, on the one hand, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{J_{-}}} on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |1,\frac{3}{2},\frac{3}{2}>}

and, on the other hand, apply on

This yields

Since . Equating above equations, we get

Following the same method-applying on on the right hand side of above equation and then equating the two results-we find


Subspace--j=1/2

We can write Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |0,\frac{1}{2},\frac{1}{2}>} as a linear combination of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_{1},j_{2},j_{3};+,+,-> } and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_{1},j_{2},j_{3};-,+,+> }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |0,\frac{1}{2},\frac{1}{2}>=\alpha |j_{1},j_{2},j_{3};+,+,-> +\beta |j_{1},j_{2},j_{3};-,+,+> }

Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |0,\frac{1}{2},\frac{1}{2}>} is normalized, while Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\hat{J_{1}}+\hat{J_{2}}+\hat{J_{3}})|j_{1},j_{2},j_{3};+,+,->} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\hat{J_{1}}+\hat{J_{2}}+\hat{J_{3}})|j_{1},j_{2},j_{3};-,+,+>} , and since Clebsch-Gordan coefficients, such as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha } and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta} , are real numbers, above equation yields

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha ^{2}+\beta ^{2}=1}

On the other hand, since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <1,\frac{3}{2},\frac{1}{2}|0,\frac{1}{2},\frac{1}{2}>=0} , hence we get,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\sqrt{3}}(\alpha +\beta )} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow } Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha =-\beta }

A substitution of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha =-\beta } gives Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha =-\beta =\frac{1}{\sqrt{2}}} , and substituting this we get,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |0,\frac{1}{2},\frac{1}{2}>= \frac{1}{\sqrt{2}}\left ( |j_{1},j_{2},j_{3};+,+,-> - |j_{1},j_{2},j_{3};-,+,+> \right )}

Following the same procedure i.e. applying Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{J_{-}}} on the left of above equation and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\hat{J_{1-}}+\hat{J_{2-}}+\hat{J_{3-}})} on the right hand side and then equating the two results-we find

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |0,\frac{1}{2},-\frac{1}{2}>=\frac{1}{\sqrt{2}}\left (- |j_{1},j_{2},j_{3};+,-,->+|j_{1},j_{2},j_{3};-,-,+> \right )}

Now to find Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |1,\frac{1}{2},\frac{1}{2}>} , we may write it as a linear combination of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_{1},j_{2},j_{3};+,+,->} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_{1},j_{2},j_{3};+,-,+>} , and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_{1},j_{2},j_{3};-,+,+>}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |1,\frac{1}{2},\frac{1}{2}>=\alpha |j_{1},j_{2},j_{3};+,+,-> +\beta |j_{1},j_{2},j_{3};+,-,+> } +\gamma |j_{1},j_{2},j_{3};-,+,+>

This state is orthogonal to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |0,\frac{1}{2},-\frac{1}{2}>} , and hence Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha =\gamma }  ; similarly, since this state is also orthogonal to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |1,\frac{3}{2},\frac{1}{2}>} , we have Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha +\beta +\gamma =0} , and hence Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2\alpha +\beta =0} or Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta =-2\alpha =-2\gamma } . Now, since all the states of above equation are orthonormal, we have Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha ^{2}+\beta ^{2}+\gamma ^{2}=1} , which when combined with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta =-2\alpha =-2\gamma } leads to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha =\gamma =-\frac{1}{\sqrt{6}} } and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta =\frac{2}{\sqrt{6}}} . We may thus write as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |1,\frac{1}{2},\frac{1}{2}>=\frac{1}{\sqrt{6}}\left ( -|j_{1},j_{2},j_{3};+,+,-> +2|j_{1},j_{2},j_{3};+,-,+> -|j_{1},j_{2},j_{3};-,+,+> \right )}

Finally, applying Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{J_{-}}} on the left side of the above equation and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\hat{J_{1-}}+\hat{J_{2-}}+\hat{J_{3-}})} on the right hand side and equating the two results, we get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |1,\frac{1}{2},-\frac{1}{2}>=\frac{1}{\sqrt{6}}\left ( |j_{1},j_{2},j_{3};+,-,-> -2|j_{1},j_{2},j_{3};-,+,-> -|j_{1},j_{2},j_{3};-,-,+> \right )}


(b) Since we have three different (nonidentical) particles, their spin angular momentum mutually commute. We may thus write Hamiltonian as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{H}=-\frac{\epsilon _{0}}{\hbar^{2}}\left ( \hat{S_{1}} + \hat{S_{2}} \right ).\hat{S_{3}}} . Due to this suggestive form of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle hat{H}} , it is appropriate, as shown in (a), to start by coupling with to obtain , and then add to to generate the total spin; . We may thus write as

since . Since the operator , and mutually commute , we may select as their joint eigenstates the kets ; we have seen in (a) how to construct these states. The eigenvalues of are thus given by

since and

As shown in above, the energy levels of this system are degenerate with respect to m, since they depend on the quantum numbers and but not m:

For instance, the energy is fourfold degenerate, since it corresponds to four different states: and . Similarly, the energy is two fold degenerate; the corresponding states are . Finally, the energy is also two fold degenerate since it corresponds to .

(Reference taken from Quantum mechanics Concepts and Applications by Nouredine Zetilli.)

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