Are the SO(3) defining representation and Spin 1 the same?












0












$begingroup$


Since the equation
$$langle j,m^prime|J_pm|j,mrangle =sqrt{(jmp m)(jpm m+1)} delta_{m^prime,m+1}$$
holds both for spin 1 (the $underline{1}$ rep for SU(2)) and angular momentum (SO(3)). Does this mean the explicit form of spin 1 generators:
$$S_x=frac{1}{sqrt{2}}
left(begin{array}{ccc}
0 & 1 & 0 \
1 & 0 & 1 \
0 & 1 & 0
end{array}right),quad
S_y=frac{1}{sqrt{2}}
left(begin{array}{ccc}
0 & -i & 0 \
i & 0 & -i \
0 & i & 0
end{array}right),quad
S_z=frac{1}{sqrt{2}}
left(begin{array}{ccc}
1 & 0 & 0 \
0 & 0 & 0\
0 & 0& -1
end{array}right),$$
can be obtained by a similar transformation from SO(3) generators?
$$
M_{23}=
left(begin{array}{ccc}
0 & 0 & 0 \
0 & 0 & -1 \
0 & 1 & 0
end{array}right),quad
M_{31}=
left(begin{array}{ccc}
0 & 0 & 1 \
0 & 0 & 0 \
-1 & 0 & 0
end{array}right),quad
M_{12}=
left(begin{array}{ccc}
0 & -1 & 0 \
1 & 0 & 0 \
0 & 0 & 0
end{array}right)
.$$
Sorry to ask such basis questions. I just don't know how to solve the matrix equation $SAS^{-1}=B$.



Further, if it is not, does this mean a same set of algebra usually does not have a unique set of solution? If it is, then they are equivalent; then how could they behave differently, in terms of the global property of two groups SU(2) and SO(3)?






group-theory lie-groups lie-algebras quantum-mechanics







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your spin matrices $S_y$ and $S_z$ are wrong. $S_z$ does not have a $1/sqrt{2}$ in front. $S_y$ is not hermitian as it stands. The $1$ should be replaced by $i$.
    $endgroup$
    – Fabian
    Dec 5 '18 at 22:19












  • $begingroup$
    Yeah Thank you @Fabian. I've corrected it.
    $endgroup$
    – Collin
    Dec 5 '18 at 22:23
















0












$begingroup$


Since the equation
$$langle j,m^prime|J_pm|j,mrangle =sqrt{(jmp m)(jpm m+1)} delta_{m^prime,m+1}$$
holds both for spin 1 (the $underline{1}$ rep for SU(2)) and angular momentum (SO(3)). Does this mean the explicit form of spin 1 generators:
$$S_x=frac{1}{sqrt{2}}
left(begin{array}{ccc}
0 & 1 & 0 \
1 & 0 & 1 \
0 & 1 & 0
end{array}right),quad
S_y=frac{1}{sqrt{2}}
left(begin{array}{ccc}
0 & -i & 0 \
i & 0 & -i \
0 & i & 0
end{array}right),quad
S_z=frac{1}{sqrt{2}}
left(begin{array}{ccc}
1 & 0 & 0 \
0 & 0 & 0\
0 & 0& -1
end{array}right),$$
can be obtained by a similar transformation from SO(3) generators?
$$
M_{23}=
left(begin{array}{ccc}
0 & 0 & 0 \
0 & 0 & -1 \
0 & 1 & 0
end{array}right),quad
M_{31}=
left(begin{array}{ccc}
0 & 0 & 1 \
0 & 0 & 0 \
-1 & 0 & 0
end{array}right),quad
M_{12}=
left(begin{array}{ccc}
0 & -1 & 0 \
1 & 0 & 0 \
0 & 0 & 0
end{array}right)
.$$
Sorry to ask such basis questions. I just don't know how to solve the matrix equation $SAS^{-1}=B$.



Further, if it is not, does this mean a same set of algebra usually does not have a unique set of solution? If it is, then they are equivalent; then how could they behave differently, in terms of the global property of two groups SU(2) and SO(3)?






group-theory lie-groups lie-algebras quantum-mechanics







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your spin matrices $S_y$ and $S_z$ are wrong. $S_z$ does not have a $1/sqrt{2}$ in front. $S_y$ is not hermitian as it stands. The $1$ should be replaced by $i$.
    $endgroup$
    – Fabian
    Dec 5 '18 at 22:19












  • $begingroup$
    Yeah Thank you @Fabian. I've corrected it.
    $endgroup$
    – Collin
    Dec 5 '18 at 22:23














0












0








0





$begingroup$


Since the equation
$$langle j,m^prime|J_pm|j,mrangle =sqrt{(jmp m)(jpm m+1)} delta_{m^prime,m+1}$$
holds both for spin 1 (the $underline{1}$ rep for SU(2)) and angular momentum (SO(3)). Does this mean the explicit form of spin 1 generators:
$$S_x=frac{1}{sqrt{2}}
left(begin{array}{ccc}
0 & 1 & 0 \
1 & 0 & 1 \
0 & 1 & 0
end{array}right),quad
S_y=frac{1}{sqrt{2}}
left(begin{array}{ccc}
0 & -i & 0 \
i & 0 & -i \
0 & i & 0
end{array}right),quad
S_z=frac{1}{sqrt{2}}
left(begin{array}{ccc}
1 & 0 & 0 \
0 & 0 & 0\
0 & 0& -1
end{array}right),$$
can be obtained by a similar transformation from SO(3) generators?
$$
M_{23}=
left(begin{array}{ccc}
0 & 0 & 0 \
0 & 0 & -1 \
0 & 1 & 0
end{array}right),quad
M_{31}=
left(begin{array}{ccc}
0 & 0 & 1 \
0 & 0 & 0 \
-1 & 0 & 0
end{array}right),quad
M_{12}=
left(begin{array}{ccc}
0 & -1 & 0 \
1 & 0 & 0 \
0 & 0 & 0
end{array}right)
.$$
Sorry to ask such basis questions. I just don't know how to solve the matrix equation $SAS^{-1}=B$.



Further, if it is not, does this mean a same set of algebra usually does not have a unique set of solution? If it is, then they are equivalent; then how could they behave differently, in terms of the global property of two groups SU(2) and SO(3)?






group-theory lie-groups lie-algebras quantum-mechanics







share|cite|improve this question











$endgroup$




Since the equation
$$langle j,m^prime|J_pm|j,mrangle =sqrt{(jmp m)(jpm m+1)} delta_{m^prime,m+1}$$
holds both for spin 1 (the $underline{1}$ rep for SU(2)) and angular momentum (SO(3)). Does this mean the explicit form of spin 1 generators:
$$S_x=frac{1}{sqrt{2}}
left(begin{array}{ccc}
0 & 1 & 0 \
1 & 0 & 1 \
0 & 1 & 0
end{array}right),quad
S_y=frac{1}{sqrt{2}}
left(begin{array}{ccc}
0 & -i & 0 \
i & 0 & -i \
0 & i & 0
end{array}right),quad
S_z=frac{1}{sqrt{2}}
left(begin{array}{ccc}
1 & 0 & 0 \
0 & 0 & 0\
0 & 0& -1
end{array}right),$$
can be obtained by a similar transformation from SO(3) generators?
$$
M_{23}=
left(begin{array}{ccc}
0 & 0 & 0 \
0 & 0 & -1 \
0 & 1 & 0
end{array}right),quad
M_{31}=
left(begin{array}{ccc}
0 & 0 & 1 \
0 & 0 & 0 \
-1 & 0 & 0
end{array}right),quad
M_{12}=
left(begin{array}{ccc}
0 & -1 & 0 \
1 & 0 & 0 \
0 & 0 & 0
end{array}right)
.$$
Sorry to ask such basis questions. I just don't know how to solve the matrix equation $SAS^{-1}=B$.



Further, if it is not, does this mean a same set of algebra usually does not have a unique set of solution? If it is, then they are equivalent; then how could they behave differently, in terms of the global property of two groups SU(2) and SO(3)?






group-theory lie-groups lie-algebras quantum-mechanics




group-theory lie-groups lie-algebras quantum-mechanics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 22:29







Collin

















asked Dec 5 '18 at 22:13









CollinCollin

1247




1247












  • $begingroup$
    Your spin matrices $S_y$ and $S_z$ are wrong. $S_z$ does not have a $1/sqrt{2}$ in front. $S_y$ is not hermitian as it stands. The $1$ should be replaced by $i$.
    $endgroup$
    – Fabian
    Dec 5 '18 at 22:19












  • $begingroup$
    Yeah Thank you @Fabian. I've corrected it.
    $endgroup$
    – Collin
    Dec 5 '18 at 22:23


















  • $begingroup$
    Your spin matrices $S_y$ and $S_z$ are wrong. $S_z$ does not have a $1/sqrt{2}$ in front. $S_y$ is not hermitian as it stands. The $1$ should be replaced by $i$.
    $endgroup$
    – Fabian
    Dec 5 '18 at 22:19












  • $begingroup$
    Yeah Thank you @Fabian. I've corrected it.
    $endgroup$
    – Collin
    Dec 5 '18 at 22:23
















$begingroup$
Your spin matrices $S_y$ and $S_z$ are wrong. $S_z$ does not have a $1/sqrt{2}$ in front. $S_y$ is not hermitian as it stands. The $1$ should be replaced by $i$.
$endgroup$
– Fabian
Dec 5 '18 at 22:19






$begingroup$
Your spin matrices $S_y$ and $S_z$ are wrong. $S_z$ does not have a $1/sqrt{2}$ in front. $S_y$ is not hermitian as it stands. The $1$ should be replaced by $i$.
$endgroup$
– Fabian
Dec 5 '18 at 22:19














$begingroup$
Yeah Thank you @Fabian. I've corrected it.
$endgroup$
– Collin
Dec 5 '18 at 22:23




$begingroup$
Yeah Thank you @Fabian. I've corrected it.
$endgroup$
– Collin
Dec 5 '18 at 22:23










1 Answer
1






active

oldest

votes


















1












$begingroup$

You are using two different conventions. While the spin-matrices are in the quantum notation (that is they are Hermitian matrices), the angular momentum matrices are in conventional representation theory notation (they are anti-Hermitian matrices). In order to compare the two, you should multiply the latter by $i$. In this case, the angular moment matrices read
$$
M_{23}=
left(begin{array}{ccc}
0 & 0 & 0 \
0 & 0 & -i \
0 & i & 0
end{array}right),quad
M_{31}=
left(begin{array}{ccc}
0 & 0 & i \
0 & 0 & 0 \
-i & 0 & 0
end{array}right),quad
M_{12}=
left(begin{array}{ccc}
0 & -i & 0 \
i & 0 & 0 \
0 & 0 & 0
end{array}right)
.
$$



With this, it is a straightforward exercise to show that
$$ U=left(
begin{array}{ccc}
-frac{1}{sqrt{2}} & frac{i}{sqrt{2}} & 0 \
0 & 0 & 1 \
frac{1}{sqrt{2}} & frac{i}{sqrt{2}} & 0 \
end{array}
right) $$
does the job; e.g., $S_z = U M_{12} U^dagger$ and so on.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Then, since they are exactly the same thing, can you tell me why would we have $SO(3)/Z_3=SU(2)$ globally? Why don't we have $SO(3)=SU(2)$? @Fabian.
    $endgroup$
    – Collin
    Dec 5 '18 at 22:39










  • $begingroup$
    Spin 1 is the same. SU(2) however also has half-integer spin representations (which SO(3) is lacking).
    $endgroup$
    – Fabian
    Dec 5 '18 at 22:44












  • $begingroup$
    By the way, it should read $SU(2)/Z_2 = SO(3)$.
    $endgroup$
    – Fabian
    Dec 5 '18 at 22:45










  • $begingroup$
    Yes Yes... I was in a rush indeed. now I know $SU(2)/Z_2=SO(3)$ is a claim depending on specific representation. Thank you sir. good day! @Fabian
    $endgroup$
    – Collin
    Dec 5 '18 at 22:49











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

You are using two different conventions. While the spin-matrices are in the quantum notation (that is they are Hermitian matrices), the angular momentum matrices are in conventional representation theory notation (they are anti-Hermitian matrices). In order to compare the two, you should multiply the latter by $i$. In this case, the angular moment matrices read
$$
M_{23}=
left(begin{array}{ccc}
0 & 0 & 0 \
0 & 0 & -i \
0 & i & 0
end{array}right),quad
M_{31}=
left(begin{array}{ccc}
0 & 0 & i \
0 & 0 & 0 \
-i & 0 & 0
end{array}right),quad
M_{12}=
left(begin{array}{ccc}
0 & -i & 0 \
i & 0 & 0 \
0 & 0 & 0
end{array}right)
.
$$



With this, it is a straightforward exercise to show that
$$ U=left(
begin{array}{ccc}
-frac{1}{sqrt{2}} & frac{i}{sqrt{2}} & 0 \
0 & 0 & 1 \
frac{1}{sqrt{2}} & frac{i}{sqrt{2}} & 0 \
end{array}
right) $$
does the job; e.g., $S_z = U M_{12} U^dagger$ and so on.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Then, since they are exactly the same thing, can you tell me why would we have $SO(3)/Z_3=SU(2)$ globally? Why don't we have $SO(3)=SU(2)$? @Fabian.
    $endgroup$
    – Collin
    Dec 5 '18 at 22:39










  • $begingroup$
    Spin 1 is the same. SU(2) however also has half-integer spin representations (which SO(3) is lacking).
    $endgroup$
    – Fabian
    Dec 5 '18 at 22:44












  • $begingroup$
    By the way, it should read $SU(2)/Z_2 = SO(3)$.
    $endgroup$
    – Fabian
    Dec 5 '18 at 22:45










  • $begingroup$
    Yes Yes... I was in a rush indeed. now I know $SU(2)/Z_2=SO(3)$ is a claim depending on specific representation. Thank you sir. good day! @Fabian
    $endgroup$
    – Collin
    Dec 5 '18 at 22:49
















1












$begingroup$

You are using two different conventions. While the spin-matrices are in the quantum notation (that is they are Hermitian matrices), the angular momentum matrices are in conventional representation theory notation (they are anti-Hermitian matrices). In order to compare the two, you should multiply the latter by $i$. In this case, the angular moment matrices read
$$
M_{23}=
left(begin{array}{ccc}
0 & 0 & 0 \
0 & 0 & -i \
0 & i & 0
end{array}right),quad
M_{31}=
left(begin{array}{ccc}
0 & 0 & i \
0 & 0 & 0 \
-i & 0 & 0
end{array}right),quad
M_{12}=
left(begin{array}{ccc}
0 & -i & 0 \
i & 0 & 0 \
0 & 0 & 0
end{array}right)
.
$$



With this, it is a straightforward exercise to show that
$$ U=left(
begin{array}{ccc}
-frac{1}{sqrt{2}} & frac{i}{sqrt{2}} & 0 \
0 & 0 & 1 \
frac{1}{sqrt{2}} & frac{i}{sqrt{2}} & 0 \
end{array}
right) $$
does the job; e.g., $S_z = U M_{12} U^dagger$ and so on.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Then, since they are exactly the same thing, can you tell me why would we have $SO(3)/Z_3=SU(2)$ globally? Why don't we have $SO(3)=SU(2)$? @Fabian.
    $endgroup$
    – Collin
    Dec 5 '18 at 22:39










  • $begingroup$
    Spin 1 is the same. SU(2) however also has half-integer spin representations (which SO(3) is lacking).
    $endgroup$
    – Fabian
    Dec 5 '18 at 22:44












  • $begingroup$
    By the way, it should read $SU(2)/Z_2 = SO(3)$.
    $endgroup$
    – Fabian
    Dec 5 '18 at 22:45










  • $begingroup$
    Yes Yes... I was in a rush indeed. now I know $SU(2)/Z_2=SO(3)$ is a claim depending on specific representation. Thank you sir. good day! @Fabian
    $endgroup$
    – Collin
    Dec 5 '18 at 22:49














1












1








1





$begingroup$

You are using two different conventions. While the spin-matrices are in the quantum notation (that is they are Hermitian matrices), the angular momentum matrices are in conventional representation theory notation (they are anti-Hermitian matrices). In order to compare the two, you should multiply the latter by $i$. In this case, the angular moment matrices read
$$
M_{23}=
left(begin{array}{ccc}
0 & 0 & 0 \
0 & 0 & -i \
0 & i & 0
end{array}right),quad
M_{31}=
left(begin{array}{ccc}
0 & 0 & i \
0 & 0 & 0 \
-i & 0 & 0
end{array}right),quad
M_{12}=
left(begin{array}{ccc}
0 & -i & 0 \
i & 0 & 0 \
0 & 0 & 0
end{array}right)
.
$$



With this, it is a straightforward exercise to show that
$$ U=left(
begin{array}{ccc}
-frac{1}{sqrt{2}} & frac{i}{sqrt{2}} & 0 \
0 & 0 & 1 \
frac{1}{sqrt{2}} & frac{i}{sqrt{2}} & 0 \
end{array}
right) $$
does the job; e.g., $S_z = U M_{12} U^dagger$ and so on.






share|cite|improve this answer











$endgroup$



You are using two different conventions. While the spin-matrices are in the quantum notation (that is they are Hermitian matrices), the angular momentum matrices are in conventional representation theory notation (they are anti-Hermitian matrices). In order to compare the two, you should multiply the latter by $i$. In this case, the angular moment matrices read
$$
M_{23}=
left(begin{array}{ccc}
0 & 0 & 0 \
0 & 0 & -i \
0 & i & 0
end{array}right),quad
M_{31}=
left(begin{array}{ccc}
0 & 0 & i \
0 & 0 & 0 \
-i & 0 & 0
end{array}right),quad
M_{12}=
left(begin{array}{ccc}
0 & -i & 0 \
i & 0 & 0 \
0 & 0 & 0
end{array}right)
.
$$



With this, it is a straightforward exercise to show that
$$ U=left(
begin{array}{ccc}
-frac{1}{sqrt{2}} & frac{i}{sqrt{2}} & 0 \
0 & 0 & 1 \
frac{1}{sqrt{2}} & frac{i}{sqrt{2}} & 0 \
end{array}
right) $$
does the job; e.g., $S_z = U M_{12} U^dagger$ and so on.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 22:34

























answered Dec 5 '18 at 22:28









FabianFabian

19.7k3674




19.7k3674












  • $begingroup$
    Then, since they are exactly the same thing, can you tell me why would we have $SO(3)/Z_3=SU(2)$ globally? Why don't we have $SO(3)=SU(2)$? @Fabian.
    $endgroup$
    – Collin
    Dec 5 '18 at 22:39










  • $begingroup$
    Spin 1 is the same. SU(2) however also has half-integer spin representations (which SO(3) is lacking).
    $endgroup$
    – Fabian
    Dec 5 '18 at 22:44












  • $begingroup$
    By the way, it should read $SU(2)/Z_2 = SO(3)$.
    $endgroup$
    – Fabian
    Dec 5 '18 at 22:45










  • $begingroup$
    Yes Yes... I was in a rush indeed. now I know $SU(2)/Z_2=SO(3)$ is a claim depending on specific representation. Thank you sir. good day! @Fabian
    $endgroup$
    – Collin
    Dec 5 '18 at 22:49


















  • $begingroup$
    Then, since they are exactly the same thing, can you tell me why would we have $SO(3)/Z_3=SU(2)$ globally? Why don't we have $SO(3)=SU(2)$? @Fabian.
    $endgroup$
    – Collin
    Dec 5 '18 at 22:39










  • $begingroup$
    Spin 1 is the same. SU(2) however also has half-integer spin representations (which SO(3) is lacking).
    $endgroup$
    – Fabian
    Dec 5 '18 at 22:44












  • $begingroup$
    By the way, it should read $SU(2)/Z_2 = SO(3)$.
    $endgroup$
    – Fabian
    Dec 5 '18 at 22:45










  • $begingroup$
    Yes Yes... I was in a rush indeed. now I know $SU(2)/Z_2=SO(3)$ is a claim depending on specific representation. Thank you sir. good day! @Fabian
    $endgroup$
    – Collin
    Dec 5 '18 at 22:49
















$begingroup$
Then, since they are exactly the same thing, can you tell me why would we have $SO(3)/Z_3=SU(2)$ globally? Why don't we have $SO(3)=SU(2)$? @Fabian.
$endgroup$
– Collin
Dec 5 '18 at 22:39




$begingroup$
Then, since they are exactly the same thing, can you tell me why would we have $SO(3)/Z_3=SU(2)$ globally? Why don't we have $SO(3)=SU(2)$? @Fabian.
$endgroup$
– Collin
Dec 5 '18 at 22:39












$begingroup$
Spin 1 is the same. SU(2) however also has half-integer spin representations (which SO(3) is lacking).
$endgroup$
– Fabian
Dec 5 '18 at 22:44






$begingroup$
Spin 1 is the same. SU(2) however also has half-integer spin representations (which SO(3) is lacking).
$endgroup$
– Fabian
Dec 5 '18 at 22:44














$begingroup$
By the way, it should read $SU(2)/Z_2 = SO(3)$.
$endgroup$
– Fabian
Dec 5 '18 at 22:45




$begingroup$
By the way, it should read $SU(2)/Z_2 = SO(3)$.
$endgroup$
– Fabian
Dec 5 '18 at 22:45












$begingroup$
Yes Yes... I was in a rush indeed. now I know $SU(2)/Z_2=SO(3)$ is a claim depending on specific representation. Thank you sir. good day! @Fabian
$endgroup$
– Collin
Dec 5 '18 at 22:49




$begingroup$
Yes Yes... I was in a rush indeed. now I know $SU(2)/Z_2=SO(3)$ is a claim depending on specific representation. Thank you sir. good day! @Fabian
$endgroup$
– Collin
Dec 5 '18 at 22:49


















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