Mixing
Evaluation of the commutator $[frac{partial}{partial x_j},frac{x_ix_j-r^2delta_{ij}}{r^3}]=-2x_i/r^3$.
$begingroup$
I am trying to evaluate a commutator of the form $[frac{partial}{partial x_j},frac{x_ix_j-r^2delta_{ij}}{r^3}]=-2x_i/r^3$. Bt acting it on a function $f(vec{r})$, I get that the commuator is equal to zero. Can you help?
group-theory quantum-mechanics
asked Jan 17 at 20:41
mithusengupta123mithusengupta123
1139
$endgroup$
add a comment |
$begingroup$
I am trying to evaluate a commutator of the form $[frac{partial}{partial x_j},frac{x_ix_j-r^2delta_{ij}}{r^3}]=-2x_i/r^3$. Bt acting it on a function $f(vec{r})$, I get that the commuator is equal to zero. Can you help?
group-theory quantum-mechanics
asked Jan 17 at 20:41
mithusengupta123mithusengupta123
1139
$endgroup$
$begingroup$
Any function $f(vec{r})$.
$endgroup$
– mithusengupta123
Jan 17 at 20:49
$begingroup$
Can you present your calculations? It's not easy to provide constructive feedback if we don't know your approach.
$endgroup$
– Luke
Jan 17 at 21:01
add a comment |
$begingroup$
I am trying to evaluate a commutator of the form $[frac{partial}{partial x_j},frac{x_ix_j-r^2delta_{ij}}{r^3}]=-2x_i/r^3$. Bt acting it on a function $f(vec{r})$, I get that the commuator is equal to zero. Can you help?
group-theory quantum-mechanics
asked Jan 17 at 20:41
mithusengupta123mithusengupta123
1139
$endgroup$
I am trying to evaluate a commutator of the form $[frac{partial}{partial x_j},frac{x_ix_j-r^2delta_{ij}}{r^3}]=-2x_i/r^3$. Bt acting it on a function $f(vec{r})$, I get that the commuator is equal to zero. Can you help?
group-theory quantum-mechanics
group-theory quantum-mechanics
asked Jan 17 at 20:41
mithusengupta123mithusengupta123
1139
asked Jan 17 at 20:41
mithusengupta123mithusengupta123
1139
edited Jan 17 at 20:49
mithusengupta123
asked Jan 17 at 20:41
mithusengupta123mithusengupta123
1139
asked Jan 17 at 20:41
mithusengupta123mithusengupta123
1139
asked Jan 17 at 20:41
mithusengupta123mithusengupta123
1139
1139
$begingroup$
Any function $f(vec{r})$.
$endgroup$
– mithusengupta123
Jan 17 at 20:49
$begingroup$
Can you present your calculations? It's not easy to provide constructive feedback if we don't know your approach.
$endgroup$
– Luke
Jan 17 at 21:01
add a comment |
$begingroup$
Any function $f(vec{r})$.
$endgroup$
– mithusengupta123
Jan 17 at 20:49
$begingroup$
Can you present your calculations? It's not easy to provide constructive feedback if we don't know your approach.
$endgroup$
– Luke
Jan 17 at 21:01
$begingroup$
Any function $f(vec{r})$.
$endgroup$
– mithusengupta123
Jan 17 at 20:49
$begingroup$
Any function $f(vec{r})$.
$endgroup$
– mithusengupta123
Jan 17 at 20:49
$begingroup$
Can you present your calculations? It's not easy to provide constructive feedback if we don't know your approach.
$endgroup$
– Luke
Jan 17 at 21:01
$begingroup$
Can you present your calculations? It's not easy to provide constructive feedback if we don't know your approach.
$endgroup$
– Luke
Jan 17 at 21:01
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Calculate first this version
begin{eqnarray}
require{cancel}
left[frac{partial}{partial x_k}, frac{x_i x_j - r^2delta_{ij}}{r^3}right]f &=&
frac{partial}{partial x_k}left(frac{x_i x_j - r^2delta_{ij}}{r^3}f right) - frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k} \
&=& left(frac{partial}{partial x_k}frac{x_i x_j - r^2delta_{ij}}{r^3}right)f + cancel{frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k}} - cancel{frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k}}
end{eqnarray}
So the problem reduces to
begin{eqnarray}
left[frac{partial}{partial x_k}, frac{x_i x_j - r^2delta_{ij}}{r^3}right] &=&frac{partial}{partial x_k}frac{x_i x_j - r^2delta_{ij}}{r^3} \
&=& frac{partial}{partial x_k} frac{x_i x_j}{r^3} - delta_{ij}frac{partial}{partial x_k}frac{1}{r} \
&=& left(frac{delta_{ik}x_j}{r^3} + frac{x_idelta_{jk}}{r^3}- 3frac{x_ix_j x_k}{r^5} right) + delta_{ij}frac{x_k}{r^3} \
&=& frac{1}{r^3}left(delta_{ik} x_j + x_idelta_{jk} + delta_{ij} x_kright) - 3 frac{x_i x_j x_k}{r^5} tag{1}
end{eqnarray}
Now replace $kto j$ and sum over $j$
begin{eqnarray}
left[frac{partial}{partial x_j}, frac{x_i x_j - r^2delta_{ij}}{r^3}right] &=& frac{1}{r^3} left(delta_{ij} x_j + x_idelta_{jj} + delta_{ij} x_jright) - 3 frac{x_i x_j x_j}{r^5} \
&=& frac{1}{r^3} left(2x_i + 3x_iright) - 3 frac{x_i r^2}{r^5} \
&=& 2frac{x_i}{r^3}
end{eqnarray}
answered Jan 17 at 21:25
caveraccaverac
14.6k31130
$endgroup$
add a comment |
$begingroup$
Consider any mathematical Expression $A$ and apply the commutator on it. Then
$[frac{partial}{partial x_j},f]A = frac{partial}{partial x_j}(fA)-f frac{partial}{partial x_j}A$
Note that in this calculation, the first term on the right Hand side is a composition of two operations:
First multiplication of $f$ with A because the $f$ is directly left next to the $A$
then applying the differential Operator,since it is left next to all other expressions.
Finally use product rule and lookwhich factor is in front of $A$. So the result of the commutator $R = [frac{partial}{partial x_j},f]$ is
$[frac{partial}{partial x_j},f]A = RA$
answered Jan 17 at 21:05
kryomaximkryomaxim
2,3571825
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Calculate first this version
begin{eqnarray}
require{cancel}
left[frac{partial}{partial x_k}, frac{x_i x_j - r^2delta_{ij}}{r^3}right]f &=&
frac{partial}{partial x_k}left(frac{x_i x_j - r^2delta_{ij}}{r^3}f right) - frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k} \
&=& left(frac{partial}{partial x_k}frac{x_i x_j - r^2delta_{ij}}{r^3}right)f + cancel{frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k}} - cancel{frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k}}
end{eqnarray}
So the problem reduces to
begin{eqnarray}
left[frac{partial}{partial x_k}, frac{x_i x_j - r^2delta_{ij}}{r^3}right] &=&frac{partial}{partial x_k}frac{x_i x_j - r^2delta_{ij}}{r^3} \
&=& frac{partial}{partial x_k} frac{x_i x_j}{r^3} - delta_{ij}frac{partial}{partial x_k}frac{1}{r} \
&=& left(frac{delta_{ik}x_j}{r^3} + frac{x_idelta_{jk}}{r^3}- 3frac{x_ix_j x_k}{r^5} right) + delta_{ij}frac{x_k}{r^3} \
&=& frac{1}{r^3}left(delta_{ik} x_j + x_idelta_{jk} + delta_{ij} x_kright) - 3 frac{x_i x_j x_k}{r^5} tag{1}
end{eqnarray}
Now replace $kto j$ and sum over $j$
begin{eqnarray}
left[frac{partial}{partial x_j}, frac{x_i x_j - r^2delta_{ij}}{r^3}right] &=& frac{1}{r^3} left(delta_{ij} x_j + x_idelta_{jj} + delta_{ij} x_jright) - 3 frac{x_i x_j x_j}{r^5} \
&=& frac{1}{r^3} left(2x_i + 3x_iright) - 3 frac{x_i r^2}{r^5} \
&=& 2frac{x_i}{r^3}
end{eqnarray}
answered Jan 17 at 21:25
caveraccaverac
14.6k31130
$endgroup$
add a comment |
$begingroup$
Calculate first this version
begin{eqnarray}
require{cancel}
left[frac{partial}{partial x_k}, frac{x_i x_j - r^2delta_{ij}}{r^3}right]f &=&
frac{partial}{partial x_k}left(frac{x_i x_j - r^2delta_{ij}}{r^3}f right) - frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k} \
&=& left(frac{partial}{partial x_k}frac{x_i x_j - r^2delta_{ij}}{r^3}right)f + cancel{frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k}} - cancel{frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k}}
end{eqnarray}
So the problem reduces to
begin{eqnarray}
left[frac{partial}{partial x_k}, frac{x_i x_j - r^2delta_{ij}}{r^3}right] &=&frac{partial}{partial x_k}frac{x_i x_j - r^2delta_{ij}}{r^3} \
&=& frac{partial}{partial x_k} frac{x_i x_j}{r^3} - delta_{ij}frac{partial}{partial x_k}frac{1}{r} \
&=& left(frac{delta_{ik}x_j}{r^3} + frac{x_idelta_{jk}}{r^3}- 3frac{x_ix_j x_k}{r^5} right) + delta_{ij}frac{x_k}{r^3} \
&=& frac{1}{r^3}left(delta_{ik} x_j + x_idelta_{jk} + delta_{ij} x_kright) - 3 frac{x_i x_j x_k}{r^5} tag{1}
end{eqnarray}
Now replace $kto j$ and sum over $j$
begin{eqnarray}
left[frac{partial}{partial x_j}, frac{x_i x_j - r^2delta_{ij}}{r^3}right] &=& frac{1}{r^3} left(delta_{ij} x_j + x_idelta_{jj} + delta_{ij} x_jright) - 3 frac{x_i x_j x_j}{r^5} \
&=& frac{1}{r^3} left(2x_i + 3x_iright) - 3 frac{x_i r^2}{r^5} \
&=& 2frac{x_i}{r^3}
end{eqnarray}
answered Jan 17 at 21:25
caveraccaverac
14.6k31130
$endgroup$
add a comment |
$begingroup$
Calculate first this version
begin{eqnarray}
require{cancel}
left[frac{partial}{partial x_k}, frac{x_i x_j - r^2delta_{ij}}{r^3}right]f &=&
frac{partial}{partial x_k}left(frac{x_i x_j - r^2delta_{ij}}{r^3}f right) - frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k} \
&=& left(frac{partial}{partial x_k}frac{x_i x_j - r^2delta_{ij}}{r^3}right)f + cancel{frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k}} - cancel{frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k}}
end{eqnarray}
So the problem reduces to
begin{eqnarray}
left[frac{partial}{partial x_k}, frac{x_i x_j - r^2delta_{ij}}{r^3}right] &=&frac{partial}{partial x_k}frac{x_i x_j - r^2delta_{ij}}{r^3} \
&=& frac{partial}{partial x_k} frac{x_i x_j}{r^3} - delta_{ij}frac{partial}{partial x_k}frac{1}{r} \
&=& left(frac{delta_{ik}x_j}{r^3} + frac{x_idelta_{jk}}{r^3}- 3frac{x_ix_j x_k}{r^5} right) + delta_{ij}frac{x_k}{r^3} \
&=& frac{1}{r^3}left(delta_{ik} x_j + x_idelta_{jk} + delta_{ij} x_kright) - 3 frac{x_i x_j x_k}{r^5} tag{1}
end{eqnarray}
Now replace $kto j$ and sum over $j$
begin{eqnarray}
left[frac{partial}{partial x_j}, frac{x_i x_j - r^2delta_{ij}}{r^3}right] &=& frac{1}{r^3} left(delta_{ij} x_j + x_idelta_{jj} + delta_{ij} x_jright) - 3 frac{x_i x_j x_j}{r^5} \
&=& frac{1}{r^3} left(2x_i + 3x_iright) - 3 frac{x_i r^2}{r^5} \
&=& 2frac{x_i}{r^3}
end{eqnarray}
answered Jan 17 at 21:25
caveraccaverac
14.6k31130
$endgroup$
Calculate first this version
begin{eqnarray}
require{cancel}
left[frac{partial}{partial x_k}, frac{x_i x_j - r^2delta_{ij}}{r^3}right]f &=&
frac{partial}{partial x_k}left(frac{x_i x_j - r^2delta_{ij}}{r^3}f right) - frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k} \
&=& left(frac{partial}{partial x_k}frac{x_i x_j - r^2delta_{ij}}{r^3}right)f + cancel{frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k}} - cancel{frac{x_i x_j - r^2delta_{ij}}{r^3}frac{partial f}{partial x_k}}
end{eqnarray}
So the problem reduces to
begin{eqnarray}
left[frac{partial}{partial x_k}, frac{x_i x_j - r^2delta_{ij}}{r^3}right] &=&frac{partial}{partial x_k}frac{x_i x_j - r^2delta_{ij}}{r^3} \
&=& frac{partial}{partial x_k} frac{x_i x_j}{r^3} - delta_{ij}frac{partial}{partial x_k}frac{1}{r} \
&=& left(frac{delta_{ik}x_j}{r^3} + frac{x_idelta_{jk}}{r^3}- 3frac{x_ix_j x_k}{r^5} right) + delta_{ij}frac{x_k}{r^3} \
&=& frac{1}{r^3}left(delta_{ik} x_j + x_idelta_{jk} + delta_{ij} x_kright) - 3 frac{x_i x_j x_k}{r^5} tag{1}
end{eqnarray}
Now replace $kto j$ and sum over $j$
begin{eqnarray}
left[frac{partial}{partial x_j}, frac{x_i x_j - r^2delta_{ij}}{r^3}right] &=& frac{1}{r^3} left(delta_{ij} x_j + x_idelta_{jj} + delta_{ij} x_jright) - 3 frac{x_i x_j x_j}{r^5} \
&=& frac{1}{r^3} left(2x_i + 3x_iright) - 3 frac{x_i r^2}{r^5} \
&=& 2frac{x_i}{r^3}
end{eqnarray}
answered Jan 17 at 21:25
caveraccaverac
14.6k31130
answered Jan 17 at 21:25
caveraccaverac
14.6k31130
answered Jan 17 at 21:25
caveraccaverac
14.6k31130
answered Jan 17 at 21:25
caveraccaverac
14.6k31130
14.6k31130
add a comment |
add a comment |
$begingroup$
Consider any mathematical Expression $A$ and apply the commutator on it. Then
$[frac{partial}{partial x_j},f]A = frac{partial}{partial x_j}(fA)-f frac{partial}{partial x_j}A$
Note that in this calculation, the first term on the right Hand side is a composition of two operations:
First multiplication of $f$ with A because the $f$ is directly left next to the $A$
then applying the differential Operator,since it is left next to all other expressions.
Finally use product rule and lookwhich factor is in front of $A$. So the result of the commutator $R = [frac{partial}{partial x_j},f]$ is
$[frac{partial}{partial x_j},f]A = RA$
answered Jan 17 at 21:05
kryomaximkryomaxim
2,3571825
$endgroup$
add a comment |
$begingroup$
Consider any mathematical Expression $A$ and apply the commutator on it. Then
$[frac{partial}{partial x_j},f]A = frac{partial}{partial x_j}(fA)-f frac{partial}{partial x_j}A$
Note that in this calculation, the first term on the right Hand side is a composition of two operations:
First multiplication of $f$ with A because the $f$ is directly left next to the $A$
then applying the differential Operator,since it is left next to all other expressions.
Finally use product rule and lookwhich factor is in front of $A$. So the result of the commutator $R = [frac{partial}{partial x_j},f]$ is
$[frac{partial}{partial x_j},f]A = RA$
answered Jan 17 at 21:05
kryomaximkryomaxim
2,3571825
$endgroup$
add a comment |
$begingroup$
Consider any mathematical Expression $A$ and apply the commutator on it. Then
$[frac{partial}{partial x_j},f]A = frac{partial}{partial x_j}(fA)-f frac{partial}{partial x_j}A$
Note that in this calculation, the first term on the right Hand side is a composition of two operations:
First multiplication of $f$ with A because the $f$ is directly left next to the $A$
then applying the differential Operator,since it is left next to all other expressions.
Finally use product rule and lookwhich factor is in front of $A$. So the result of the commutator $R = [frac{partial}{partial x_j},f]$ is
$[frac{partial}{partial x_j},f]A = RA$
answered Jan 17 at 21:05
kryomaximkryomaxim
2,3571825
$endgroup$
Consider any mathematical Expression $A$ and apply the commutator on it. Then
$[frac{partial}{partial x_j},f]A = frac{partial}{partial x_j}(fA)-f frac{partial}{partial x_j}A$
Note that in this calculation, the first term on the right Hand side is a composition of two operations:
First multiplication of $f$ with A because the $f$ is directly left next to the $A$
then applying the differential Operator,since it is left next to all other expressions.
Finally use product rule and lookwhich factor is in front of $A$. So the result of the commutator $R = [frac{partial}{partial x_j},f]$ is
$[frac{partial}{partial x_j},f]A = RA$
answered Jan 17 at 21:05
kryomaximkryomaxim
2,3571825
answered Jan 17 at 21:05
kryomaximkryomaxim
2,3571825
answered Jan 17 at 21:05
kryomaximkryomaxim
2,3571825
answered Jan 17 at 21:05
kryomaximkryomaxim
2,3571825
2,3571825
add a comment |
add a comment |
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$begingroup$
Any function $f(vec{r})$.
$endgroup$
– mithusengupta123
Jan 17 at 20:49
$begingroup$
Can you present your calculations? It's not easy to provide constructive feedback if we don't know your approach.
$endgroup$
– Luke
Jan 17 at 21:01