Mixing
Proof of a property of the trace of a product of two operators, using Dirac bra-ket notation
$begingroup$
I want to show that Tr(ΩΛ)=Tr(ΛΩ). My attempt:
$$mathrm{Tr}(OmegaLambda)=sum_{i}(OmegaLambda)_{ii}=sum_{i}(sum_{k} Omega_{ik}Lambda_{ki})=sum_{i}(sum_{k}langle i|Omega |kranglelangle k|Lambda |irangle)=sum_{i}(sum_{k}langle k|Lambda| iranglelangle i|Omega |krangle)= sum_{k}(sum_{i}langle k|Lambda| iranglelangle i|Omega |krangle)=sum_{k}langle k|Lambdamathbb{I}Omega |krangle=sum_{k}langle k|LambdaOmega |krangle=sum_{k}(LambdaOmega)_{kk}=mathrm{Tr}(LambdaOmega)$$
The commutation of the fourth equality is because they are scalars and scalars can commute. Then I changed the order of summation but I'm not pretty sure that this is correct. Could you give some advice to improve my proof ? Note: This exercise belongs to a chapter titled Active and Passive Transformations, which says something about unitary operators, I don't see how to use them in order to verify the above relation, though.
linear-algebra proof-verification
asked Jan 31 at 3:11
SamaSama
357
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add a comment |
$begingroup$
I want to show that Tr(ΩΛ)=Tr(ΛΩ). My attempt:
$$mathrm{Tr}(OmegaLambda)=sum_{i}(OmegaLambda)_{ii}=sum_{i}(sum_{k} Omega_{ik}Lambda_{ki})=sum_{i}(sum_{k}langle i|Omega |kranglelangle k|Lambda |irangle)=sum_{i}(sum_{k}langle k|Lambda| iranglelangle i|Omega |krangle)= sum_{k}(sum_{i}langle k|Lambda| iranglelangle i|Omega |krangle)=sum_{k}langle k|Lambdamathbb{I}Omega |krangle=sum_{k}langle k|LambdaOmega |krangle=sum_{k}(LambdaOmega)_{kk}=mathrm{Tr}(LambdaOmega)$$
The commutation of the fourth equality is because they are scalars and scalars can commute. Then I changed the order of summation but I'm not pretty sure that this is correct. Could you give some advice to improve my proof ? Note: This exercise belongs to a chapter titled Active and Passive Transformations, which says something about unitary operators, I don't see how to use them in order to verify the above relation, though.
linear-algebra proof-verification
asked Jan 31 at 3:11
SamaSama
357
$endgroup$
add a comment |
$begingroup$
I want to show that Tr(ΩΛ)=Tr(ΛΩ). My attempt:
$$mathrm{Tr}(OmegaLambda)=sum_{i}(OmegaLambda)_{ii}=sum_{i}(sum_{k} Omega_{ik}Lambda_{ki})=sum_{i}(sum_{k}langle i|Omega |kranglelangle k|Lambda |irangle)=sum_{i}(sum_{k}langle k|Lambda| iranglelangle i|Omega |krangle)= sum_{k}(sum_{i}langle k|Lambda| iranglelangle i|Omega |krangle)=sum_{k}langle k|Lambdamathbb{I}Omega |krangle=sum_{k}langle k|LambdaOmega |krangle=sum_{k}(LambdaOmega)_{kk}=mathrm{Tr}(LambdaOmega)$$
The commutation of the fourth equality is because they are scalars and scalars can commute. Then I changed the order of summation but I'm not pretty sure that this is correct. Could you give some advice to improve my proof ? Note: This exercise belongs to a chapter titled Active and Passive Transformations, which says something about unitary operators, I don't see how to use them in order to verify the above relation, though.
linear-algebra proof-verification
asked Jan 31 at 3:11
SamaSama
357
$endgroup$
I want to show that Tr(ΩΛ)=Tr(ΛΩ). My attempt:
$$mathrm{Tr}(OmegaLambda)=sum_{i}(OmegaLambda)_{ii}=sum_{i}(sum_{k} Omega_{ik}Lambda_{ki})=sum_{i}(sum_{k}langle i|Omega |kranglelangle k|Lambda |irangle)=sum_{i}(sum_{k}langle k|Lambda| iranglelangle i|Omega |krangle)= sum_{k}(sum_{i}langle k|Lambda| iranglelangle i|Omega |krangle)=sum_{k}langle k|Lambdamathbb{I}Omega |krangle=sum_{k}langle k|LambdaOmega |krangle=sum_{k}(LambdaOmega)_{kk}=mathrm{Tr}(LambdaOmega)$$
The commutation of the fourth equality is because they are scalars and scalars can commute. Then I changed the order of summation but I'm not pretty sure that this is correct. Could you give some advice to improve my proof ? Note: This exercise belongs to a chapter titled Active and Passive Transformations, which says something about unitary operators, I don't see how to use them in order to verify the above relation, though.
linear-algebra proof-verification
linear-algebra proof-verification
asked Jan 31 at 3:11
SamaSama
357
asked Jan 31 at 3:11
SamaSama
357
asked Jan 31 at 3:11
SamaSama
357
asked Jan 31 at 3:11
SamaSama
357
asked Jan 31 at 3:11
SamaSama
357
357
add a comment |
add a comment |
2 Answers
2
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Let $A=(a_{ij})$ be an $n times n$ matrix.
Then, the sum of elements of $i$-th row of $A$ is $sum_jlimits a_{ij}$.
Thus the sum of all elements of $A$ is $sumlimits_i sumlimits_j a_{ij}$.
Similarly, the sum of elements of $j$-th columm of $A$ is $sumlimits_i a_{ij}$.
Thus the sum of all elements of $A$ is $sumlimits_j sumlimits_i a_{ij}$.
Hence,
$$sumlimits_i sumlimits_j a_{ij} = sumlimits_j sumlimits_i a_{ij}.$$
answered Jan 31 at 3:38
Doyun NamDoyun Nam
66619
$endgroup$
$begingroup$
Thank you! What do you think of the rest of the proof?
$endgroup$
– Sama
Jan 31 at 3:48
$begingroup$
Specially about the commutation
$endgroup$
– Sama
Jan 31 at 3:50
$begingroup$
@Sama I'm sorry for I'm not familiar with bra-ket notation. So, I can't say that your notation is rigorously correct. However, I think your proof is correct.
$endgroup$
– Doyun Nam
Jan 31 at 4:03
add a comment |
$begingroup$
Your proof is correct, although you can directly conclude $Omega_{ik} Lambda_{ki} = Lambda_{ki} Omega_{ik}$ without any bra-kets!
(This then boils down to the proof given by Doyun).
You might also want to place some parentheses to indicate how you obtain the identity matrix at step 6.
(And some mathematicians might object to the lack of ranges on your indices.)
answered Jan 31 at 4:51
ffffforallffffforall
36028
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2 Answers
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oldest
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2 Answers
2
active
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$begingroup$
Let $A=(a_{ij})$ be an $n times n$ matrix.
Then, the sum of elements of $i$-th row of $A$ is $sum_jlimits a_{ij}$.
Thus the sum of all elements of $A$ is $sumlimits_i sumlimits_j a_{ij}$.
Similarly, the sum of elements of $j$-th columm of $A$ is $sumlimits_i a_{ij}$.
Thus the sum of all elements of $A$ is $sumlimits_j sumlimits_i a_{ij}$.
Hence,
$$sumlimits_i sumlimits_j a_{ij} = sumlimits_j sumlimits_i a_{ij}.$$
answered Jan 31 at 3:38
Doyun NamDoyun Nam
66619
$endgroup$
$begingroup$
Thank you! What do you think of the rest of the proof?
$endgroup$
– Sama
Jan 31 at 3:48
$begingroup$
Specially about the commutation
$endgroup$
– Sama
Jan 31 at 3:50
$begingroup$
@Sama I'm sorry for I'm not familiar with bra-ket notation. So, I can't say that your notation is rigorously correct. However, I think your proof is correct.
$endgroup$
– Doyun Nam
Jan 31 at 4:03
add a comment |
$begingroup$
Let $A=(a_{ij})$ be an $n times n$ matrix.
Then, the sum of elements of $i$-th row of $A$ is $sum_jlimits a_{ij}$.
Thus the sum of all elements of $A$ is $sumlimits_i sumlimits_j a_{ij}$.
Similarly, the sum of elements of $j$-th columm of $A$ is $sumlimits_i a_{ij}$.
Thus the sum of all elements of $A$ is $sumlimits_j sumlimits_i a_{ij}$.
Hence,
$$sumlimits_i sumlimits_j a_{ij} = sumlimits_j sumlimits_i a_{ij}.$$
answered Jan 31 at 3:38
Doyun NamDoyun Nam
66619
$endgroup$
$begingroup$
Thank you! What do you think of the rest of the proof?
$endgroup$
– Sama
Jan 31 at 3:48
$begingroup$
Specially about the commutation
$endgroup$
– Sama
Jan 31 at 3:50
$begingroup$
@Sama I'm sorry for I'm not familiar with bra-ket notation. So, I can't say that your notation is rigorously correct. However, I think your proof is correct.
$endgroup$
– Doyun Nam
Jan 31 at 4:03
add a comment |
$begingroup$
Let $A=(a_{ij})$ be an $n times n$ matrix.
Then, the sum of elements of $i$-th row of $A$ is $sum_jlimits a_{ij}$.
Thus the sum of all elements of $A$ is $sumlimits_i sumlimits_j a_{ij}$.
Similarly, the sum of elements of $j$-th columm of $A$ is $sumlimits_i a_{ij}$.
Thus the sum of all elements of $A$ is $sumlimits_j sumlimits_i a_{ij}$.
Hence,
$$sumlimits_i sumlimits_j a_{ij} = sumlimits_j sumlimits_i a_{ij}.$$
answered Jan 31 at 3:38
Doyun NamDoyun Nam
66619
$endgroup$
Let $A=(a_{ij})$ be an $n times n$ matrix.
Then, the sum of elements of $i$-th row of $A$ is $sum_jlimits a_{ij}$.
Thus the sum of all elements of $A$ is $sumlimits_i sumlimits_j a_{ij}$.
Similarly, the sum of elements of $j$-th columm of $A$ is $sumlimits_i a_{ij}$.
Thus the sum of all elements of $A$ is $sumlimits_j sumlimits_i a_{ij}$.
Hence,
$$sumlimits_i sumlimits_j a_{ij} = sumlimits_j sumlimits_i a_{ij}.$$
answered Jan 31 at 3:38
Doyun NamDoyun Nam
66619
answered Jan 31 at 3:38
Doyun NamDoyun Nam
66619
answered Jan 31 at 3:38
Doyun NamDoyun Nam
66619
answered Jan 31 at 3:38
Doyun NamDoyun Nam
66619
66619
$begingroup$
Thank you! What do you think of the rest of the proof?
$endgroup$
– Sama
Jan 31 at 3:48
$begingroup$
Specially about the commutation
$endgroup$
– Sama
Jan 31 at 3:50
$begingroup$
@Sama I'm sorry for I'm not familiar with bra-ket notation. So, I can't say that your notation is rigorously correct. However, I think your proof is correct.
$endgroup$
– Doyun Nam
Jan 31 at 4:03
add a comment |
$begingroup$
Thank you! What do you think of the rest of the proof?
$endgroup$
– Sama
Jan 31 at 3:48
$begingroup$
Specially about the commutation
$endgroup$
– Sama
Jan 31 at 3:50
$begingroup$
@Sama I'm sorry for I'm not familiar with bra-ket notation. So, I can't say that your notation is rigorously correct. However, I think your proof is correct.
$endgroup$
– Doyun Nam
Jan 31 at 4:03
$begingroup$
Thank you! What do you think of the rest of the proof?
$endgroup$
– Sama
Jan 31 at 3:48
$begingroup$
Thank you! What do you think of the rest of the proof?
$endgroup$
– Sama
Jan 31 at 3:48
$begingroup$
Specially about the commutation
$endgroup$
– Sama
Jan 31 at 3:50
$begingroup$
Specially about the commutation
$endgroup$
– Sama
Jan 31 at 3:50
$begingroup$
@Sama I'm sorry for I'm not familiar with bra-ket notation. So, I can't say that your notation is rigorously correct. However, I think your proof is correct.
$endgroup$
– Doyun Nam
Jan 31 at 4:03
$begingroup$
@Sama I'm sorry for I'm not familiar with bra-ket notation. So, I can't say that your notation is rigorously correct. However, I think your proof is correct.
$endgroup$
– Doyun Nam
Jan 31 at 4:03
add a comment |
$begingroup$
Your proof is correct, although you can directly conclude $Omega_{ik} Lambda_{ki} = Lambda_{ki} Omega_{ik}$ without any bra-kets!
(This then boils down to the proof given by Doyun).
You might also want to place some parentheses to indicate how you obtain the identity matrix at step 6.
(And some mathematicians might object to the lack of ranges on your indices.)
answered Jan 31 at 4:51
ffffforallffffforall
36028
$endgroup$
add a comment |
$begingroup$
Your proof is correct, although you can directly conclude $Omega_{ik} Lambda_{ki} = Lambda_{ki} Omega_{ik}$ without any bra-kets!
(This then boils down to the proof given by Doyun).
You might also want to place some parentheses to indicate how you obtain the identity matrix at step 6.
(And some mathematicians might object to the lack of ranges on your indices.)
answered Jan 31 at 4:51
ffffforallffffforall
36028
$endgroup$
add a comment |
$begingroup$
Your proof is correct, although you can directly conclude $Omega_{ik} Lambda_{ki} = Lambda_{ki} Omega_{ik}$ without any bra-kets!
(This then boils down to the proof given by Doyun).
You might also want to place some parentheses to indicate how you obtain the identity matrix at step 6.
(And some mathematicians might object to the lack of ranges on your indices.)
answered Jan 31 at 4:51
ffffforallffffforall
36028
$endgroup$
Your proof is correct, although you can directly conclude $Omega_{ik} Lambda_{ki} = Lambda_{ki} Omega_{ik}$ without any bra-kets!
(This then boils down to the proof given by Doyun).
You might also want to place some parentheses to indicate how you obtain the identity matrix at step 6.
(And some mathematicians might object to the lack of ranges on your indices.)
answered Jan 31 at 4:51
ffffforallffffforall
36028
answered Jan 31 at 4:51
ffffforallffffforall
36028
answered Jan 31 at 4:51
ffffforallffffforall
36028
answered Jan 31 at 4:51
ffffforallffffforall
36028
36028
add a comment |
add a comment |
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