Are projective measurement bases always orthonormal?
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Are projective measurement bases always orthonormal?
measurement
edited 36 mins ago
Blue♦
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asked 58 mins ago
ahelwer
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up vote
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Are projective measurement bases always orthonormal?
measurement
edited 36 mins ago
Blue♦
5,60511250
asked 58 mins ago
ahelwer
1,143112
Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
– Blue♦
18 mins ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Are projective measurement bases always orthonormal?
measurement
edited 36 mins ago
Blue♦
5,60511250
asked 58 mins ago
ahelwer
1,143112
Are projective measurement bases always orthonormal?
measurement
measurement
edited 36 mins ago
Blue♦
5,60511250
asked 58 mins ago
ahelwer
1,143112
edited 36 mins ago
Blue♦
5,60511250
asked 58 mins ago
ahelwer
1,143112
edited 36 mins ago
Blue♦
5,60511250
edited 36 mins ago
Blue♦
5,60511250
edited 36 mins ago
Blue♦
5,60511250
5,60511250
asked 58 mins ago
ahelwer
1,143112
asked 58 mins ago
ahelwer
1,143112
asked 58 mins ago
ahelwer
1,143112
1,143112
Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
– Blue♦
18 mins ago
add a comment |
Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
– Blue♦
18 mins ago
Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
– Blue♦
18 mins ago
Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
– Blue♦
18 mins ago
add a comment |
1 Answer
1
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oldest
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up vote
2
down vote
Yes.
Remember that you require several properties including $P_i^2=P_i$ and
$$
sum_iP_i=mathbb{I}.
$$
The first of these show you that the $P_i$ have eigenvalues 0 and 1. Now take a $|phirangle$ that is an eigenvalue 1 of a particular projector $P_i$.
$$
left(sum_jP_jright)|phirangle=mathbb{I}|phirangle
$$
Clearly, this simplifies to
$$
|phirangle+sum_{jneq i}P_j|phirangle=|phirangle.
$$
Hence,
$$
sum_{jneq i}P_j|phirangle=0.
$$
The $P_j$ are all non-negative, so the only way that this can be 0 is if $P_j|phirangle=0$ for all $jneq i$. (To expand upon this, assume there's a $P_k$ such that $P_k|phirangle=|psirangleneq 0$. This means that
$$
sum_{jneq i,k}langlepsi|P_j|phirangle=-langlepsi|P_k|phirangle,
$$
so some terms must be negative, which is impossible if the eigenvalues are all 0 and 1.)
answered 14 mins ago
DaftWullie
11.2k1536
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Yes.
Remember that you require several properties including $P_i^2=P_i$ and
$$
sum_iP_i=mathbb{I}.
$$
The first of these show you that the $P_i$ have eigenvalues 0 and 1. Now take a $|phirangle$ that is an eigenvalue 1 of a particular projector $P_i$.
$$
left(sum_jP_jright)|phirangle=mathbb{I}|phirangle
$$
Clearly, this simplifies to
$$
|phirangle+sum_{jneq i}P_j|phirangle=|phirangle.
$$
Hence,
$$
sum_{jneq i}P_j|phirangle=0.
$$
The $P_j$ are all non-negative, so the only way that this can be 0 is if $P_j|phirangle=0$ for all $jneq i$. (To expand upon this, assume there's a $P_k$ such that $P_k|phirangle=|psirangleneq 0$. This means that
$$
sum_{jneq i,k}langlepsi|P_j|phirangle=-langlepsi|P_k|phirangle,
$$
so some terms must be negative, which is impossible if the eigenvalues are all 0 and 1.)
answered 14 mins ago
DaftWullie
11.2k1536
add a comment |
up vote
2
down vote
Yes.
Remember that you require several properties including $P_i^2=P_i$ and
$$
sum_iP_i=mathbb{I}.
$$
The first of these show you that the $P_i$ have eigenvalues 0 and 1. Now take a $|phirangle$ that is an eigenvalue 1 of a particular projector $P_i$.
$$
left(sum_jP_jright)|phirangle=mathbb{I}|phirangle
$$
Clearly, this simplifies to
$$
|phirangle+sum_{jneq i}P_j|phirangle=|phirangle.
$$
Hence,
$$
sum_{jneq i}P_j|phirangle=0.
$$
The $P_j$ are all non-negative, so the only way that this can be 0 is if $P_j|phirangle=0$ for all $jneq i$. (To expand upon this, assume there's a $P_k$ such that $P_k|phirangle=|psirangleneq 0$. This means that
$$
sum_{jneq i,k}langlepsi|P_j|phirangle=-langlepsi|P_k|phirangle,
$$
so some terms must be negative, which is impossible if the eigenvalues are all 0 and 1.)
answered 14 mins ago
DaftWullie
11.2k1536
add a comment |
up vote
2
down vote
up vote
2
down vote
Yes.
Remember that you require several properties including $P_i^2=P_i$ and
$$
sum_iP_i=mathbb{I}.
$$
The first of these show you that the $P_i$ have eigenvalues 0 and 1. Now take a $|phirangle$ that is an eigenvalue 1 of a particular projector $P_i$.
$$
left(sum_jP_jright)|phirangle=mathbb{I}|phirangle
$$
Clearly, this simplifies to
$$
|phirangle+sum_{jneq i}P_j|phirangle=|phirangle.
$$
Hence,
$$
sum_{jneq i}P_j|phirangle=0.
$$
The $P_j$ are all non-negative, so the only way that this can be 0 is if $P_j|phirangle=0$ for all $jneq i$. (To expand upon this, assume there's a $P_k$ such that $P_k|phirangle=|psirangleneq 0$. This means that
$$
sum_{jneq i,k}langlepsi|P_j|phirangle=-langlepsi|P_k|phirangle,
$$
so some terms must be negative, which is impossible if the eigenvalues are all 0 and 1.)
answered 14 mins ago
DaftWullie
11.2k1536
Yes.
Remember that you require several properties including $P_i^2=P_i$ and
$$
sum_iP_i=mathbb{I}.
$$
The first of these show you that the $P_i$ have eigenvalues 0 and 1. Now take a $|phirangle$ that is an eigenvalue 1 of a particular projector $P_i$.
$$
left(sum_jP_jright)|phirangle=mathbb{I}|phirangle
$$
Clearly, this simplifies to
$$
|phirangle+sum_{jneq i}P_j|phirangle=|phirangle.
$$
Hence,
$$
sum_{jneq i}P_j|phirangle=0.
$$
The $P_j$ are all non-negative, so the only way that this can be 0 is if $P_j|phirangle=0$ for all $jneq i$. (To expand upon this, assume there's a $P_k$ such that $P_k|phirangle=|psirangleneq 0$. This means that
$$
sum_{jneq i,k}langlepsi|P_j|phirangle=-langlepsi|P_k|phirangle,
$$
so some terms must be negative, which is impossible if the eigenvalues are all 0 and 1.)
answered 14 mins ago
DaftWullie
11.2k1536
answered 14 mins ago
DaftWullie
11.2k1536
answered 14 mins ago
DaftWullie
11.2k1536
answered 14 mins ago
DaftWullie
11.2k1536
11.2k1536
add a comment |
add a comment |
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Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
– Blue♦
18 mins ago