Are projective measurement bases always orthonormal?
Are projective measurement bases always orthonormal?
measurement
edited Dec 3 at 7:21
Blue♦
5,66521354
asked Dec 3 at 6:58
ahelwer
1,270112
add a comment |
Are projective measurement bases always orthonormal?
measurement
edited Dec 3 at 7:21
Blue♦
5,66521354
asked Dec 3 at 6:58
ahelwer
1,270112
2
Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
– Blue♦
Dec 3 at 7:39
add a comment |
Are projective measurement bases always orthonormal?
measurement
edited Dec 3 at 7:21
Blue♦
5,66521354
asked Dec 3 at 6:58
ahelwer
1,270112
Are projective measurement bases always orthonormal?
measurement
measurement
edited Dec 3 at 7:21
Blue♦
5,66521354
asked Dec 3 at 6:58
ahelwer
1,270112
edited Dec 3 at 7:21
Blue♦
5,66521354
asked Dec 3 at 6:58
ahelwer
1,270112
edited Dec 3 at 7:21
Blue♦
5,66521354
edited Dec 3 at 7:21
Blue♦
5,66521354
edited Dec 3 at 7:21
Blue♦
5,66521354
5,66521354
asked Dec 3 at 6:58
ahelwer
1,270112
asked Dec 3 at 6:58
ahelwer
1,270112
asked Dec 3 at 6:58
ahelwer
1,270112
1,270112
2
Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
– Blue♦
Dec 3 at 7:39
add a comment |
2
Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
– Blue♦
Dec 3 at 7:39
2
2
Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
– Blue♦
Dec 3 at 7:39
Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
– Blue♦
Dec 3 at 7:39
add a comment |
2 Answers
2
active
oldest
votes
Yes.
Remember that you require several properties of a projective measurement including $P_i^2=P_i$ for each projector, 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 eigenvector of eigenvalue 1 of a particular projector $P_i$. Use this in the identity relation:
$$
left(sum_jP_jright)|phirangle=mathbb{I}|phirangle
$$
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 Dec 3 at 7:42
DaftWullie
11.9k1536
add a comment |
Here is another way to see this.
A projection $P$ is an operator such that $P^2=P$.
This directly implies that we can attach to each projector $P$ a set of orthonormal states that represent it, by choosing any orthonormal base for its range. More precisely, if $P_i$ has trace $operatorname{tr}(P_i)=n$, then we can represent $P_i$ as a set of orthonormal states ${lvertpsi_{i,j}rangle}_{j=1}^n$.
Note in particular that if $operatorname{tr}(P_i)=1$ then this choice is unique, meaning that there is always a bijection between trace-1 projections and states.
The projector $P_i$ and the corresponding states are connected through
$$P_i=sum_{j=1}^n lvertpsi_{ij}rangle!langle psi_{ij}rvert.$$
In the simpler case of $operatorname{tr}(P_i)=1$ this reads $P_i=lvertpsi_irangle!langlepsi_irvert$.
Now, if you are asking for a projective measurement basis, then you require a set of operators which describes every possible outcome of your state.
This condition is expressed mathematically by requiring $$sum_i P_i=I,$$
which in terms of the associated ket states reads
$$sum_{ij}lvertpsi_{ij}rangle!langlepsi_{ij}rvert=I,$$
which is the completeness relation for the vectors ${lvertpsi_{ij}rangle}_{ij}$.
This immediately implies that this is also an orthonormal set (to see it, take for example the sandwich of this expression with any $lvertpsi_{ij}rangle$).
Orthogonality of $P_i$ is equivalent to orthogonality of the corresponding $lvertpsi_{ij}rangle$, thus the conclusion.
answered Dec 3 at 19:04
glS
3,546437
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Yes.
Remember that you require several properties of a projective measurement including $P_i^2=P_i$ for each projector, 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 eigenvector of eigenvalue 1 of a particular projector $P_i$. Use this in the identity relation:
$$
left(sum_jP_jright)|phirangle=mathbb{I}|phirangle
$$
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 Dec 3 at 7:42
DaftWullie
11.9k1536
add a comment |
Yes.
Remember that you require several properties of a projective measurement including $P_i^2=P_i$ for each projector, 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 eigenvector of eigenvalue 1 of a particular projector $P_i$. Use this in the identity relation:
$$
left(sum_jP_jright)|phirangle=mathbb{I}|phirangle
$$
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 Dec 3 at 7:42
DaftWullie
11.9k1536
add a comment |
Yes.
Remember that you require several properties of a projective measurement including $P_i^2=P_i$ for each projector, 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 eigenvector of eigenvalue 1 of a particular projector $P_i$. Use this in the identity relation:
$$
left(sum_jP_jright)|phirangle=mathbb{I}|phirangle
$$
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 Dec 3 at 7:42
DaftWullie
11.9k1536
Yes.
Remember that you require several properties of a projective measurement including $P_i^2=P_i$ for each projector, 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 eigenvector of eigenvalue 1 of a particular projector $P_i$. Use this in the identity relation:
$$
left(sum_jP_jright)|phirangle=mathbb{I}|phirangle
$$
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 Dec 3 at 7:42
DaftWullie
11.9k1536
edited Dec 3 at 16:20
answered Dec 3 at 7:42
DaftWullie
11.9k1536
answered Dec 3 at 7:42
DaftWullie
11.9k1536
answered Dec 3 at 7:42
DaftWullie
11.9k1536
11.9k1536
add a comment |
add a comment |
Here is another way to see this.
A projection $P$ is an operator such that $P^2=P$.
This directly implies that we can attach to each projector $P$ a set of orthonormal states that represent it, by choosing any orthonormal base for its range. More precisely, if $P_i$ has trace $operatorname{tr}(P_i)=n$, then we can represent $P_i$ as a set of orthonormal states ${lvertpsi_{i,j}rangle}_{j=1}^n$.
Note in particular that if $operatorname{tr}(P_i)=1$ then this choice is unique, meaning that there is always a bijection between trace-1 projections and states.
The projector $P_i$ and the corresponding states are connected through
$$P_i=sum_{j=1}^n lvertpsi_{ij}rangle!langle psi_{ij}rvert.$$
In the simpler case of $operatorname{tr}(P_i)=1$ this reads $P_i=lvertpsi_irangle!langlepsi_irvert$.
Now, if you are asking for a projective measurement basis, then you require a set of operators which describes every possible outcome of your state.
This condition is expressed mathematically by requiring $$sum_i P_i=I,$$
which in terms of the associated ket states reads
$$sum_{ij}lvertpsi_{ij}rangle!langlepsi_{ij}rvert=I,$$
which is the completeness relation for the vectors ${lvertpsi_{ij}rangle}_{ij}$.
This immediately implies that this is also an orthonormal set (to see it, take for example the sandwich of this expression with any $lvertpsi_{ij}rangle$).
Orthogonality of $P_i$ is equivalent to orthogonality of the corresponding $lvertpsi_{ij}rangle$, thus the conclusion.
answered Dec 3 at 19:04
glS
3,546437
add a comment |
Here is another way to see this.
A projection $P$ is an operator such that $P^2=P$.
This directly implies that we can attach to each projector $P$ a set of orthonormal states that represent it, by choosing any orthonormal base for its range. More precisely, if $P_i$ has trace $operatorname{tr}(P_i)=n$, then we can represent $P_i$ as a set of orthonormal states ${lvertpsi_{i,j}rangle}_{j=1}^n$.
Note in particular that if $operatorname{tr}(P_i)=1$ then this choice is unique, meaning that there is always a bijection between trace-1 projections and states.
The projector $P_i$ and the corresponding states are connected through
$$P_i=sum_{j=1}^n lvertpsi_{ij}rangle!langle psi_{ij}rvert.$$
In the simpler case of $operatorname{tr}(P_i)=1$ this reads $P_i=lvertpsi_irangle!langlepsi_irvert$.
Now, if you are asking for a projective measurement basis, then you require a set of operators which describes every possible outcome of your state.
This condition is expressed mathematically by requiring $$sum_i P_i=I,$$
which in terms of the associated ket states reads
$$sum_{ij}lvertpsi_{ij}rangle!langlepsi_{ij}rvert=I,$$
which is the completeness relation for the vectors ${lvertpsi_{ij}rangle}_{ij}$.
This immediately implies that this is also an orthonormal set (to see it, take for example the sandwich of this expression with any $lvertpsi_{ij}rangle$).
Orthogonality of $P_i$ is equivalent to orthogonality of the corresponding $lvertpsi_{ij}rangle$, thus the conclusion.
answered Dec 3 at 19:04
glS
3,546437
add a comment |
Here is another way to see this.
A projection $P$ is an operator such that $P^2=P$.
This directly implies that we can attach to each projector $P$ a set of orthonormal states that represent it, by choosing any orthonormal base for its range. More precisely, if $P_i$ has trace $operatorname{tr}(P_i)=n$, then we can represent $P_i$ as a set of orthonormal states ${lvertpsi_{i,j}rangle}_{j=1}^n$.
Note in particular that if $operatorname{tr}(P_i)=1$ then this choice is unique, meaning that there is always a bijection between trace-1 projections and states.
The projector $P_i$ and the corresponding states are connected through
$$P_i=sum_{j=1}^n lvertpsi_{ij}rangle!langle psi_{ij}rvert.$$
In the simpler case of $operatorname{tr}(P_i)=1$ this reads $P_i=lvertpsi_irangle!langlepsi_irvert$.
Now, if you are asking for a projective measurement basis, then you require a set of operators which describes every possible outcome of your state.
This condition is expressed mathematically by requiring $$sum_i P_i=I,$$
which in terms of the associated ket states reads
$$sum_{ij}lvertpsi_{ij}rangle!langlepsi_{ij}rvert=I,$$
which is the completeness relation for the vectors ${lvertpsi_{ij}rangle}_{ij}$.
This immediately implies that this is also an orthonormal set (to see it, take for example the sandwich of this expression with any $lvertpsi_{ij}rangle$).
Orthogonality of $P_i$ is equivalent to orthogonality of the corresponding $lvertpsi_{ij}rangle$, thus the conclusion.
answered Dec 3 at 19:04
glS
3,546437
Here is another way to see this.
A projection $P$ is an operator such that $P^2=P$.
This directly implies that we can attach to each projector $P$ a set of orthonormal states that represent it, by choosing any orthonormal base for its range. More precisely, if $P_i$ has trace $operatorname{tr}(P_i)=n$, then we can represent $P_i$ as a set of orthonormal states ${lvertpsi_{i,j}rangle}_{j=1}^n$.
Note in particular that if $operatorname{tr}(P_i)=1$ then this choice is unique, meaning that there is always a bijection between trace-1 projections and states.
The projector $P_i$ and the corresponding states are connected through
$$P_i=sum_{j=1}^n lvertpsi_{ij}rangle!langle psi_{ij}rvert.$$
In the simpler case of $operatorname{tr}(P_i)=1$ this reads $P_i=lvertpsi_irangle!langlepsi_irvert$.
Now, if you are asking for a projective measurement basis, then you require a set of operators which describes every possible outcome of your state.
This condition is expressed mathematically by requiring $$sum_i P_i=I,$$
which in terms of the associated ket states reads
$$sum_{ij}lvertpsi_{ij}rangle!langlepsi_{ij}rvert=I,$$
which is the completeness relation for the vectors ${lvertpsi_{ij}rangle}_{ij}$.
This immediately implies that this is also an orthonormal set (to see it, take for example the sandwich of this expression with any $lvertpsi_{ij}rangle$).
Orthogonality of $P_i$ is equivalent to orthogonality of the corresponding $lvertpsi_{ij}rangle$, thus the conclusion.
answered Dec 3 at 19:04
glS
3,546437
answered Dec 3 at 19:04
glS
3,546437
answered Dec 3 at 19:04
glS
3,546437
answered Dec 3 at 19:04
glS
3,546437
3,546437
add a comment |
add a comment |
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Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
– Blue♦
Dec 3 at 7:39