Are projective measurement bases always orthonormal?
Are projective measurement bases always orthonormal?
measurement
edited Dec 3 '18 at 7:21
Blue♦
5,67221354
asked Dec 3 '18 at 6:58
ahelwer
1,270112
add a comment |
Are projective measurement bases always orthonormal?
measurement
edited Dec 3 '18 at 7:21
Blue♦
5,67221354
asked Dec 3 '18 at 6:58
ahelwer
1,270112
2
Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
– Blue♦
Dec 3 '18 at 7:39
add a comment |
Are projective measurement bases always orthonormal?
measurement
edited Dec 3 '18 at 7:21
Blue♦
5,67221354
asked Dec 3 '18 at 6:58
ahelwer
1,270112
Are projective measurement bases always orthonormal?
measurement
measurement
edited Dec 3 '18 at 7:21
Blue♦
5,67221354
asked Dec 3 '18 at 6:58
ahelwer
1,270112
edited Dec 3 '18 at 7:21
Blue♦
5,67221354
asked Dec 3 '18 at 6:58
ahelwer
1,270112
edited Dec 3 '18 at 7:21
Blue♦
5,67221354
edited Dec 3 '18 at 7:21
Blue♦
5,67221354
edited Dec 3 '18 at 7:21
Blue♦
5,67221354
5,67221354
asked Dec 3 '18 at 6:58
ahelwer
1,270112
asked Dec 3 '18 at 6:58
ahelwer
1,270112
asked Dec 3 '18 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 '18 at 7:39
add a comment |
2
Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
– Blue♦
Dec 3 '18 at 7:39
2
2
Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
– Blue♦
Dec 3 '18 at 7:39
Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
– Blue♦
Dec 3 '18 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 '18 at 7:42
DaftWullie
12.1k1537
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 '18 at 19:04
glS
3,566437
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "694"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fquantumcomputing.stackexchange.com%2fquestions%2f4846%2fare-projective-measurement-bases-always-orthonormal%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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 '18 at 7:42
DaftWullie
12.1k1537
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 '18 at 7:42
DaftWullie
12.1k1537
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 '18 at 7:42
DaftWullie
12.1k1537
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 '18 at 7:42
DaftWullie
12.1k1537
edited Dec 3 '18 at 16:20
answered Dec 3 '18 at 7:42
DaftWullie
12.1k1537
answered Dec 3 '18 at 7:42
DaftWullie
12.1k1537
answered Dec 3 '18 at 7:42
DaftWullie
12.1k1537
12.1k1537
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 '18 at 19:04
glS
3,566437
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 '18 at 19:04
glS
3,566437
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 '18 at 19:04
glS
3,566437
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 '18 at 19:04
glS
3,566437
answered Dec 3 '18 at 19:04
glS
3,566437
answered Dec 3 '18 at 19:04
glS
3,566437
answered Dec 3 '18 at 19:04
glS
3,566437
3,566437
add a comment |
add a comment |
Thanks for contributing an answer to Quantum Computing Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fquantumcomputing.stackexchange.com%2fquestions%2f4846%2fare-projective-measurement-bases-always-orthonormal%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
Related: What is the difference between general measurement and projective measurement? & Wikipedia: POVM
– Blue♦
Dec 3 '18 at 7:39