Mixing
Inequality relating the probabilities of a quantum state to the euclidean distance of states.
$begingroup$
My professor has provided us with the following proposition (without proof).
I am trying to prove this. i'm having quite some trouble proving the first inequality, right under the first sentence. Ive tried using the triangle inequality, Cauchy-Schwartz and brute forcing several times but to no avail.
I would appreciate any recommendations on how to attempt solving this, while still not providing the entire proof.
Note: We use "states" equivalently with "unit vector". To clarify, we are in a finite dimensional Hilbert space (as per Adrian Keister's comment).
linear-algebra inequality quantum-mechanics quantum-computation
asked Jan 25 at 22:04
Alexander H. R.Alexander H. R.
798
$endgroup$
add a comment |
$begingroup$
My professor has provided us with the following proposition (without proof).
I am trying to prove this. i'm having quite some trouble proving the first inequality, right under the first sentence. Ive tried using the triangle inequality, Cauchy-Schwartz and brute forcing several times but to no avail.
I would appreciate any recommendations on how to attempt solving this, while still not providing the entire proof.
Note: We use "states" equivalently with "unit vector". To clarify, we are in a finite dimensional Hilbert space (as per Adrian Keister's comment).
linear-algebra inequality quantum-mechanics quantum-computation
asked Jan 25 at 22:04
Alexander H. R.Alexander H. R.
798
$endgroup$
$begingroup$
I assume you're in a finite-dimensional space $H?$ Otherwise Inequality 3.1 wouldn't mean a whole lot.
$endgroup$
– Adrian Keister
Jan 25 at 22:22
$begingroup$
Yes, I will edit the question to clarify.
$endgroup$
– Alexander H. R.
Jan 25 at 22:40
$begingroup$
@KeithMcClary I'm not entirely sure I understand what you are trying to say or how to use it. Would you be able to elaborate?
$endgroup$
– Alexander H. R.
Jan 25 at 23:38
$begingroup$
Disregard. I see they are unit vectors.
$endgroup$
– Keith McClary
Jan 25 at 23:40
add a comment |
$begingroup$
My professor has provided us with the following proposition (without proof).
I am trying to prove this. i'm having quite some trouble proving the first inequality, right under the first sentence. Ive tried using the triangle inequality, Cauchy-Schwartz and brute forcing several times but to no avail.
I would appreciate any recommendations on how to attempt solving this, while still not providing the entire proof.
Note: We use "states" equivalently with "unit vector". To clarify, we are in a finite dimensional Hilbert space (as per Adrian Keister's comment).
linear-algebra inequality quantum-mechanics quantum-computation
asked Jan 25 at 22:04
Alexander H. R.Alexander H. R.
798
$endgroup$
My professor has provided us with the following proposition (without proof).
I am trying to prove this. i'm having quite some trouble proving the first inequality, right under the first sentence. Ive tried using the triangle inequality, Cauchy-Schwartz and brute forcing several times but to no avail.
I would appreciate any recommendations on how to attempt solving this, while still not providing the entire proof.
Note: We use "states" equivalently with "unit vector". To clarify, we are in a finite dimensional Hilbert space (as per Adrian Keister's comment).
linear-algebra inequality quantum-mechanics quantum-computation
linear-algebra inequality quantum-mechanics quantum-computation
asked Jan 25 at 22:04
Alexander H. R.Alexander H. R.
798
asked Jan 25 at 22:04
Alexander H. R.Alexander H. R.
798
edited Jan 25 at 22:43
Alexander H. R.
asked Jan 25 at 22:04
Alexander H. R.Alexander H. R.
798
asked Jan 25 at 22:04
Alexander H. R.Alexander H. R.
798
asked Jan 25 at 22:04
Alexander H. R.Alexander H. R.
798
798
$begingroup$
I assume you're in a finite-dimensional space $H?$ Otherwise Inequality 3.1 wouldn't mean a whole lot.
$endgroup$
– Adrian Keister
Jan 25 at 22:22
$begingroup$
Yes, I will edit the question to clarify.
$endgroup$
– Alexander H. R.
Jan 25 at 22:40
$begingroup$
@KeithMcClary I'm not entirely sure I understand what you are trying to say or how to use it. Would you be able to elaborate?
$endgroup$
– Alexander H. R.
Jan 25 at 23:38
$begingroup$
Disregard. I see they are unit vectors.
$endgroup$
– Keith McClary
Jan 25 at 23:40
add a comment |
$begingroup$
I assume you're in a finite-dimensional space $H?$ Otherwise Inequality 3.1 wouldn't mean a whole lot.
$endgroup$
– Adrian Keister
Jan 25 at 22:22
$begingroup$
Yes, I will edit the question to clarify.
$endgroup$
– Alexander H. R.
Jan 25 at 22:40
$begingroup$
@KeithMcClary I'm not entirely sure I understand what you are trying to say or how to use it. Would you be able to elaborate?
$endgroup$
– Alexander H. R.
Jan 25 at 23:38
$begingroup$
Disregard. I see they are unit vectors.
$endgroup$
– Keith McClary
Jan 25 at 23:40
$begingroup$
I assume you're in a finite-dimensional space $H?$ Otherwise Inequality 3.1 wouldn't mean a whole lot.
$endgroup$
– Adrian Keister
Jan 25 at 22:22
$begingroup$
I assume you're in a finite-dimensional space $H?$ Otherwise Inequality 3.1 wouldn't mean a whole lot.
$endgroup$
– Adrian Keister
Jan 25 at 22:22
$begingroup$
Yes, I will edit the question to clarify.
$endgroup$
– Alexander H. R.
Jan 25 at 22:40
$begingroup$
Yes, I will edit the question to clarify.
$endgroup$
– Alexander H. R.
Jan 25 at 22:40
$begingroup$
@KeithMcClary I'm not entirely sure I understand what you are trying to say or how to use it. Would you be able to elaborate?
$endgroup$
– Alexander H. R.
Jan 25 at 23:38
$begingroup$
@KeithMcClary I'm not entirely sure I understand what you are trying to say or how to use it. Would you be able to elaborate?
$endgroup$
– Alexander H. R.
Jan 25 at 23:38
$begingroup$
Disregard. I see they are unit vectors.
$endgroup$
– Keith McClary
Jan 25 at 23:40
$begingroup$
Disregard. I see they are unit vectors.
$endgroup$
– Keith McClary
Jan 25 at 23:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $psi^perp$, $phi^perp$ be the projections of $psi$, $phi$ on the subspace perpendicular to $x$.
$ |langle x|psirangle |^2 = 1- lVert psi^perp rVert ^2 $ and $ |langle x|phirangle|^2 = 1- lVert phi^perp rVert ^2 $
so the LHS is
$$ lvert lVert psi^perp rVert ^2 - lVert phi^perp rVert ^2 rvert = lvert lVert psi^perp rVert + lVert phi^perp rVert rvert quad lvert lVert psi^perp rVert - lVert phi^perp rVert rvert le 2lvert lVert psi^perp rVert - lVert phi^perp rVert rvert$$
For given lengths, $lVert psi^perp - phi^perp rVert$ is smallest when the vectors are in the same direction, when it equals $lvert lVert psi^perp rVert - lVert phi^perp rVert rvert$, and
$lVert psi^perp - phi^perp rVert le lVert psi - phi rVert$
answered Jan 26 at 1:40
Keith McClaryKeith McClary
8381412
$endgroup$
$begingroup$
Thank you! What prompted you to use projections? Is there something that I'm missing that should have led me to using the projections of $psi$ and $phi$?
$endgroup$
– Alexander H. R.
Jan 26 at 18:11
$begingroup$
I couldn't do anything with the projections on $x$ so I tried to visualise what it looked like in 3D, which led to $lVert psi^perp - phi^perp rVert le lVert psi - phi rVert$.
$endgroup$
– Keith McClary
Jan 27 at 4:53
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: "69"
};
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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%2fmath.stackexchange.com%2fquestions%2f3087665%2finequality-relating-the-probabilities-of-a-quantum-state-to-the-euclidean-distan%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $psi^perp$, $phi^perp$ be the projections of $psi$, $phi$ on the subspace perpendicular to $x$.
$ |langle x|psirangle |^2 = 1- lVert psi^perp rVert ^2 $ and $ |langle x|phirangle|^2 = 1- lVert phi^perp rVert ^2 $
so the LHS is
$$ lvert lVert psi^perp rVert ^2 - lVert phi^perp rVert ^2 rvert = lvert lVert psi^perp rVert + lVert phi^perp rVert rvert quad lvert lVert psi^perp rVert - lVert phi^perp rVert rvert le 2lvert lVert psi^perp rVert - lVert phi^perp rVert rvert$$
For given lengths, $lVert psi^perp - phi^perp rVert$ is smallest when the vectors are in the same direction, when it equals $lvert lVert psi^perp rVert - lVert phi^perp rVert rvert$, and
$lVert psi^perp - phi^perp rVert le lVert psi - phi rVert$
answered Jan 26 at 1:40
Keith McClaryKeith McClary
8381412
$endgroup$
$begingroup$
Thank you! What prompted you to use projections? Is there something that I'm missing that should have led me to using the projections of $psi$ and $phi$?
$endgroup$
– Alexander H. R.
Jan 26 at 18:11
$begingroup$
I couldn't do anything with the projections on $x$ so I tried to visualise what it looked like in 3D, which led to $lVert psi^perp - phi^perp rVert le lVert psi - phi rVert$.
$endgroup$
– Keith McClary
Jan 27 at 4:53
add a comment |
$begingroup$
Let $psi^perp$, $phi^perp$ be the projections of $psi$, $phi$ on the subspace perpendicular to $x$.
$ |langle x|psirangle |^2 = 1- lVert psi^perp rVert ^2 $ and $ |langle x|phirangle|^2 = 1- lVert phi^perp rVert ^2 $
so the LHS is
$$ lvert lVert psi^perp rVert ^2 - lVert phi^perp rVert ^2 rvert = lvert lVert psi^perp rVert + lVert phi^perp rVert rvert quad lvert lVert psi^perp rVert - lVert phi^perp rVert rvert le 2lvert lVert psi^perp rVert - lVert phi^perp rVert rvert$$
For given lengths, $lVert psi^perp - phi^perp rVert$ is smallest when the vectors are in the same direction, when it equals $lvert lVert psi^perp rVert - lVert phi^perp rVert rvert$, and
$lVert psi^perp - phi^perp rVert le lVert psi - phi rVert$
answered Jan 26 at 1:40
Keith McClaryKeith McClary
8381412
$endgroup$
$begingroup$
Thank you! What prompted you to use projections? Is there something that I'm missing that should have led me to using the projections of $psi$ and $phi$?
$endgroup$
– Alexander H. R.
Jan 26 at 18:11
$begingroup$
I couldn't do anything with the projections on $x$ so I tried to visualise what it looked like in 3D, which led to $lVert psi^perp - phi^perp rVert le lVert psi - phi rVert$.
$endgroup$
– Keith McClary
Jan 27 at 4:53
add a comment |
$begingroup$
Let $psi^perp$, $phi^perp$ be the projections of $psi$, $phi$ on the subspace perpendicular to $x$.
$ |langle x|psirangle |^2 = 1- lVert psi^perp rVert ^2 $ and $ |langle x|phirangle|^2 = 1- lVert phi^perp rVert ^2 $
so the LHS is
$$ lvert lVert psi^perp rVert ^2 - lVert phi^perp rVert ^2 rvert = lvert lVert psi^perp rVert + lVert phi^perp rVert rvert quad lvert lVert psi^perp rVert - lVert phi^perp rVert rvert le 2lvert lVert psi^perp rVert - lVert phi^perp rVert rvert$$
For given lengths, $lVert psi^perp - phi^perp rVert$ is smallest when the vectors are in the same direction, when it equals $lvert lVert psi^perp rVert - lVert phi^perp rVert rvert$, and
$lVert psi^perp - phi^perp rVert le lVert psi - phi rVert$
answered Jan 26 at 1:40
Keith McClaryKeith McClary
8381412
$endgroup$
Let $psi^perp$, $phi^perp$ be the projections of $psi$, $phi$ on the subspace perpendicular to $x$.
$ |langle x|psirangle |^2 = 1- lVert psi^perp rVert ^2 $ and $ |langle x|phirangle|^2 = 1- lVert phi^perp rVert ^2 $
so the LHS is
$$ lvert lVert psi^perp rVert ^2 - lVert phi^perp rVert ^2 rvert = lvert lVert psi^perp rVert + lVert phi^perp rVert rvert quad lvert lVert psi^perp rVert - lVert phi^perp rVert rvert le 2lvert lVert psi^perp rVert - lVert phi^perp rVert rvert$$
For given lengths, $lVert psi^perp - phi^perp rVert$ is smallest when the vectors are in the same direction, when it equals $lvert lVert psi^perp rVert - lVert phi^perp rVert rvert$, and
$lVert psi^perp - phi^perp rVert le lVert psi - phi rVert$
answered Jan 26 at 1:40
Keith McClaryKeith McClary
8381412
edited Jan 26 at 1:48
answered Jan 26 at 1:40
Keith McClaryKeith McClary
8381412
answered Jan 26 at 1:40
Keith McClaryKeith McClary
8381412
answered Jan 26 at 1:40
Keith McClaryKeith McClary
8381412
8381412
$begingroup$
Thank you! What prompted you to use projections? Is there something that I'm missing that should have led me to using the projections of $psi$ and $phi$?
$endgroup$
– Alexander H. R.
Jan 26 at 18:11
$begingroup$
I couldn't do anything with the projections on $x$ so I tried to visualise what it looked like in 3D, which led to $lVert psi^perp - phi^perp rVert le lVert psi - phi rVert$.
$endgroup$
– Keith McClary
Jan 27 at 4:53
add a comment |
$begingroup$
Thank you! What prompted you to use projections? Is there something that I'm missing that should have led me to using the projections of $psi$ and $phi$?
$endgroup$
– Alexander H. R.
Jan 26 at 18:11
$begingroup$
I couldn't do anything with the projections on $x$ so I tried to visualise what it looked like in 3D, which led to $lVert psi^perp - phi^perp rVert le lVert psi - phi rVert$.
$endgroup$
– Keith McClary
Jan 27 at 4:53
$begingroup$
Thank you! What prompted you to use projections? Is there something that I'm missing that should have led me to using the projections of $psi$ and $phi$?
$endgroup$
– Alexander H. R.
Jan 26 at 18:11
$begingroup$
Thank you! What prompted you to use projections? Is there something that I'm missing that should have led me to using the projections of $psi$ and $phi$?
$endgroup$
– Alexander H. R.
Jan 26 at 18:11
$begingroup$
I couldn't do anything with the projections on $x$ so I tried to visualise what it looked like in 3D, which led to $lVert psi^perp - phi^perp rVert le lVert psi - phi rVert$.
$endgroup$
– Keith McClary
Jan 27 at 4:53
$begingroup$
I couldn't do anything with the projections on $x$ so I tried to visualise what it looked like in 3D, which led to $lVert psi^perp - phi^perp rVert le lVert psi - phi rVert$.
$endgroup$
– Keith McClary
Jan 27 at 4:53
add a comment |
Thanks for contributing an answer to Mathematics 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.
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%2fmath.stackexchange.com%2fquestions%2f3087665%2finequality-relating-the-probabilities-of-a-quantum-state-to-the-euclidean-distan%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
$begingroup$
I assume you're in a finite-dimensional space $H?$ Otherwise Inequality 3.1 wouldn't mean a whole lot.
$endgroup$
– Adrian Keister
Jan 25 at 22:22
$begingroup$
Yes, I will edit the question to clarify.
$endgroup$
– Alexander H. R.
Jan 25 at 22:40
$begingroup$
@KeithMcClary I'm not entirely sure I understand what you are trying to say or how to use it. Would you be able to elaborate?
$endgroup$
– Alexander H. R.
Jan 25 at 23:38
$begingroup$
Disregard. I see they are unit vectors.
$endgroup$
– Keith McClary
Jan 25 at 23:40