Is this a “good enough” statement of Wigner's theorem from Quantum Mechanics?
$begingroup$
I'm a fourth year physics and math student who is writing up a report on some quantum mechanical symmetries and their consequences. The "audience" for my paper are either senior physics majors or first year graduate students in physics. Wigner's theorem's mathematical content is unfortunately beyond the scope of my report so I decided to "water it down". My question is this: is this an acceptably simplified version of Wigner's theorem?
Theorem (Wigner)
Let $Psi, Phi$ be arbitrary state vectors in a Hilbert space $mathscr{H}$. Suppose $delta: mathscr{H} mapsto mathscr{H}$ is bijective and Wigner symmetric (that is, under the mapping $delta$, the transition probability $langle Psi | Phi rangle$ is unchanged). Then
- there exists a unitary or anti-unitary operator $U$ such that
$$delta(psi) = U psi U^{-1},$$
$U$ is unique up to an arbitrary phase factor.
Is there anything I'm misunderstanding with this formulation of Wigner's theorem?
mathematical-physics quantum-mechanics
asked Dec 6 '18 at 22:19
talrefaetalrefae
446
$endgroup$
add a comment |
$begingroup$
I'm a fourth year physics and math student who is writing up a report on some quantum mechanical symmetries and their consequences. The "audience" for my paper are either senior physics majors or first year graduate students in physics. Wigner's theorem's mathematical content is unfortunately beyond the scope of my report so I decided to "water it down". My question is this: is this an acceptably simplified version of Wigner's theorem?
Theorem (Wigner)
Let $Psi, Phi$ be arbitrary state vectors in a Hilbert space $mathscr{H}$. Suppose $delta: mathscr{H} mapsto mathscr{H}$ is bijective and Wigner symmetric (that is, under the mapping $delta$, the transition probability $langle Psi | Phi rangle$ is unchanged). Then
- there exists a unitary or anti-unitary operator $U$ such that
$$delta(psi) = U psi U^{-1},$$
$U$ is unique up to an arbitrary phase factor.
Is there anything I'm misunderstanding with this formulation of Wigner's theorem?
mathematical-physics quantum-mechanics
asked Dec 6 '18 at 22:19
talrefaetalrefae
446
$endgroup$
1
$begingroup$
Cross-posted to Physics.
$endgroup$
– rob
Dec 9 '18 at 22:23
add a comment |
$begingroup$
I'm a fourth year physics and math student who is writing up a report on some quantum mechanical symmetries and their consequences. The "audience" for my paper are either senior physics majors or first year graduate students in physics. Wigner's theorem's mathematical content is unfortunately beyond the scope of my report so I decided to "water it down". My question is this: is this an acceptably simplified version of Wigner's theorem?
Theorem (Wigner)
Let $Psi, Phi$ be arbitrary state vectors in a Hilbert space $mathscr{H}$. Suppose $delta: mathscr{H} mapsto mathscr{H}$ is bijective and Wigner symmetric (that is, under the mapping $delta$, the transition probability $langle Psi | Phi rangle$ is unchanged). Then
- there exists a unitary or anti-unitary operator $U$ such that
$$delta(psi) = U psi U^{-1},$$
$U$ is unique up to an arbitrary phase factor.
Is there anything I'm misunderstanding with this formulation of Wigner's theorem?
mathematical-physics quantum-mechanics
asked Dec 6 '18 at 22:19
talrefaetalrefae
446
$endgroup$
I'm a fourth year physics and math student who is writing up a report on some quantum mechanical symmetries and their consequences. The "audience" for my paper are either senior physics majors or first year graduate students in physics. Wigner's theorem's mathematical content is unfortunately beyond the scope of my report so I decided to "water it down". My question is this: is this an acceptably simplified version of Wigner's theorem?
Theorem (Wigner)
Let $Psi, Phi$ be arbitrary state vectors in a Hilbert space $mathscr{H}$. Suppose $delta: mathscr{H} mapsto mathscr{H}$ is bijective and Wigner symmetric (that is, under the mapping $delta$, the transition probability $langle Psi | Phi rangle$ is unchanged). Then
- there exists a unitary or anti-unitary operator $U$ such that
$$delta(psi) = U psi U^{-1},$$
$U$ is unique up to an arbitrary phase factor.
Is there anything I'm misunderstanding with this formulation of Wigner's theorem?
mathematical-physics quantum-mechanics
mathematical-physics quantum-mechanics
asked Dec 6 '18 at 22:19
talrefaetalrefae
446
asked Dec 6 '18 at 22:19
talrefaetalrefae
446
asked Dec 6 '18 at 22:19
talrefaetalrefae
446
asked Dec 6 '18 at 22:19
talrefaetalrefae
446
asked Dec 6 '18 at 22:19
talrefaetalrefae
446
446
1
$begingroup$
Cross-posted to Physics.
$endgroup$
– rob
Dec 9 '18 at 22:23
add a comment |
1
$begingroup$
Cross-posted to Physics.
$endgroup$
– rob
Dec 9 '18 at 22:23
1
1
$begingroup$
Cross-posted to Physics.
$endgroup$
– rob
Dec 9 '18 at 22:23
$begingroup$
Cross-posted to Physics.
$endgroup$
– rob
Dec 9 '18 at 22:23
add a comment |
0
active
oldest
votes
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%2f3029130%2fis-this-a-good-enough-statement-of-wigners-theorem-from-quantum-mechanics%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3029130%2fis-this-a-good-enough-statement-of-wigners-theorem-from-quantum-mechanics%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
1
$begingroup$
Cross-posted to Physics.
$endgroup$
– rob
Dec 9 '18 at 22:23