Vjylku

Manipulating a general length function

Quenching swords in dragon blood; why?

Using loops to create tables

I am on the US no-fly list. What can I do in order to be allowed on flights which go through US airspace?

page split between longtable caption and table

Why zero tolerance on nudity in space?

Could flying insects re-enter the Earth's atmosphere from space without burning up?

Why did Jodrell Bank assist the Soviet Union to collect data from their spacecraft in the mid 1960's?

How should I handle players who ignore the session zero agreement?

Rear brake cable temporary fix possible?

Issues with new Macs: hardware makes them difficult to use … what options might be available in the future?

It took me a lot of time to make this, pls like. (YouTube Comments #1)

Slow moving projectiles from a hand-held weapon - how do they reach the target?

What is this metal M-shaped device for?

Number of FLOP (Floating Point Operations) for exponentiation

How do we succintly describe a boxing match?

Dilemma of explaining to interviewer that he is the reason for declining second interview

Is it a fallacy if someone claims they need an explanation for every word of your argument to the point where they don't understand common terms?

Finding radius of circle

Is there any differences between “gucken” and “schauen”?

"On one hand" vs "on the one hand."

Do authors have to be politically correct in article-writing?

Why do members of Congress in committee hearings ask witnesses the same question multiple times?

Does my logo design convey the right feelings for a University Student's Council?

Does the “particle exchange” operator have any validity?

If wavefunction is just a probability function, how does an electron interfere with itselfIs it accurate to say “a wavefunction is a function of particle positions or momenta”?In Bohmian mechanics, how does the particle's position affect where a particle is detected?What is meant by the term “single particle state”Does quantum mechanics imply that particles have no trajectories?Why is the phase picked up during identical particle exchange a topological invariant?Are quantum mechanics non-deterministic?Creation and annihilation operators for any operator which has discrete and countable eigenvalues?Do the exchange operator and Hamiltonian commute for non-identical particles?Would every particle in the universe not have some form of measurement occurring at any given time?

2

$begingroup$

In introductory quantum mechanics books the topic of identical particles often introduces a "particle exchange" operator. This operator, when applied to a multi-particle wave-function, exchanges the positions of two identical particles.

However, it seems to me that this is a non-physical thing. Particles can't really "exchange positions" can they? Does such an operator really have validity?

quantum-mechanics

share|cite|improve this question

asked 53 mins ago

Paul YoungPaul Young

1,049112

$endgroup$

add a comment | 

2

$begingroup$

In introductory quantum mechanics books the topic of identical particles often introduces a "particle exchange" operator. This operator, when applied to a multi-particle wave-function, exchanges the positions of two identical particles.

However, it seems to me that this is a non-physical thing. Particles can't really "exchange positions" can they? Does such an operator really have validity?

quantum-mechanics

share|cite|improve this question

asked 53 mins ago

Paul YoungPaul Young

1,049112

$endgroup$

add a comment | 

$begingroup$

In introductory quantum mechanics books the topic of identical particles often introduces a "particle exchange" operator. This operator, when applied to a multi-particle wave-function, exchanges the positions of two identical particles.

However, it seems to me that this is a non-physical thing. Particles can't really "exchange positions" can they? Does such an operator really have validity?

quantum-mechanics

share|cite|improve this question

asked 53 mins ago

Paul YoungPaul Young

1,049112

$endgroup$

share|cite|improve this question

asked 53 mins ago

Paul YoungPaul Young

1,049112

share|cite|improve this question

asked 53 mins ago

Paul YoungPaul Young

1,049112

asked 53 mins ago

Paul YoungPaul Young

1,049112

asked 53 mins ago

Paul YoungPaul Young

1,049112

1 Answer
1

active

oldest

votes

4

$begingroup$

If I'm reading your question right, I think you're having a relatively common issue. Feynman himself had the same issue when confronted with a creation operator for an electron. "How can an electron really be created? It violates the conservation of charge!"

The point is that not every operator needs to represent a physical realizable time evolution. Quite often, you just want to consider the formal properties of an abstract operator, regardless of whether it has any direct meaning itself. (In fact, at the level of quantum field theory, it certainly doesn't -- the particle exchange operator does nothing whatsoever, because the particles are identical. The simpler formalism of nonrelativistic quantum mechanics doesn't have that built in, which is why it's useful to consider the exchange operator.)

You've already seen this attitude earlier in quantum mechanics. For example, the position operator $hat{x}$ is not a "physical thing", in the sense that if you have a state $|psi rangle$, $hat{x} |psi rangle$ is not necessarily a sensible physical state. Instead $hat{x}$ is useful because it represents an observable quantity. In quantum mechanics, operators are our basic tools. Some represent observables, some represent physical time evolutions, and some represent neither, and there's nothing wrong with those last ones.

share|cite|improve this answer

answered 37 mins ago

knzhouknzhou

44.6k11121215

$endgroup$

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: "151"
};
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
});


}
});

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f464019%2fdoes-the-particle-exchange-operator-have-any-validity%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

4

$begingroup$

If I'm reading your question right, I think you're having a relatively common issue. Feynman himself had the same issue when confronted with a creation operator for an electron. "How can an electron really be created? It violates the conservation of charge!"

The point is that not every operator needs to represent a physical realizable time evolution. Quite often, you just want to consider the formal properties of an abstract operator, regardless of whether it has any direct meaning itself. (In fact, at the level of quantum field theory, it certainly doesn't -- the particle exchange operator does nothing whatsoever, because the particles are identical. The simpler formalism of nonrelativistic quantum mechanics doesn't have that built in, which is why it's useful to consider the exchange operator.)

You've already seen this attitude earlier in quantum mechanics. For example, the position operator $hat{x}$ is not a "physical thing", in the sense that if you have a state $|psi rangle$, $hat{x} |psi rangle$ is not necessarily a sensible physical state. Instead $hat{x}$ is useful because it represents an observable quantity. In quantum mechanics, operators are our basic tools. Some represent observables, some represent physical time evolutions, and some represent neither, and there's nothing wrong with those last ones.

share|cite|improve this answer

answered 37 mins ago

knzhouknzhou

44.6k11121215

$endgroup$

add a comment | 

4

$begingroup$

If I'm reading your question right, I think you're having a relatively common issue. Feynman himself had the same issue when confronted with a creation operator for an electron. "How can an electron really be created? It violates the conservation of charge!"

The point is that not every operator needs to represent a physical realizable time evolution. Quite often, you just want to consider the formal properties of an abstract operator, regardless of whether it has any direct meaning itself. (In fact, at the level of quantum field theory, it certainly doesn't -- the particle exchange operator does nothing whatsoever, because the particles are identical. The simpler formalism of nonrelativistic quantum mechanics doesn't have that built in, which is why it's useful to consider the exchange operator.)

You've already seen this attitude earlier in quantum mechanics. For example, the position operator $hat{x}$ is not a "physical thing", in the sense that if you have a state $|psi rangle$, $hat{x} |psi rangle$ is not necessarily a sensible physical state. Instead $hat{x}$ is useful because it represents an observable quantity. In quantum mechanics, operators are our basic tools. Some represent observables, some represent physical time evolutions, and some represent neither, and there's nothing wrong with those last ones.

share|cite|improve this answer

answered 37 mins ago

knzhouknzhou

44.6k11121215

$endgroup$

add a comment | 

$begingroup$

If I'm reading your question right, I think you're having a relatively common issue. Feynman himself had the same issue when confronted with a creation operator for an electron. "How can an electron really be created? It violates the conservation of charge!"

The point is that not every operator needs to represent a physical realizable time evolution. Quite often, you just want to consider the formal properties of an abstract operator, regardless of whether it has any direct meaning itself. (In fact, at the level of quantum field theory, it certainly doesn't -- the particle exchange operator does nothing whatsoever, because the particles are identical. The simpler formalism of nonrelativistic quantum mechanics doesn't have that built in, which is why it's useful to consider the exchange operator.)

You've already seen this attitude earlier in quantum mechanics. For example, the position operator $hat{x}$ is not a "physical thing", in the sense that if you have a state $|psi rangle$, $hat{x} |psi rangle$ is not necessarily a sensible physical state. Instead $hat{x}$ is useful because it represents an observable quantity. In quantum mechanics, operators are our basic tools. Some represent observables, some represent physical time evolutions, and some represent neither, and there's nothing wrong with those last ones.

share|cite|improve this answer

answered 37 mins ago

knzhouknzhou

44.6k11121215

$endgroup$

share|cite|improve this answer

answered 37 mins ago

knzhouknzhou

44.6k11121215

answered 37 mins ago

knzhouknzhou

44.6k11121215

answered 37 mins ago

knzhouknzhou

44.6k11121215