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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 1 hour ago

Paul YoungPaul Young

1,049112

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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 1 hour 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 1 hour ago

Paul YoungPaul Young

1,049112

$endgroup$

share|cite|improve this question

asked 1 hour ago

Paul YoungPaul Young

1,049112

share|cite|improve this question

asked 1 hour ago

Paul YoungPaul Young

1,049112

asked 1 hour ago

Paul YoungPaul Young

1,049112

asked 1 hour ago

Paul YoungPaul Young

1,049112

1 Answer
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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 1 hour ago

knzhouknzhou

44.6k11121215

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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 1 hour 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 1 hour 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 1 hour ago

knzhouknzhou

44.6k11121215

$endgroup$

share|cite|improve this answer

answered 1 hour ago

knzhouknzhou

44.6k11121215

answered 1 hour ago

knzhouknzhou

44.6k11121215

answered 1 hour ago

knzhouknzhou

44.6k11121215