Does the “particle exchange” operator have any validity?
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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
asked 8 hours ago
Paul YoungPaul Young
1,074112
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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
asked 8 hours ago
Paul YoungPaul Young
1,074112
$endgroup$
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
asked 8 hours ago
Paul YoungPaul Young
1,074112
$endgroup$
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
quantum-mechanics
asked 8 hours ago
Paul YoungPaul Young
1,074112
asked 8 hours ago
Paul YoungPaul Young
1,074112
asked 8 hours ago
Paul YoungPaul Young
1,074112
asked 8 hours ago
Paul YoungPaul Young
1,074112
asked 8 hours ago
Paul YoungPaul Young
1,074112
1,074112
add a comment |
add a comment |
1 Answer
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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.
answered 7 hours ago
knzhouknzhou
44.7k11122216
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1 Answer
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1 Answer
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$begingroup$
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.
answered 7 hours ago
knzhouknzhou
44.7k11122216
$endgroup$
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.
answered 7 hours ago
knzhouknzhou
44.7k11122216
$endgroup$
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.
answered 7 hours ago
knzhouknzhou
44.7k11122216
$endgroup$
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.
answered 7 hours ago
knzhouknzhou
44.7k11122216
answered 7 hours ago
knzhouknzhou
44.7k11122216
answered 7 hours ago
knzhouknzhou
44.7k11122216
answered 7 hours ago
knzhouknzhou
44.7k11122216
44.7k11122216
add a comment |
add a comment |
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