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







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    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









    $endgroup$















      2












      2








      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









      $endgroup$




      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






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      share|cite|improve this question











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      asked 55 mins ago









      Paul YoungPaul Young

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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.






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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.






            share|cite|improve this answer









            $endgroup$


















              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









              $endgroup$
















                4












                4








                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









                $endgroup$



                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 39 mins ago









                knzhouknzhou

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                44.6k11121215






























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