How can two electrons repel if it's impossible for free electrons to absorb or emit energy?











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There is no acceptable/viable mechanism for a free electron to absorb or emit energy, without violating energy or momentum conservation. So its wavefunction cannot collapse into becoming a particle, right? How do 2 free electrons repel each other then?










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  • 3




    "There is no known mechanism for free electron to absorb or emit energy"? Who said that?
    – knzhou
    4 hours ago










  • @knzhou: it will violate simultaneous conservation of momentum and energy
    – user6760
    4 hours ago










  • Free electron can't absorb a photon and can a free particle absorb emit photons
    – user6760
    2 hours ago












  • if your concern is that repelling is cause by the exchange of virtual photons, and a free electron can neither absorb nor emit a photon with violating conversation laws, then don't worry: virtual photons are off shell--they have the energy and momentum required to converse energy and momentum. Unless you are using old fashioned perturbation theory, then energy is violated, but only briefly.
    – JEB
    1 min ago

















up vote
3
down vote

favorite
2












There is no acceptable/viable mechanism for a free electron to absorb or emit energy, without violating energy or momentum conservation. So its wavefunction cannot collapse into becoming a particle, right? How do 2 free electrons repel each other then?










share|cite|improve this question




















  • 3




    "There is no known mechanism for free electron to absorb or emit energy"? Who said that?
    – knzhou
    4 hours ago










  • @knzhou: it will violate simultaneous conservation of momentum and energy
    – user6760
    4 hours ago










  • Free electron can't absorb a photon and can a free particle absorb emit photons
    – user6760
    2 hours ago












  • if your concern is that repelling is cause by the exchange of virtual photons, and a free electron can neither absorb nor emit a photon with violating conversation laws, then don't worry: virtual photons are off shell--they have the energy and momentum required to converse energy and momentum. Unless you are using old fashioned perturbation theory, then energy is violated, but only briefly.
    – JEB
    1 min ago















up vote
3
down vote

favorite
2









up vote
3
down vote

favorite
2






2





There is no acceptable/viable mechanism for a free electron to absorb or emit energy, without violating energy or momentum conservation. So its wavefunction cannot collapse into becoming a particle, right? How do 2 free electrons repel each other then?










share|cite|improve this question















There is no acceptable/viable mechanism for a free electron to absorb or emit energy, without violating energy or momentum conservation. So its wavefunction cannot collapse into becoming a particle, right? How do 2 free electrons repel each other then?







electrons wavefunction-collapse






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edited 1 hour ago

























asked 4 hours ago









user6760

2,23511734




2,23511734








  • 3




    "There is no known mechanism for free electron to absorb or emit energy"? Who said that?
    – knzhou
    4 hours ago










  • @knzhou: it will violate simultaneous conservation of momentum and energy
    – user6760
    4 hours ago










  • Free electron can't absorb a photon and can a free particle absorb emit photons
    – user6760
    2 hours ago












  • if your concern is that repelling is cause by the exchange of virtual photons, and a free electron can neither absorb nor emit a photon with violating conversation laws, then don't worry: virtual photons are off shell--they have the energy and momentum required to converse energy and momentum. Unless you are using old fashioned perturbation theory, then energy is violated, but only briefly.
    – JEB
    1 min ago
















  • 3




    "There is no known mechanism for free electron to absorb or emit energy"? Who said that?
    – knzhou
    4 hours ago










  • @knzhou: it will violate simultaneous conservation of momentum and energy
    – user6760
    4 hours ago










  • Free electron can't absorb a photon and can a free particle absorb emit photons
    – user6760
    2 hours ago












  • if your concern is that repelling is cause by the exchange of virtual photons, and a free electron can neither absorb nor emit a photon with violating conversation laws, then don't worry: virtual photons are off shell--they have the energy and momentum required to converse energy and momentum. Unless you are using old fashioned perturbation theory, then energy is violated, but only briefly.
    – JEB
    1 min ago










3




3




"There is no known mechanism for free electron to absorb or emit energy"? Who said that?
– knzhou
4 hours ago




"There is no known mechanism for free electron to absorb or emit energy"? Who said that?
– knzhou
4 hours ago












@knzhou: it will violate simultaneous conservation of momentum and energy
– user6760
4 hours ago




@knzhou: it will violate simultaneous conservation of momentum and energy
– user6760
4 hours ago












Free electron can't absorb a photon and can a free particle absorb emit photons
– user6760
2 hours ago






Free electron can't absorb a photon and can a free particle absorb emit photons
– user6760
2 hours ago














if your concern is that repelling is cause by the exchange of virtual photons, and a free electron can neither absorb nor emit a photon with violating conversation laws, then don't worry: virtual photons are off shell--they have the energy and momentum required to converse energy and momentum. Unless you are using old fashioned perturbation theory, then energy is violated, but only briefly.
– JEB
1 min ago






if your concern is that repelling is cause by the exchange of virtual photons, and a free electron can neither absorb nor emit a photon with violating conversation laws, then don't worry: virtual photons are off shell--they have the energy and momentum required to converse energy and momentum. Unless you are using old fashioned perturbation theory, then energy is violated, but only briefly.
– JEB
1 min ago












4 Answers
4






active

oldest

votes

















up vote
4
down vote













It is true that the reactions
$$e + gamma to e, quad e to e + gamma$$
cannot occur without violating energy or momentum conservation. But that doesn't mean that electrons can't interact with anything! For example, scattering
$$e + gamma to e + gamma$$
is perfectly allowed. And a classical electromagnetic field is built out of many photons, so the interaction of an electron with such a field can be thought of as an interaction with many photons at once. There are plenty of ways a free electron can interact without violating energy or momentum conservation, so there's no problem here.






share|cite|improve this answer




























    up vote
    1
    down vote













    Your first statement is false: energy can indeed be added at will to electrons by accelerating them with electrostatic charge distributions, as for example in the case of rapidly varying radio frequency (electromagnetic) fields. Neither energy nor momentum conservation is violated in this case. Search on SLAC for more details about this.



    Your other questions are unclear. I recommend you do the search, read a bit, and return here if you have further questions.






    share|cite|improve this answer





















    • i know when electron in an atom absorb energy it becomes excited, as it goes from excited to lower state energy is being released as photon and in ur case it is radiowave. However I'm at a loss when I try imagine 2 free electrons (particle/wave) can repell each other?
      – user6760
      2 hours ago










    • @user6760 The free electron can also gain energy from photon. That also leads to a measurement problem.
      – Bill Alsept
      1 hour ago










    • @BillAlsept: I'm referring to free electron not valence electron
      – user6760
      1 hour ago










    • @user6760 if you fire free electrons through a double slit experiment and try to observe them, the photons will physically interfere.
      – Bill Alsept
      55 mins ago












    • @BillAlsept: of course 1 free electron interacts with one atom
      – user6760
      48 mins ago


















    up vote
    0
    down vote













    To resolve this paradox requires study of time dependent perturbation theory; solving Schrodinger's equation with a time dependent perturbation corresponding to the interaction time of two particles.



    If you do this you arrive at the following conclusions:



    A single free electron cannot absorb a free photon ( $e + gamma to e$ is not a valid interaction)



    A single free electron cannot emit a free photon ( $e to e + gamma$ is not a valid interaction)



    However, two electrons can scatter by exchange of energy ( $ e + e to e + e$ is a valid interaction)



    In this later case it is common to refer to this process being due to exchange of "a virtual photon" between the two electrons. But this is just a description of the calculation of time dependent perturbation theory.






    share|cite|improve this answer




























      up vote
      0
      down vote













      Basically, because the process doesn't simultaneously conserve energy and momentum; this is why we say that it's mediated by a virtual photon.



      In more technical language, this means that the photons that are exchanged between two interacting electrons are allowed to be "off shell", where the "shell" is the relationship $E^2 = p^2c^2$ (for a massless particle). Real particles are required to be on-shell, but virtual particles are allowed to stray from that condition, at least by some amount.



      In the Feynman-diagram description of the scattering between two electrons you often find diagrams where one electron emits a photon which is then absorbed by the other; this is a virtual photon and as such both the 'emission' and 'absorption' processes are exempt from these considerations.



      As always, though, it bears repeating that Feynman diagrams are calculational tools, and none of the virtual particles that appear in those diagrams actually physically exist in any definable sense. The fact that the 'emission' and 'absorption' processes, as well as the virtual photon itself, seem to defy energy or momentum conservation is purely a quirk of the way in which we've chosen to interpret limited chunks of our calculation.






      share|cite|improve this answer





















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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

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        active

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        up vote
        4
        down vote













        It is true that the reactions
        $$e + gamma to e, quad e to e + gamma$$
        cannot occur without violating energy or momentum conservation. But that doesn't mean that electrons can't interact with anything! For example, scattering
        $$e + gamma to e + gamma$$
        is perfectly allowed. And a classical electromagnetic field is built out of many photons, so the interaction of an electron with such a field can be thought of as an interaction with many photons at once. There are plenty of ways a free electron can interact without violating energy or momentum conservation, so there's no problem here.






        share|cite|improve this answer

























          up vote
          4
          down vote













          It is true that the reactions
          $$e + gamma to e, quad e to e + gamma$$
          cannot occur without violating energy or momentum conservation. But that doesn't mean that electrons can't interact with anything! For example, scattering
          $$e + gamma to e + gamma$$
          is perfectly allowed. And a classical electromagnetic field is built out of many photons, so the interaction of an electron with such a field can be thought of as an interaction with many photons at once. There are plenty of ways a free electron can interact without violating energy or momentum conservation, so there's no problem here.






          share|cite|improve this answer























            up vote
            4
            down vote










            up vote
            4
            down vote









            It is true that the reactions
            $$e + gamma to e, quad e to e + gamma$$
            cannot occur without violating energy or momentum conservation. But that doesn't mean that electrons can't interact with anything! For example, scattering
            $$e + gamma to e + gamma$$
            is perfectly allowed. And a classical electromagnetic field is built out of many photons, so the interaction of an electron with such a field can be thought of as an interaction with many photons at once. There are plenty of ways a free electron can interact without violating energy or momentum conservation, so there's no problem here.






            share|cite|improve this answer












            It is true that the reactions
            $$e + gamma to e, quad e to e + gamma$$
            cannot occur without violating energy or momentum conservation. But that doesn't mean that electrons can't interact with anything! For example, scattering
            $$e + gamma to e + gamma$$
            is perfectly allowed. And a classical electromagnetic field is built out of many photons, so the interaction of an electron with such a field can be thought of as an interaction with many photons at once. There are plenty of ways a free electron can interact without violating energy or momentum conservation, so there's no problem here.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 4 hours ago









            knzhou

            40.5k11113194




            40.5k11113194






















                up vote
                1
                down vote













                Your first statement is false: energy can indeed be added at will to electrons by accelerating them with electrostatic charge distributions, as for example in the case of rapidly varying radio frequency (electromagnetic) fields. Neither energy nor momentum conservation is violated in this case. Search on SLAC for more details about this.



                Your other questions are unclear. I recommend you do the search, read a bit, and return here if you have further questions.






                share|cite|improve this answer





















                • i know when electron in an atom absorb energy it becomes excited, as it goes from excited to lower state energy is being released as photon and in ur case it is radiowave. However I'm at a loss when I try imagine 2 free electrons (particle/wave) can repell each other?
                  – user6760
                  2 hours ago










                • @user6760 The free electron can also gain energy from photon. That also leads to a measurement problem.
                  – Bill Alsept
                  1 hour ago










                • @BillAlsept: I'm referring to free electron not valence electron
                  – user6760
                  1 hour ago










                • @user6760 if you fire free electrons through a double slit experiment and try to observe them, the photons will physically interfere.
                  – Bill Alsept
                  55 mins ago












                • @BillAlsept: of course 1 free electron interacts with one atom
                  – user6760
                  48 mins ago















                up vote
                1
                down vote













                Your first statement is false: energy can indeed be added at will to electrons by accelerating them with electrostatic charge distributions, as for example in the case of rapidly varying radio frequency (electromagnetic) fields. Neither energy nor momentum conservation is violated in this case. Search on SLAC for more details about this.



                Your other questions are unclear. I recommend you do the search, read a bit, and return here if you have further questions.






                share|cite|improve this answer





















                • i know when electron in an atom absorb energy it becomes excited, as it goes from excited to lower state energy is being released as photon and in ur case it is radiowave. However I'm at a loss when I try imagine 2 free electrons (particle/wave) can repell each other?
                  – user6760
                  2 hours ago










                • @user6760 The free electron can also gain energy from photon. That also leads to a measurement problem.
                  – Bill Alsept
                  1 hour ago










                • @BillAlsept: I'm referring to free electron not valence electron
                  – user6760
                  1 hour ago










                • @user6760 if you fire free electrons through a double slit experiment and try to observe them, the photons will physically interfere.
                  – Bill Alsept
                  55 mins ago












                • @BillAlsept: of course 1 free electron interacts with one atom
                  – user6760
                  48 mins ago













                up vote
                1
                down vote










                up vote
                1
                down vote









                Your first statement is false: energy can indeed be added at will to electrons by accelerating them with electrostatic charge distributions, as for example in the case of rapidly varying radio frequency (electromagnetic) fields. Neither energy nor momentum conservation is violated in this case. Search on SLAC for more details about this.



                Your other questions are unclear. I recommend you do the search, read a bit, and return here if you have further questions.






                share|cite|improve this answer












                Your first statement is false: energy can indeed be added at will to electrons by accelerating them with electrostatic charge distributions, as for example in the case of rapidly varying radio frequency (electromagnetic) fields. Neither energy nor momentum conservation is violated in this case. Search on SLAC for more details about this.



                Your other questions are unclear. I recommend you do the search, read a bit, and return here if you have further questions.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 3 hours ago









                niels nielsen

                15.3k42649




                15.3k42649












                • i know when electron in an atom absorb energy it becomes excited, as it goes from excited to lower state energy is being released as photon and in ur case it is radiowave. However I'm at a loss when I try imagine 2 free electrons (particle/wave) can repell each other?
                  – user6760
                  2 hours ago










                • @user6760 The free electron can also gain energy from photon. That also leads to a measurement problem.
                  – Bill Alsept
                  1 hour ago










                • @BillAlsept: I'm referring to free electron not valence electron
                  – user6760
                  1 hour ago










                • @user6760 if you fire free electrons through a double slit experiment and try to observe them, the photons will physically interfere.
                  – Bill Alsept
                  55 mins ago












                • @BillAlsept: of course 1 free electron interacts with one atom
                  – user6760
                  48 mins ago


















                • i know when electron in an atom absorb energy it becomes excited, as it goes from excited to lower state energy is being released as photon and in ur case it is radiowave. However I'm at a loss when I try imagine 2 free electrons (particle/wave) can repell each other?
                  – user6760
                  2 hours ago










                • @user6760 The free electron can also gain energy from photon. That also leads to a measurement problem.
                  – Bill Alsept
                  1 hour ago










                • @BillAlsept: I'm referring to free electron not valence electron
                  – user6760
                  1 hour ago










                • @user6760 if you fire free electrons through a double slit experiment and try to observe them, the photons will physically interfere.
                  – Bill Alsept
                  55 mins ago












                • @BillAlsept: of course 1 free electron interacts with one atom
                  – user6760
                  48 mins ago
















                i know when electron in an atom absorb energy it becomes excited, as it goes from excited to lower state energy is being released as photon and in ur case it is radiowave. However I'm at a loss when I try imagine 2 free electrons (particle/wave) can repell each other?
                – user6760
                2 hours ago




                i know when electron in an atom absorb energy it becomes excited, as it goes from excited to lower state energy is being released as photon and in ur case it is radiowave. However I'm at a loss when I try imagine 2 free electrons (particle/wave) can repell each other?
                – user6760
                2 hours ago












                @user6760 The free electron can also gain energy from photon. That also leads to a measurement problem.
                – Bill Alsept
                1 hour ago




                @user6760 The free electron can also gain energy from photon. That also leads to a measurement problem.
                – Bill Alsept
                1 hour ago












                @BillAlsept: I'm referring to free electron not valence electron
                – user6760
                1 hour ago




                @BillAlsept: I'm referring to free electron not valence electron
                – user6760
                1 hour ago












                @user6760 if you fire free electrons through a double slit experiment and try to observe them, the photons will physically interfere.
                – Bill Alsept
                55 mins ago






                @user6760 if you fire free electrons through a double slit experiment and try to observe them, the photons will physically interfere.
                – Bill Alsept
                55 mins ago














                @BillAlsept: of course 1 free electron interacts with one atom
                – user6760
                48 mins ago




                @BillAlsept: of course 1 free electron interacts with one atom
                – user6760
                48 mins ago










                up vote
                0
                down vote













                To resolve this paradox requires study of time dependent perturbation theory; solving Schrodinger's equation with a time dependent perturbation corresponding to the interaction time of two particles.



                If you do this you arrive at the following conclusions:



                A single free electron cannot absorb a free photon ( $e + gamma to e$ is not a valid interaction)



                A single free electron cannot emit a free photon ( $e to e + gamma$ is not a valid interaction)



                However, two electrons can scatter by exchange of energy ( $ e + e to e + e$ is a valid interaction)



                In this later case it is common to refer to this process being due to exchange of "a virtual photon" between the two electrons. But this is just a description of the calculation of time dependent perturbation theory.






                share|cite|improve this answer

























                  up vote
                  0
                  down vote













                  To resolve this paradox requires study of time dependent perturbation theory; solving Schrodinger's equation with a time dependent perturbation corresponding to the interaction time of two particles.



                  If you do this you arrive at the following conclusions:



                  A single free electron cannot absorb a free photon ( $e + gamma to e$ is not a valid interaction)



                  A single free electron cannot emit a free photon ( $e to e + gamma$ is not a valid interaction)



                  However, two electrons can scatter by exchange of energy ( $ e + e to e + e$ is a valid interaction)



                  In this later case it is common to refer to this process being due to exchange of "a virtual photon" between the two electrons. But this is just a description of the calculation of time dependent perturbation theory.






                  share|cite|improve this answer























                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    To resolve this paradox requires study of time dependent perturbation theory; solving Schrodinger's equation with a time dependent perturbation corresponding to the interaction time of two particles.



                    If you do this you arrive at the following conclusions:



                    A single free electron cannot absorb a free photon ( $e + gamma to e$ is not a valid interaction)



                    A single free electron cannot emit a free photon ( $e to e + gamma$ is not a valid interaction)



                    However, two electrons can scatter by exchange of energy ( $ e + e to e + e$ is a valid interaction)



                    In this later case it is common to refer to this process being due to exchange of "a virtual photon" between the two electrons. But this is just a description of the calculation of time dependent perturbation theory.






                    share|cite|improve this answer












                    To resolve this paradox requires study of time dependent perturbation theory; solving Schrodinger's equation with a time dependent perturbation corresponding to the interaction time of two particles.



                    If you do this you arrive at the following conclusions:



                    A single free electron cannot absorb a free photon ( $e + gamma to e$ is not a valid interaction)



                    A single free electron cannot emit a free photon ( $e to e + gamma$ is not a valid interaction)



                    However, two electrons can scatter by exchange of energy ( $ e + e to e + e$ is a valid interaction)



                    In this later case it is common to refer to this process being due to exchange of "a virtual photon" between the two electrons. But this is just a description of the calculation of time dependent perturbation theory.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    Bruce Greetham

                    1,1751416




                    1,1751416






















                        up vote
                        0
                        down vote













                        Basically, because the process doesn't simultaneously conserve energy and momentum; this is why we say that it's mediated by a virtual photon.



                        In more technical language, this means that the photons that are exchanged between two interacting electrons are allowed to be "off shell", where the "shell" is the relationship $E^2 = p^2c^2$ (for a massless particle). Real particles are required to be on-shell, but virtual particles are allowed to stray from that condition, at least by some amount.



                        In the Feynman-diagram description of the scattering between two electrons you often find diagrams where one electron emits a photon which is then absorbed by the other; this is a virtual photon and as such both the 'emission' and 'absorption' processes are exempt from these considerations.



                        As always, though, it bears repeating that Feynman diagrams are calculational tools, and none of the virtual particles that appear in those diagrams actually physically exist in any definable sense. The fact that the 'emission' and 'absorption' processes, as well as the virtual photon itself, seem to defy energy or momentum conservation is purely a quirk of the way in which we've chosen to interpret limited chunks of our calculation.






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          Basically, because the process doesn't simultaneously conserve energy and momentum; this is why we say that it's mediated by a virtual photon.



                          In more technical language, this means that the photons that are exchanged between two interacting electrons are allowed to be "off shell", where the "shell" is the relationship $E^2 = p^2c^2$ (for a massless particle). Real particles are required to be on-shell, but virtual particles are allowed to stray from that condition, at least by some amount.



                          In the Feynman-diagram description of the scattering between two electrons you often find diagrams where one electron emits a photon which is then absorbed by the other; this is a virtual photon and as such both the 'emission' and 'absorption' processes are exempt from these considerations.



                          As always, though, it bears repeating that Feynman diagrams are calculational tools, and none of the virtual particles that appear in those diagrams actually physically exist in any definable sense. The fact that the 'emission' and 'absorption' processes, as well as the virtual photon itself, seem to defy energy or momentum conservation is purely a quirk of the way in which we've chosen to interpret limited chunks of our calculation.






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            Basically, because the process doesn't simultaneously conserve energy and momentum; this is why we say that it's mediated by a virtual photon.



                            In more technical language, this means that the photons that are exchanged between two interacting electrons are allowed to be "off shell", where the "shell" is the relationship $E^2 = p^2c^2$ (for a massless particle). Real particles are required to be on-shell, but virtual particles are allowed to stray from that condition, at least by some amount.



                            In the Feynman-diagram description of the scattering between two electrons you often find diagrams where one electron emits a photon which is then absorbed by the other; this is a virtual photon and as such both the 'emission' and 'absorption' processes are exempt from these considerations.



                            As always, though, it bears repeating that Feynman diagrams are calculational tools, and none of the virtual particles that appear in those diagrams actually physically exist in any definable sense. The fact that the 'emission' and 'absorption' processes, as well as the virtual photon itself, seem to defy energy or momentum conservation is purely a quirk of the way in which we've chosen to interpret limited chunks of our calculation.






                            share|cite|improve this answer












                            Basically, because the process doesn't simultaneously conserve energy and momentum; this is why we say that it's mediated by a virtual photon.



                            In more technical language, this means that the photons that are exchanged between two interacting electrons are allowed to be "off shell", where the "shell" is the relationship $E^2 = p^2c^2$ (for a massless particle). Real particles are required to be on-shell, but virtual particles are allowed to stray from that condition, at least by some amount.



                            In the Feynman-diagram description of the scattering between two electrons you often find diagrams where one electron emits a photon which is then absorbed by the other; this is a virtual photon and as such both the 'emission' and 'absorption' processes are exempt from these considerations.



                            As always, though, it bears repeating that Feynman diagrams are calculational tools, and none of the virtual particles that appear in those diagrams actually physically exist in any definable sense. The fact that the 'emission' and 'absorption' processes, as well as the virtual photon itself, seem to defy energy or momentum conservation is purely a quirk of the way in which we've chosen to interpret limited chunks of our calculation.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 45 mins ago









                            Emilio Pisanty

                            81.6k21194406




                            81.6k21194406






























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