What happens to a wavefunction upon measurement when there's degeneracy?












2












$begingroup$


I'll use a hydrogen atom as an example. A hydrogen atom has multiple energy eigenstates for all but one of its energy levels. Suppose I measure a hydrogen atom to have an energy $E_n$ where $n > 1$. What's the state of the hydrogen atom right after my measurement?






quantum-mechanics quantum-states







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    I'll use a hydrogen atom as an example. A hydrogen atom has multiple energy eigenstates for all but one of its energy levels. Suppose I measure a hydrogen atom to have an energy $E_n$ where $n > 1$. What's the state of the hydrogen atom right after my measurement?






    quantum-mechanics quantum-states







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      I'll use a hydrogen atom as an example. A hydrogen atom has multiple energy eigenstates for all but one of its energy levels. Suppose I measure a hydrogen atom to have an energy $E_n$ where $n > 1$. What's the state of the hydrogen atom right after my measurement?






      quantum-mechanics quantum-states







      share|cite|improve this question









      $endgroup$




      I'll use a hydrogen atom as an example. A hydrogen atom has multiple energy eigenstates for all but one of its energy levels. Suppose I measure a hydrogen atom to have an energy $E_n$ where $n > 1$. What's the state of the hydrogen atom right after my measurement?






      quantum-mechanics quantum-states




      quantum-mechanics quantum-states






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 6 hours ago









      PiKindOfGuyPiKindOfGuy

      488515




      488515






















          2 Answers
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          3












          $begingroup$

          This occurs in all measurements, because you don't measure the quantum number of every particle in the universe.



          If you believe in the Copenhagen interpretation, then in your example the state after measurement is the projection of the original state onto the subspace with energy $E_n$. For example, let's use the quantum numbers $n$, $l$, where $l$ is the angular momentum. If the original state was $|1,0rangle+i|2,0rangle-|2,1rangle$, and you measure $n=2$, then after measurement the state becomes $i|2,0rangle-|2,1rangle$.



          If you believe instead in standard quantum mechanics, then nothing happens to the wavefunction you're observing, but you become entangled with the wavefunction in a way that leads to the same facts about the results of your future measurements.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I'm a beginner. Should I concern myself with what you call "standard quantum mechanics"? Entanglement is still pretty foreign to me.
            $endgroup$
            – PiKindOfGuy
            6 hours ago










          • $begingroup$
            Normalize your states so I can accept your answer.
            $endgroup$
            – PiKindOfGuy
            6 hours ago



















          2












          $begingroup$

          The measurement is a projection so the appropriate projection operator would be
          $$
          hatPi= sum_{r=1}^k vert E_n;alpha_rranglelangle E_n;alpha_rvert
          $$
          where $vert E_n;alpha_rrangle$ is an eigenstate of $hat H$ with energy $E_n$ and $alpha_r$ denotes all the other eigenvalues required to fully label your states, given a complete set of commuting observables.



          Thus, you’d get
          $$
          vertPhirangle=hat Pivert psirangle = sum_r vert E_n;alpha_rrangle c_{r}, ,tag{1}
          $$
          where $c_r=langle E_n;alpha rvertpsirangle$. Note that projection does not preserve the norm so (1) must be normalized by dividing “manually” by
          $sqrt{sum_r vert c_rvert^2}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So Ben's answer is correct, except that the states ought to be normalized (for proper use)?
            $endgroup$
            – PiKindOfGuy
            6 hours ago






          • 1




            $begingroup$
            Yes his is correct albeit a little less general, and the normalization is missing.
            $endgroup$
            – ZeroTheHero
            6 hours ago










          • $begingroup$
            Should I pick the answer that I think is most complete or the one that helped me the most?
            $endgroup$
            – PiKindOfGuy
            6 hours ago










          • $begingroup$
            Frankly it doesn’t matter much. You can just let it happen and decide in 24hrs.
            $endgroup$
            – ZeroTheHero
            6 hours ago











          Your Answer





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          2 Answers
          2






          active

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

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          3












          $begingroup$

          This occurs in all measurements, because you don't measure the quantum number of every particle in the universe.



          If you believe in the Copenhagen interpretation, then in your example the state after measurement is the projection of the original state onto the subspace with energy $E_n$. For example, let's use the quantum numbers $n$, $l$, where $l$ is the angular momentum. If the original state was $|1,0rangle+i|2,0rangle-|2,1rangle$, and you measure $n=2$, then after measurement the state becomes $i|2,0rangle-|2,1rangle$.



          If you believe instead in standard quantum mechanics, then nothing happens to the wavefunction you're observing, but you become entangled with the wavefunction in a way that leads to the same facts about the results of your future measurements.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I'm a beginner. Should I concern myself with what you call "standard quantum mechanics"? Entanglement is still pretty foreign to me.
            $endgroup$
            – PiKindOfGuy
            6 hours ago










          • $begingroup$
            Normalize your states so I can accept your answer.
            $endgroup$
            – PiKindOfGuy
            6 hours ago
















          3












          $begingroup$

          This occurs in all measurements, because you don't measure the quantum number of every particle in the universe.



          If you believe in the Copenhagen interpretation, then in your example the state after measurement is the projection of the original state onto the subspace with energy $E_n$. For example, let's use the quantum numbers $n$, $l$, where $l$ is the angular momentum. If the original state was $|1,0rangle+i|2,0rangle-|2,1rangle$, and you measure $n=2$, then after measurement the state becomes $i|2,0rangle-|2,1rangle$.



          If you believe instead in standard quantum mechanics, then nothing happens to the wavefunction you're observing, but you become entangled with the wavefunction in a way that leads to the same facts about the results of your future measurements.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I'm a beginner. Should I concern myself with what you call "standard quantum mechanics"? Entanglement is still pretty foreign to me.
            $endgroup$
            – PiKindOfGuy
            6 hours ago










          • $begingroup$
            Normalize your states so I can accept your answer.
            $endgroup$
            – PiKindOfGuy
            6 hours ago














          3












          3








          3





          $begingroup$

          This occurs in all measurements, because you don't measure the quantum number of every particle in the universe.



          If you believe in the Copenhagen interpretation, then in your example the state after measurement is the projection of the original state onto the subspace with energy $E_n$. For example, let's use the quantum numbers $n$, $l$, where $l$ is the angular momentum. If the original state was $|1,0rangle+i|2,0rangle-|2,1rangle$, and you measure $n=2$, then after measurement the state becomes $i|2,0rangle-|2,1rangle$.



          If you believe instead in standard quantum mechanics, then nothing happens to the wavefunction you're observing, but you become entangled with the wavefunction in a way that leads to the same facts about the results of your future measurements.






          share|cite|improve this answer









          $endgroup$



          This occurs in all measurements, because you don't measure the quantum number of every particle in the universe.



          If you believe in the Copenhagen interpretation, then in your example the state after measurement is the projection of the original state onto the subspace with energy $E_n$. For example, let's use the quantum numbers $n$, $l$, where $l$ is the angular momentum. If the original state was $|1,0rangle+i|2,0rangle-|2,1rangle$, and you measure $n=2$, then after measurement the state becomes $i|2,0rangle-|2,1rangle$.



          If you believe instead in standard quantum mechanics, then nothing happens to the wavefunction you're observing, but you become entangled with the wavefunction in a way that leads to the same facts about the results of your future measurements.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 6 hours ago









          Ben CrowellBen Crowell

          50.2k5155296




          50.2k5155296












          • $begingroup$
            I'm a beginner. Should I concern myself with what you call "standard quantum mechanics"? Entanglement is still pretty foreign to me.
            $endgroup$
            – PiKindOfGuy
            6 hours ago










          • $begingroup$
            Normalize your states so I can accept your answer.
            $endgroup$
            – PiKindOfGuy
            6 hours ago


















          • $begingroup$
            I'm a beginner. Should I concern myself with what you call "standard quantum mechanics"? Entanglement is still pretty foreign to me.
            $endgroup$
            – PiKindOfGuy
            6 hours ago










          • $begingroup$
            Normalize your states so I can accept your answer.
            $endgroup$
            – PiKindOfGuy
            6 hours ago
















          $begingroup$
          I'm a beginner. Should I concern myself with what you call "standard quantum mechanics"? Entanglement is still pretty foreign to me.
          $endgroup$
          – PiKindOfGuy
          6 hours ago




          $begingroup$
          I'm a beginner. Should I concern myself with what you call "standard quantum mechanics"? Entanglement is still pretty foreign to me.
          $endgroup$
          – PiKindOfGuy
          6 hours ago












          $begingroup$
          Normalize your states so I can accept your answer.
          $endgroup$
          – PiKindOfGuy
          6 hours ago




          $begingroup$
          Normalize your states so I can accept your answer.
          $endgroup$
          – PiKindOfGuy
          6 hours ago











          2












          $begingroup$

          The measurement is a projection so the appropriate projection operator would be
          $$
          hatPi= sum_{r=1}^k vert E_n;alpha_rranglelangle E_n;alpha_rvert
          $$
          where $vert E_n;alpha_rrangle$ is an eigenstate of $hat H$ with energy $E_n$ and $alpha_r$ denotes all the other eigenvalues required to fully label your states, given a complete set of commuting observables.



          Thus, you’d get
          $$
          vertPhirangle=hat Pivert psirangle = sum_r vert E_n;alpha_rrangle c_{r}, ,tag{1}
          $$
          where $c_r=langle E_n;alpha rvertpsirangle$. Note that projection does not preserve the norm so (1) must be normalized by dividing “manually” by
          $sqrt{sum_r vert c_rvert^2}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So Ben's answer is correct, except that the states ought to be normalized (for proper use)?
            $endgroup$
            – PiKindOfGuy
            6 hours ago






          • 1




            $begingroup$
            Yes his is correct albeit a little less general, and the normalization is missing.
            $endgroup$
            – ZeroTheHero
            6 hours ago










          • $begingroup$
            Should I pick the answer that I think is most complete or the one that helped me the most?
            $endgroup$
            – PiKindOfGuy
            6 hours ago










          • $begingroup$
            Frankly it doesn’t matter much. You can just let it happen and decide in 24hrs.
            $endgroup$
            – ZeroTheHero
            6 hours ago
















          2












          $begingroup$

          The measurement is a projection so the appropriate projection operator would be
          $$
          hatPi= sum_{r=1}^k vert E_n;alpha_rranglelangle E_n;alpha_rvert
          $$
          where $vert E_n;alpha_rrangle$ is an eigenstate of $hat H$ with energy $E_n$ and $alpha_r$ denotes all the other eigenvalues required to fully label your states, given a complete set of commuting observables.



          Thus, you’d get
          $$
          vertPhirangle=hat Pivert psirangle = sum_r vert E_n;alpha_rrangle c_{r}, ,tag{1}
          $$
          where $c_r=langle E_n;alpha rvertpsirangle$. Note that projection does not preserve the norm so (1) must be normalized by dividing “manually” by
          $sqrt{sum_r vert c_rvert^2}$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So Ben's answer is correct, except that the states ought to be normalized (for proper use)?
            $endgroup$
            – PiKindOfGuy
            6 hours ago






          • 1




            $begingroup$
            Yes his is correct albeit a little less general, and the normalization is missing.
            $endgroup$
            – ZeroTheHero
            6 hours ago










          • $begingroup$
            Should I pick the answer that I think is most complete or the one that helped me the most?
            $endgroup$
            – PiKindOfGuy
            6 hours ago










          • $begingroup$
            Frankly it doesn’t matter much. You can just let it happen and decide in 24hrs.
            $endgroup$
            – ZeroTheHero
            6 hours ago














          2












          2








          2





          $begingroup$

          The measurement is a projection so the appropriate projection operator would be
          $$
          hatPi= sum_{r=1}^k vert E_n;alpha_rranglelangle E_n;alpha_rvert
          $$
          where $vert E_n;alpha_rrangle$ is an eigenstate of $hat H$ with energy $E_n$ and $alpha_r$ denotes all the other eigenvalues required to fully label your states, given a complete set of commuting observables.



          Thus, you’d get
          $$
          vertPhirangle=hat Pivert psirangle = sum_r vert E_n;alpha_rrangle c_{r}, ,tag{1}
          $$
          where $c_r=langle E_n;alpha rvertpsirangle$. Note that projection does not preserve the norm so (1) must be normalized by dividing “manually” by
          $sqrt{sum_r vert c_rvert^2}$.






          share|cite|improve this answer









          $endgroup$



          The measurement is a projection so the appropriate projection operator would be
          $$
          hatPi= sum_{r=1}^k vert E_n;alpha_rranglelangle E_n;alpha_rvert
          $$
          where $vert E_n;alpha_rrangle$ is an eigenstate of $hat H$ with energy $E_n$ and $alpha_r$ denotes all the other eigenvalues required to fully label your states, given a complete set of commuting observables.



          Thus, you’d get
          $$
          vertPhirangle=hat Pivert psirangle = sum_r vert E_n;alpha_rrangle c_{r}, ,tag{1}
          $$
          where $c_r=langle E_n;alpha rvertpsirangle$. Note that projection does not preserve the norm so (1) must be normalized by dividing “manually” by
          $sqrt{sum_r vert c_rvert^2}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 6 hours ago









          ZeroTheHeroZeroTheHero

          19.6k53159




          19.6k53159












          • $begingroup$
            So Ben's answer is correct, except that the states ought to be normalized (for proper use)?
            $endgroup$
            – PiKindOfGuy
            6 hours ago






          • 1




            $begingroup$
            Yes his is correct albeit a little less general, and the normalization is missing.
            $endgroup$
            – ZeroTheHero
            6 hours ago










          • $begingroup$
            Should I pick the answer that I think is most complete or the one that helped me the most?
            $endgroup$
            – PiKindOfGuy
            6 hours ago










          • $begingroup$
            Frankly it doesn’t matter much. You can just let it happen and decide in 24hrs.
            $endgroup$
            – ZeroTheHero
            6 hours ago


















          • $begingroup$
            So Ben's answer is correct, except that the states ought to be normalized (for proper use)?
            $endgroup$
            – PiKindOfGuy
            6 hours ago






          • 1




            $begingroup$
            Yes his is correct albeit a little less general, and the normalization is missing.
            $endgroup$
            – ZeroTheHero
            6 hours ago










          • $begingroup$
            Should I pick the answer that I think is most complete or the one that helped me the most?
            $endgroup$
            – PiKindOfGuy
            6 hours ago










          • $begingroup$
            Frankly it doesn’t matter much. You can just let it happen and decide in 24hrs.
            $endgroup$
            – ZeroTheHero
            6 hours ago
















          $begingroup$
          So Ben's answer is correct, except that the states ought to be normalized (for proper use)?
          $endgroup$
          – PiKindOfGuy
          6 hours ago




          $begingroup$
          So Ben's answer is correct, except that the states ought to be normalized (for proper use)?
          $endgroup$
          – PiKindOfGuy
          6 hours ago




          1




          1




          $begingroup$
          Yes his is correct albeit a little less general, and the normalization is missing.
          $endgroup$
          – ZeroTheHero
          6 hours ago




          $begingroup$
          Yes his is correct albeit a little less general, and the normalization is missing.
          $endgroup$
          – ZeroTheHero
          6 hours ago












          $begingroup$
          Should I pick the answer that I think is most complete or the one that helped me the most?
          $endgroup$
          – PiKindOfGuy
          6 hours ago




          $begingroup$
          Should I pick the answer that I think is most complete or the one that helped me the most?
          $endgroup$
          – PiKindOfGuy
          6 hours ago












          $begingroup$
          Frankly it doesn’t matter much. You can just let it happen and decide in 24hrs.
          $endgroup$
          – ZeroTheHero
          6 hours ago




          $begingroup$
          Frankly it doesn’t matter much. You can just let it happen and decide in 24hrs.
          $endgroup$
          – ZeroTheHero
          6 hours ago


















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