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
          2






          active

          oldest

          votes


















          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





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "151"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: false,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: null,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f458360%2fwhat-happens-to-a-wavefunction-upon-measurement-when-theres-degeneracy%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          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


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Physics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f458360%2fwhat-happens-to-a-wavefunction-upon-measurement-when-theres-degeneracy%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Усть-Каменогорск

          Image
          У этого термина существуют и другие значения, см. Усть-Каменогорск. Город Усть-Каменогорск каз. Өскемен Флаг Герб 49°57′ с. ш. 82°37′ в. д. H G Я O L Страна Казахстан   Казахстан Статус областной центр Область Восточно-Казахстанская Аким Жаксылык Омар [1] История и география Основан 1720 Город с 1868 Площадь 303,15 [2] км² Высота центра 283 ± 1 м Тип климата резко-континентальный Часовой пояс UTC+6 Население Население 329 090 [3]  человек ( 2018 ) Национальности русские 53,56 % казахи 42,99 % немцы 0,82 % украинцы 0,57 % татары 0,79 % белорусы 0,14 % азербайджанцы 0,23 % корейцы 0,20 % прочие 1,08 % -1 шт [4] Этнохороним усть-каменогорцы, усть-каменогорец Цифровые идентификаторы Телефонный код +7 (7232) Почтовый индекс F0***** [5] Автомобильный код 16 (ранее F) oskemen.kz Усть-Каменогорск Усть-Каменогорск ...
          Read more

          Халкинская богословская школа

          Image
          Халкинская богословская школа тур. Rum Ortodoks Ruhban Okulu греч. Ιερά Θεολογική Σχολή της Χάλκης Главное здание Халкинской школы Международное название Theological School of Halki Год основания 1844 Конфессия православие Церковь Константинопольская Ректор Елпидифор (Ламбриниадис) Расположение о. Хейбелиада, Турция Юридический адрес Ümit tepesi No 5, 34973 Heybeliada, Istanbul, Turkey Сайт www.greece.org  [[commons:Category:Halki seminary|Халкинская богословская школа тур. Rum Ortodoks Ruhban Okulu греч. Ιερά Θεολογική Σχολή της Χάλκης на Викискладе]] Ха́лкинская богосло́вская шко́ла (или Духовная семинария на Халки , тур. Rum Ortodoks Ruhban Okulu , греч. Ιερά Θεολογική Σχολή της Χάλκης ) — недействующее среднее специальное учебное заведение Константинопольской православной церкви, расположенное на территории Свято-Троицкого мужского монастыря на вершине холма Надежды (тур. Ümit Tepesi ) острова Хейбелиада, в Турции....
          Read more

          Высокополье (Харьковская область)

          Image
          У этого термина существуют и другие значения, см. Высокополье. Село Высокополье укр. Високопілля Герб 49°55′07″ с. ш. 35°25′13″ в. д. H G Я O L Страна Украина   Украина Область Харьковская Район Валковский Сельский совет Высокопольский История и география Основан 1670 Прежние названия Хмелевое Площадь 55,56 км² Часовой пояс UTC+2, летом UTC+3 Население Население 1957 человек ( 2001 ) Плотность 35,220 чел./км² Цифровые идентификаторы Телефонный код +380 5753 Почтовый индекс 63020 Автомобильный код AX, КХ / 21 КОАТУУ 6321281501 Высокополье Высокополье Высокополье  Аудио, фото и видео на Викискладе Высокопо́лье (укр. Високопілля ) — село, Высокопольский сельский совет, Валковский район, Харьковская область, Украина. Код КОАТУУ — 6321281501. Население по переписи 2001 г. составляет 1957 (866/1091 м/ж) человек. Является административным центр...
          Read more