How come particles are not constantly “measured” by the whole universe?












4












$begingroup$


Let's say we are doing the double slit experiment with electrons. We get an interference pattern, and if we put detectors at slits, then we get two piles pattern because we measure electrons' positions when going through slits. But an electron interacts with other particles in a lot of different ways, e.g. electric field, gravity. Seems like the whole universe is receiving information about the electron's position. Why is it not the case and the electron goes through slits "unmeasured"?



Bonus question: in real experiments do we face the problem of not "shielding" particles from "measurement" good enough and thus getting a mix of both patterns on the screen?










share|cite|improve this question







New contributor




FunkyLoiso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Very closely related question here. It's basically the same idea, but specialized to a particular apparatus.
    $endgroup$
    – knzhou
    1 hour ago
















4












$begingroup$


Let's say we are doing the double slit experiment with electrons. We get an interference pattern, and if we put detectors at slits, then we get two piles pattern because we measure electrons' positions when going through slits. But an electron interacts with other particles in a lot of different ways, e.g. electric field, gravity. Seems like the whole universe is receiving information about the electron's position. Why is it not the case and the electron goes through slits "unmeasured"?



Bonus question: in real experiments do we face the problem of not "shielding" particles from "measurement" good enough and thus getting a mix of both patterns on the screen?










share|cite|improve this question







New contributor




FunkyLoiso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Very closely related question here. It's basically the same idea, but specialized to a particular apparatus.
    $endgroup$
    – knzhou
    1 hour ago














4












4








4


2



$begingroup$


Let's say we are doing the double slit experiment with electrons. We get an interference pattern, and if we put detectors at slits, then we get two piles pattern because we measure electrons' positions when going through slits. But an electron interacts with other particles in a lot of different ways, e.g. electric field, gravity. Seems like the whole universe is receiving information about the electron's position. Why is it not the case and the electron goes through slits "unmeasured"?



Bonus question: in real experiments do we face the problem of not "shielding" particles from "measurement" good enough and thus getting a mix of both patterns on the screen?










share|cite|improve this question







New contributor




FunkyLoiso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Let's say we are doing the double slit experiment with electrons. We get an interference pattern, and if we put detectors at slits, then we get two piles pattern because we measure electrons' positions when going through slits. But an electron interacts with other particles in a lot of different ways, e.g. electric field, gravity. Seems like the whole universe is receiving information about the electron's position. Why is it not the case and the electron goes through slits "unmeasured"?



Bonus question: in real experiments do we face the problem of not "shielding" particles from "measurement" good enough and thus getting a mix of both patterns on the screen?







quantum-mechanics double-slit-experiment measurement-problem






share|cite|improve this question







New contributor




FunkyLoiso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




FunkyLoiso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




FunkyLoiso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 1 hour ago









FunkyLoisoFunkyLoiso

211




211




New contributor




FunkyLoiso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





FunkyLoiso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






FunkyLoiso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    Very closely related question here. It's basically the same idea, but specialized to a particular apparatus.
    $endgroup$
    – knzhou
    1 hour ago


















  • $begingroup$
    Very closely related question here. It's basically the same idea, but specialized to a particular apparatus.
    $endgroup$
    – knzhou
    1 hour ago
















$begingroup$
Very closely related question here. It's basically the same idea, but specialized to a particular apparatus.
$endgroup$
– knzhou
1 hour ago




$begingroup$
Very closely related question here. It's basically the same idea, but specialized to a particular apparatus.
$endgroup$
– knzhou
1 hour ago










3 Answers
3






active

oldest

votes


















2












$begingroup$

There are time-scales related to interactions, or, equivalently, interaction rates. An experiment that measures electron interference needs to make sure that the time-of-flight of the electrons from the electron source to the observation screen is much shorter than any of the time-scales of possible interactions.



In interference experiments, we therefore define a coherence time for the interfering particles.



In real experiments, we do indeed face the problem of shielding particles from being measured by the environment, before they interfere. For example, in electron interferometers realized in solid-state devices, we have to go to very low temperatures, where the interactions between electrons and phonons become very 'slow' (their rate becomes very small). We also have to make sure that the devices are small enough that the Coulomb-interaction between electrons, which persists even at the lowest temperatures, does not spoil the interference (the decoherence rate due to electron-electron interaction does also depend on temperature: the rate becomes smaller with decreasing temperature).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Could you please explain the interaction rates? Let's think about gravitational interaction between the electron and the slits themselves. Naive thinking tells me that the electron exerts gravitational pull on the slits. The slit the electron went through experiences more pull and gets more deformation, which can potentially be measured had we good enough tools. This looks to me like an infinite interaction rate, the pull is smoothly increasing over time. Is the pull actually quantified and is there a probability of 0 quanta exchanged? Or maybe the pull is in superposition itself?
    $endgroup$
    – FunkyLoiso
    51 mins ago










  • $begingroup$
    Nice, +1. And to address the OP's question about whether this means that particles are "measured" -- the notion of measurement and wavefunction collapse in the Copenhagen interpretation never made much sense, because there was no way to say what was a "measurement." This has been clarified by understanding of decoherence, which is basically what flaudemus is describing.
    $endgroup$
    – Ben Crowell
    18 mins ago



















0












$begingroup$

As long as these interactions are weak and do not distinguish between the two slits, they can be disregarded.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Distinguishing which slit is which is the factor that causes the wavelike interference pattern to disappear. Experiments show that the more the path can be determined the more they look like single photons.



    Here's some notes on a course where this is worked out explicitly for a Mach-Zender quantum interference experiment, where this continuum between "classical" and "quantum" is made mathematically explicit.



    So yes, the more the experiment's electrons interacts with the "universe" in a way that the "universe" can gain information about which slit it went through, the more the "quantum interference pattern" disappears. This is a good intuition for why things at a macroscopic level behave classically: because the individual quantum pieces are interacting with the environment so much that all of this "quantum perserving" information leaks out.






    share|cite|improve this answer









    $endgroup$













      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
      });


      }
      });






      FunkyLoiso is a new contributor. Be nice, and check out our Code of Conduct.










      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f460855%2fhow-come-particles-are-not-constantly-measured-by-the-whole-universe%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      There are time-scales related to interactions, or, equivalently, interaction rates. An experiment that measures electron interference needs to make sure that the time-of-flight of the electrons from the electron source to the observation screen is much shorter than any of the time-scales of possible interactions.



      In interference experiments, we therefore define a coherence time for the interfering particles.



      In real experiments, we do indeed face the problem of shielding particles from being measured by the environment, before they interfere. For example, in electron interferometers realized in solid-state devices, we have to go to very low temperatures, where the interactions between electrons and phonons become very 'slow' (their rate becomes very small). We also have to make sure that the devices are small enough that the Coulomb-interaction between electrons, which persists even at the lowest temperatures, does not spoil the interference (the decoherence rate due to electron-electron interaction does also depend on temperature: the rate becomes smaller with decreasing temperature).






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Could you please explain the interaction rates? Let's think about gravitational interaction between the electron and the slits themselves. Naive thinking tells me that the electron exerts gravitational pull on the slits. The slit the electron went through experiences more pull and gets more deformation, which can potentially be measured had we good enough tools. This looks to me like an infinite interaction rate, the pull is smoothly increasing over time. Is the pull actually quantified and is there a probability of 0 quanta exchanged? Or maybe the pull is in superposition itself?
        $endgroup$
        – FunkyLoiso
        51 mins ago










      • $begingroup$
        Nice, +1. And to address the OP's question about whether this means that particles are "measured" -- the notion of measurement and wavefunction collapse in the Copenhagen interpretation never made much sense, because there was no way to say what was a "measurement." This has been clarified by understanding of decoherence, which is basically what flaudemus is describing.
        $endgroup$
        – Ben Crowell
        18 mins ago
















      2












      $begingroup$

      There are time-scales related to interactions, or, equivalently, interaction rates. An experiment that measures electron interference needs to make sure that the time-of-flight of the electrons from the electron source to the observation screen is much shorter than any of the time-scales of possible interactions.



      In interference experiments, we therefore define a coherence time for the interfering particles.



      In real experiments, we do indeed face the problem of shielding particles from being measured by the environment, before they interfere. For example, in electron interferometers realized in solid-state devices, we have to go to very low temperatures, where the interactions between electrons and phonons become very 'slow' (their rate becomes very small). We also have to make sure that the devices are small enough that the Coulomb-interaction between electrons, which persists even at the lowest temperatures, does not spoil the interference (the decoherence rate due to electron-electron interaction does also depend on temperature: the rate becomes smaller with decreasing temperature).






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Could you please explain the interaction rates? Let's think about gravitational interaction between the electron and the slits themselves. Naive thinking tells me that the electron exerts gravitational pull on the slits. The slit the electron went through experiences more pull and gets more deformation, which can potentially be measured had we good enough tools. This looks to me like an infinite interaction rate, the pull is smoothly increasing over time. Is the pull actually quantified and is there a probability of 0 quanta exchanged? Or maybe the pull is in superposition itself?
        $endgroup$
        – FunkyLoiso
        51 mins ago










      • $begingroup$
        Nice, +1. And to address the OP's question about whether this means that particles are "measured" -- the notion of measurement and wavefunction collapse in the Copenhagen interpretation never made much sense, because there was no way to say what was a "measurement." This has been clarified by understanding of decoherence, which is basically what flaudemus is describing.
        $endgroup$
        – Ben Crowell
        18 mins ago














      2












      2








      2





      $begingroup$

      There are time-scales related to interactions, or, equivalently, interaction rates. An experiment that measures electron interference needs to make sure that the time-of-flight of the electrons from the electron source to the observation screen is much shorter than any of the time-scales of possible interactions.



      In interference experiments, we therefore define a coherence time for the interfering particles.



      In real experiments, we do indeed face the problem of shielding particles from being measured by the environment, before they interfere. For example, in electron interferometers realized in solid-state devices, we have to go to very low temperatures, where the interactions between electrons and phonons become very 'slow' (their rate becomes very small). We also have to make sure that the devices are small enough that the Coulomb-interaction between electrons, which persists even at the lowest temperatures, does not spoil the interference (the decoherence rate due to electron-electron interaction does also depend on temperature: the rate becomes smaller with decreasing temperature).






      share|cite|improve this answer









      $endgroup$



      There are time-scales related to interactions, or, equivalently, interaction rates. An experiment that measures electron interference needs to make sure that the time-of-flight of the electrons from the electron source to the observation screen is much shorter than any of the time-scales of possible interactions.



      In interference experiments, we therefore define a coherence time for the interfering particles.



      In real experiments, we do indeed face the problem of shielding particles from being measured by the environment, before they interfere. For example, in electron interferometers realized in solid-state devices, we have to go to very low temperatures, where the interactions between electrons and phonons become very 'slow' (their rate becomes very small). We also have to make sure that the devices are small enough that the Coulomb-interaction between electrons, which persists even at the lowest temperatures, does not spoil the interference (the decoherence rate due to electron-electron interaction does also depend on temperature: the rate becomes smaller with decreasing temperature).







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 1 hour ago









      flaudemusflaudemus

      2425




      2425












      • $begingroup$
        Could you please explain the interaction rates? Let's think about gravitational interaction between the electron and the slits themselves. Naive thinking tells me that the electron exerts gravitational pull on the slits. The slit the electron went through experiences more pull and gets more deformation, which can potentially be measured had we good enough tools. This looks to me like an infinite interaction rate, the pull is smoothly increasing over time. Is the pull actually quantified and is there a probability of 0 quanta exchanged? Or maybe the pull is in superposition itself?
        $endgroup$
        – FunkyLoiso
        51 mins ago










      • $begingroup$
        Nice, +1. And to address the OP's question about whether this means that particles are "measured" -- the notion of measurement and wavefunction collapse in the Copenhagen interpretation never made much sense, because there was no way to say what was a "measurement." This has been clarified by understanding of decoherence, which is basically what flaudemus is describing.
        $endgroup$
        – Ben Crowell
        18 mins ago


















      • $begingroup$
        Could you please explain the interaction rates? Let's think about gravitational interaction between the electron and the slits themselves. Naive thinking tells me that the electron exerts gravitational pull on the slits. The slit the electron went through experiences more pull and gets more deformation, which can potentially be measured had we good enough tools. This looks to me like an infinite interaction rate, the pull is smoothly increasing over time. Is the pull actually quantified and is there a probability of 0 quanta exchanged? Or maybe the pull is in superposition itself?
        $endgroup$
        – FunkyLoiso
        51 mins ago










      • $begingroup$
        Nice, +1. And to address the OP's question about whether this means that particles are "measured" -- the notion of measurement and wavefunction collapse in the Copenhagen interpretation never made much sense, because there was no way to say what was a "measurement." This has been clarified by understanding of decoherence, which is basically what flaudemus is describing.
        $endgroup$
        – Ben Crowell
        18 mins ago
















      $begingroup$
      Could you please explain the interaction rates? Let's think about gravitational interaction between the electron and the slits themselves. Naive thinking tells me that the electron exerts gravitational pull on the slits. The slit the electron went through experiences more pull and gets more deformation, which can potentially be measured had we good enough tools. This looks to me like an infinite interaction rate, the pull is smoothly increasing over time. Is the pull actually quantified and is there a probability of 0 quanta exchanged? Or maybe the pull is in superposition itself?
      $endgroup$
      – FunkyLoiso
      51 mins ago




      $begingroup$
      Could you please explain the interaction rates? Let's think about gravitational interaction between the electron and the slits themselves. Naive thinking tells me that the electron exerts gravitational pull on the slits. The slit the electron went through experiences more pull and gets more deformation, which can potentially be measured had we good enough tools. This looks to me like an infinite interaction rate, the pull is smoothly increasing over time. Is the pull actually quantified and is there a probability of 0 quanta exchanged? Or maybe the pull is in superposition itself?
      $endgroup$
      – FunkyLoiso
      51 mins ago












      $begingroup$
      Nice, +1. And to address the OP's question about whether this means that particles are "measured" -- the notion of measurement and wavefunction collapse in the Copenhagen interpretation never made much sense, because there was no way to say what was a "measurement." This has been clarified by understanding of decoherence, which is basically what flaudemus is describing.
      $endgroup$
      – Ben Crowell
      18 mins ago




      $begingroup$
      Nice, +1. And to address the OP's question about whether this means that particles are "measured" -- the notion of measurement and wavefunction collapse in the Copenhagen interpretation never made much sense, because there was no way to say what was a "measurement." This has been clarified by understanding of decoherence, which is basically what flaudemus is describing.
      $endgroup$
      – Ben Crowell
      18 mins ago











      0












      $begingroup$

      As long as these interactions are weak and do not distinguish between the two slits, they can be disregarded.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        As long as these interactions are weak and do not distinguish between the two slits, they can be disregarded.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          As long as these interactions are weak and do not distinguish between the two slits, they can be disregarded.






          share|cite|improve this answer









          $endgroup$



          As long as these interactions are weak and do not distinguish between the two slits, they can be disregarded.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          my2ctsmy2cts

          5,2782618




          5,2782618























              0












              $begingroup$

              Distinguishing which slit is which is the factor that causes the wavelike interference pattern to disappear. Experiments show that the more the path can be determined the more they look like single photons.



              Here's some notes on a course where this is worked out explicitly for a Mach-Zender quantum interference experiment, where this continuum between "classical" and "quantum" is made mathematically explicit.



              So yes, the more the experiment's electrons interacts with the "universe" in a way that the "universe" can gain information about which slit it went through, the more the "quantum interference pattern" disappears. This is a good intuition for why things at a macroscopic level behave classically: because the individual quantum pieces are interacting with the environment so much that all of this "quantum perserving" information leaks out.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Distinguishing which slit is which is the factor that causes the wavelike interference pattern to disappear. Experiments show that the more the path can be determined the more they look like single photons.



                Here's some notes on a course where this is worked out explicitly for a Mach-Zender quantum interference experiment, where this continuum between "classical" and "quantum" is made mathematically explicit.



                So yes, the more the experiment's electrons interacts with the "universe" in a way that the "universe" can gain information about which slit it went through, the more the "quantum interference pattern" disappears. This is a good intuition for why things at a macroscopic level behave classically: because the individual quantum pieces are interacting with the environment so much that all of this "quantum perserving" information leaks out.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Distinguishing which slit is which is the factor that causes the wavelike interference pattern to disappear. Experiments show that the more the path can be determined the more they look like single photons.



                  Here's some notes on a course where this is worked out explicitly for a Mach-Zender quantum interference experiment, where this continuum between "classical" and "quantum" is made mathematically explicit.



                  So yes, the more the experiment's electrons interacts with the "universe" in a way that the "universe" can gain information about which slit it went through, the more the "quantum interference pattern" disappears. This is a good intuition for why things at a macroscopic level behave classically: because the individual quantum pieces are interacting with the environment so much that all of this "quantum perserving" information leaks out.






                  share|cite|improve this answer









                  $endgroup$



                  Distinguishing which slit is which is the factor that causes the wavelike interference pattern to disappear. Experiments show that the more the path can be determined the more they look like single photons.



                  Here's some notes on a course where this is worked out explicitly for a Mach-Zender quantum interference experiment, where this continuum between "classical" and "quantum" is made mathematically explicit.



                  So yes, the more the experiment's electrons interacts with the "universe" in a way that the "universe" can gain information about which slit it went through, the more the "quantum interference pattern" disappears. This is a good intuition for why things at a macroscopic level behave classically: because the individual quantum pieces are interacting with the environment so much that all of this "quantum perserving" information leaks out.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  Steven SagonaSteven Sagona

                  200217




                  200217






















                      FunkyLoiso is a new contributor. Be nice, and check out our Code of Conduct.










                      draft saved

                      draft discarded


















                      FunkyLoiso is a new contributor. Be nice, and check out our Code of Conduct.













                      FunkyLoiso is a new contributor. Be nice, and check out our Code of Conduct.












                      FunkyLoiso is a new contributor. Be nice, and check out our Code of Conduct.
















                      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%2f460855%2fhow-come-particles-are-not-constantly-measured-by-the-whole-universe%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

                      Callistus I

                      Tabula Rosettana

                      How to label and detect the document text images