Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Clarification on wavefunction collapse

  1. Oct 16, 2007 #1
    I am just beginning to understand this concept. Some help would be appreciated.
    Let me know if I am wrong in saying the following:
    "The wave function (say [tex]\Psi[/tex]] collapses to an eigen vector of the operator corresponding to the physical quantity(say [tex]\lambda[/tex]) being measured. This is because the act of measuring interferes with the system"

    Now what confuses me (further) is that, as a result of the wavefunction collapsing to an Eigen vector, the subsequent measurements give the values with a probability 1.
    I understand that if [tex]\Psi[/tex]=[tex]\Sigma[/tex] ai Ni and [tex]\Psi[/tex] collapses to some Ni, ai=1 => probability is 1.
    But doesnt this eigen function Ni now act as a new [tex]\Psi[/tex] !
    We should be able to find a basis set of vectors corresponding to Ni in which case it should again collapse to another smaller eigenvector. This doesnt seem to happen.
    Why?
    Hope the question is clear.
     
    Last edited: Oct 16, 2007
  2. jcsd
  3. Oct 16, 2007 #2
  4. Oct 16, 2007 #3
    I did notice those. They dont (atleast directly) answer this question. Kindly clarify my understanding first and then the question. I am pretty confused at the moment and the forums you mentioned only confuse me further. I am still learning this concept! Those forums are for pros like you!
     
  5. Oct 16, 2007 #4

    f95toli

    User Avatar
    Science Advisor
    Gold Member

    In one of the threads I gave a reference to an article by Zurek:

    www.arxiv.org/abs/quant-ph/0306072

    Look in the section on einselecton and pointer states; you will find the answer there.
    Essentially, it has to do with the fact that pointer states are (almost by
    definition) "robust" to new measurements which is why they won't change.
    Also, note that the idea of a "collaps" is somewhat old fashioned, it is a terminology which is rarely used nowadays. The main problem with the word "collapse" is that it somehow implies an instantatious process, which as it happens is not quite right.

    If you want to take a look at what a modern paper that deals with "practical" measurement theory looks like you can e.g. download the following paper

    www.arxiv.org/abs/0709.4264

    (There are many others but I hade the link available since I was reading this paper at work today).
    The main point is that there is no "collapse process" as such and it is also possible to model how the measurement process affects the properties of the quantum system.
     
  6. Oct 16, 2007 #5

    JesseM

    User Avatar
    Science Advisor

    Yes, immediately after the "collapse" the eigenvector is just the new quantum state of the system.
    Using what basis? If you use the basis corresponding to the same operator that caused the collapse to Ni, then decomposing this into a sum of eigenvectors means that the coefficient of all the other basis vectors in the sum will be zero, so a second "collapse" with the same operator immediately after the first one will just give Ni again.
     
    Last edited: Oct 16, 2007
  7. Oct 17, 2007 #6
    I suggest in addition the semi popular paper (readable for the experimentalists also) by S.L. Adler “Why Decoherence has not Solved the Measurement Problem” (quant-ph/0112095). You will find the answers there.

    You will improve substantially your knowledge of statmech reading John von Neumann, Zs.f. Phys., 57, 30 (1929). It will help you to understand the content of the quoted paper by J.Gambetta et al.

    Regards, Dany.

    P.S.
    The main problem is compliance with the SR and not with the words.It happens that SR somehow appears quite right.
     
    Last edited: Oct 17, 2007
  8. Oct 17, 2007 #7

    JesseM

    User Avatar
    Science Advisor

    I think you guys are going way beyond what the original poster was asking. The question wasn't about the measurement problem or decoherence, it was just about the basic mathematics of the projection postulate and how it alters the state vector.
     
  9. Oct 17, 2007 #8
    Sure and your post provide the complete answer.

    Regards, Dany.
     
  10. Oct 17, 2007 #9

    JesseM

    User Avatar
    Science Advisor

    Fair enough, threads often spin off into new directions, I just wanted to make sure people weren't misunderstanding the question in the OP.
     
  11. Oct 17, 2007 #10
    I guess my post #157 in “cat in a box paradox” session also provide fairly complete explanation of that phenomenon (I use C.E.Shannon Theory of Communication).

    Regards, Dany.
     
    Last edited: Oct 17, 2007
  12. Oct 17, 2007 #11

    f95toli

    User Avatar
    Science Advisor
    Gold Member

    What is it you think I don't understand?
    I refered to this paper for two reasons: It is a good example how measurement problems are handled in a "practical" context (i.e. prediction of measurement time and other parameters which are important for experiments) and, secondly, it is also a good example of how it is possible to make these predictions without having to introduce an instantaneous "collapse" that makes the system completely classical.
     
    Last edited: Oct 17, 2007
  13. Oct 17, 2007 #12

    cks

    User Avatar

    Let me put it this way to see whether it gives you some lights over your problems.

    First, assume a system that is not measured yet. From the Schrodinger equation, you can find the general wavefunction of the system without doing experiment on it first(numerical or analytical). that is |phi> = sigma ai|xi>

    Now, you know that by measuring the system, the system will collapse to one of its eigenvectors , that is either x1, x2, x3, x4 and so on and you're pretty sure that it will definitely collapse to one of this eigenvector.

    Suppose, by measuring this system and it collapse to the eigenvector x2. The probability of obtaining x2 is (a2)^2. (I guess you may misunderstand the meaning of probability here, here means out of let's say 1000 SAME SYSTEM YOU'RE MEASURING, you will get (a2)^2 *1000 systems that will produce eigenvectors x2!!! This is a very important concept)

    But, let's say you didn't try out on the 1000 same system, but you do an experiment on the previous system that has already collapsed to x2, then the system that has already collapsed will make you to only give you an eigenvalue of a2 forever.
     
  14. Oct 17, 2007 #13

    cks

    User Avatar

    This means your probability of obtaining x2 for the system that has already collapsed is definitely one because you will forever get the eigenvalue a2.
     
  15. Oct 17, 2007 #14

    cks

    User Avatar

    Now, i'm pretty sure you know what I mean.

    I understand what you mean by smaller eigenvector. If the system you're measuring is of non-degenerate(means only one eigenvector corresponding to one eigenvalue), then you'll never be able to get "smaller eigenvector" for example, infinite squre well.
     
  16. Oct 17, 2007 #15

    cks

    User Avatar

    If you want to get smaller eigenvector, you're to measure those system with degeneracy for example hydrogen atom.
     
  17. Oct 17, 2007 #16

    cks

    User Avatar

    If I say the wrong things, please correct me.
     
  18. Oct 17, 2007 #17
    thank you

    Thank you all for your replies. I found the answer and some more enthusiasm to study quantum!
     
  19. Oct 17, 2007 #18

    f95toli

    User Avatar
    Science Advisor
    Gold Member

    What is not physics?
    And what do you mean by "ref on Zurek"?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Clarification on wavefunction collapse
  1. Wavefunction collapse (Replies: 8)

  2. Wavefunction Collapse (Replies: 5)

  3. Collapse Wavefunction (Replies: 2)

  4. Wavefunction collapse (Replies: 1)

Loading...