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Most basic question about psi ever

  1. Aug 21, 2014 #1

    fluidistic

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    Hi guys,
    I thought that |psi|^2 was a probability density function and that when integrated over a region this would give me the probability to find the particle in that region. (assuming that psi was normalized)
    But some ph.d. in physics told me that no, that's wrong and when I use the word particle I'm already using an interpretation of QM.
    I notice that Landau never mention the word "particle" when he defines psi.

    Now that I think about it, if I have a system of many particles and a single psi representating the probability density funtion of all those particles, I do not really know what the integral of |psi|^2 would represent in that case. The prob. to measure a single one of these many particles inside the integrated region?

    Am I completely off?
     
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  3. Aug 21, 2014 #2

    atyy

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    The Born rule says that the probability to find the system with state |ψ> in a state |ø> is |<ø|ψ>|2.

    Let's do a system with one particle first. The state |x> means the particle has a definite position x. So the probability to find the particle at position x is the probability to find the particle in state |x>. By the Born rule, that probability is |<x|ψ>|2 = |ψ(x)|2.

    For a system with two non-identical particles, the state |x,y> means particle 1 has position x and particle 2 has position y. Then the probability to find patricle 1 at position x and particle 2 at position y is |<x,y|ψ>|2 = |ψ(x,y)|2

    http://galileo.phys.virginia.edu/classes/252/symmetry/Symmetry.html
    http://www.eng.fsu.edu/~dommelen/quantum/style_a/complex.html
     
    Last edited: Aug 21, 2014
  4. Aug 22, 2014 #3
    Here is an example about "A wave function for a single electron". Only shows the places where the electron's probability density is above a certain value (here 0.02 nm−3) : this is calculated from the probability amplitude.

    They are also very careful in the dialectic.

    Patrick
     
  5. Aug 22, 2014 #4

    Nugatory

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    Not so much wrong as incomplete. The way you were thinking about it is fine for a system consisting of a single particle constrained to move only on the x-axis, but it doesn't work for anything more complicated - as you've discovered when you found yourself asking:

    Consider the two-particle case: we have two particles, both constrained to move only on the x-axis. Now we write the wave function as ##\psi(p,q)## where ##p## and ##q## are the positions of the two particles. The probability of finding the first particle in the region bounded by ##A## and ##A+\Delta{A}## and the second particle in the region bounded by ##B## and ##B+\Delta{B}## is given by
    [tex]
    \int_A^{A+\Delta{A}}\int_B^{B+\Delta{B}}\psi^*(p,q)\psi(p,q)\;\; dpdq
    [/tex]
    There are two things to notice here.
    First, this is not an integration across physical space. Instead, it's an integration across a region of an abstract two-dimensional space in which the two axes correspond to the two positions of the two particles on the physical x-axis.
    Second, this solution is formally identical to the solution for a single particle that is free to move in about in two dimensions. In fact, it works for any system with two degrees of freedom; ##p## and ##q## don't even have to represent physical positions. They can be whatever dynamical variables are needed for the problem at hand.
     
  6. Aug 22, 2014 #5
    Perhaps when you read this kind of articule "There are no particles, there are only fields" on arxiv. Calculate the position is not it already an interpretation ?

    This is how I understand the sentence : "But some ph.d. in physics told me that no, that's wrong and when I use the word particle I'm already using an interpretation of QM".


    However, I may be wrong.

    Patrick
     
  7. Aug 22, 2014 #6

    bhobba

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    Its not an interpretation - its an assumption - namely position is an observable.

    The link you gave is correct - there are no particles - only fields.

    From those fields position may or may not be an observable:
    http://www.mat.univie.ac.at/~neum/physfaq/topics/position.html

    Thanks
    Bill
     
  8. Aug 22, 2014 #7

    Nugatory

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    It might be time to start a new thread here - we're moving way beyond "the most basic question about psi ever"
     
  9. Aug 22, 2014 #8
    Is it the same in QFT ? Is there a position operator in second quantization ?

    Patrick
     
  10. Aug 22, 2014 #9

    bhobba

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    Well, second quantisation is a rather bad name.

    There is a basic difficulty with normal QM and relativity.

    Position is an operator, and time a parameter, yet relativity says they should be treated on the same footing.

    QFT treats both as a parameter ie a field.

    So, no, position is not inherently an observable in QFT, and that is by design.

    Thanks
    Bill
     
  11. Aug 22, 2014 #10

    stevendaryl

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    That was something that occurred to me just recently, and was briefly discussed in some thread. It seems that the physics of 3-dimensional space with N particles is almost the same as the physics of 3N-dimensional space with 1 particle. The difference is locality of interactions. For some interactions that seem local in 3N-dimensional configuration space, the corresponding interaction in 3-dimensional space would be nonlocal.
     
    Last edited: Aug 22, 2014
  12. Aug 22, 2014 #11
    In this article the author write : "Localization via position operators is in conflict with causality".

    He give also comments on the question whether these mathematical differences have significant consequences for the physical interpretation of basic concepts of QM.

    Patrick
     
  13. Aug 22, 2014 #12

    atyy

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  14. Aug 22, 2014 #13

    bhobba

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    I had a quick look - it looked fine to me.

    What was your issue.

    Thanks
    Bill
     
  15. Aug 22, 2014 #14

    atyy

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    In non-relativistic quantum mechanics, there are certainly systems with a fixed number of particles. These are not classical particles. They are quantum particles, which do not simultaneously have definite position and momentum. Thus they do not have classical trajectories.

    The matter of interpretation is whether they have non-classical trajectories. In Bohmian Mechanics, they do.

    In exact relativistic quantum field theory, it can be argued that fields are more fundamental. Here the quantum particle is an excitation of the field. However, no one can guarantee that relativity is exact, so even to explain relativistic phenomena, quantum mechanics with a fixed number of entities is viable, as in lattice gauge theory.
     
  16. Aug 22, 2014 #15

    atyy

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    The abstract claims "As this paper shows, experiment and theory imply unbounded fields, not bounded particles, are fundamental." That is wrong. Experiment only goes up to a certain energy, and has a certain precision. Claiming to know the final theory is as much pseudoscience as all the things he criticizes.

    For example, can he distinguish exact relativistic theories defined in infinite volume from a lattice gauge theory with fine but finite lattice spacing in large but finite volume? The Hilbert space of the latter is finite dimensional, and so more like quantum mechanics than quantum field theory.
     
    Last edited: Aug 22, 2014
  17. Aug 22, 2014 #16

    bhobba

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    Yea - fair point.

    But I am sure the author would agree and point out you are reading more into it than intended.

    QFT is a deeper more exact theory than standard QM - that's all that's meant - not that its the final theory.

    Thanks
    Bill
     
  18. Aug 22, 2014 #17

    atyy

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    If he is talking about QFT, then it is roughly a theory with an infinite number of particles, or at least a theory in which particles can be created and destroyed. So it is a theory with more than one particle. But he writes "Thus the Schroedinger field is a space-filling physical field whose value at any spatial point is the probability amplitude for an interaction to occur at that point." and "It follows that the Schroedinger matter field, the analogue of the classical EM field, is a physical, space-filling field." That is of course fine for one non-relativistic particle, and a very useful mental image. But given his concern about quantum field theory, this picture of the single particle wave function is one that does not generalize to more than one particle. The wave function of two particles is not in general "space filling" in the sense that he means.
     
  19. Aug 22, 2014 #18

    Demystifier

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    Two (hopefully non-controversial) points:

    First, in non-relativistic QFT (used e.g. in solid state physics), wave function is still the probability amplitude for the position of the particle.

    Second, in fermionic QFT, the field is not even an observable.

    In addition, 3 different concepts are often mixed up: relativistic QM, QFT (which may be non-relativistic), relativistic QFT.
     
  20. Aug 22, 2014 #19

    bhobba

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    Indeed, as the Fock space formalism shows. And even deeper, when deriving QFT, one often uses a model of a large number of interacting particles that you take the limit in the continuum.

    That does not contradict the infinite particle picture.

    A quantum field can be viewed as a field of operators. The space it operates on is spanned by the creation and annihilation operators. Its two sides of exactly the same coin.

    I don't think its fine for a single particle. A matter field, just like any quantum field, is a field of quantum operators acting on the creation and annihilation operators. You cant speak of single particles in that formalism.

    The paper is not meant to be a rigorous development of QFT. It is meant to be a review article of the idea that quantum fields are a deeper theory than normal QM, and as such explains many of the issues people get worried about with QM.

    He freely admits it doesn't solve the measurement problem, but many things like the double slit experiment become clearer.

    Also it soon will be published, if it already hasn't been, in the AJP. Articles with glaringly obvious issues are normally not published there, they are vetted thoroughly to ensure that doesn't happen. Of course no process is perfect and some incorrect things can creep through - but I don't think that's what's going on here. Since its explaining highly complex ideas that should really be expressed mathematically in English subtleties are bound to arise. IMHO that all that's going on - overall it is correct.

    It's at a bit more advanced level to its beginner companion book - Fields Of Color:
    https://www.amazon.com/Fields-Color-theory-escaped-Einstein/dp/0473179768

    But if you want a more advanced look again check out:
    https://www.amazon.com/An-Introduction-Realistic-Quantum-Physics/dp/9812381767

    The author may have a copy on his site as well - but don't hold me to it.

    That QFT explains many of the issues (not all - but many) with standard QM is for me not controversial at all.

    Thanks
    Bill
     
    Last edited by a moderator: May 6, 2017
  21. Aug 22, 2014 #20

    atyy

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  22. Aug 23, 2014 #21

    fluidistic

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    Thank you guys.
    If I understood well I wasn't wrong for non-relativistic QFT (and non relativistic QM?) but things get much more complicated when dealing with relativistic QFT, in other words theories that are supposed to be more precise/accurate than the one I'm currently learning.
     
  23. Aug 23, 2014 #22

    stevendaryl

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    What are you calling "the wave function" in QFT? In QFT, there are two different things that might be considered analogous to the wave function of single-particle, non-relativistic quantum mechanics: (1) The state (which is a kind of many-body wave function), (2) the field operator, which obeys an equation similar to the Schrodinger equation, but is an operator, rather than a complex-valued function.

    The state is an element of Fock space. For a fixed number of particles, it's basically a many-particle wave function. But in working with QFT, people tend not to deal with the state, and mostly work with the field operators. The field operator [itex]\phi(x)[/itex] is an operator for creating a particle at location x, rather than the probability amplitude for a particle being at x.

    ...Oh, actually, now that I think about it, the expression [itex]\phi^\dagger(x) \phi(x)[/itex], does give the probability (density) of finding a particle at x. So I guess that is analogous to the same expression in non-relativistic quantum mechanics. It seems very different to me, since it's an operator, rather than a number.
     
  24. Aug 23, 2014 #23

    atyy

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    Yes. Non-relativistic QFT is exactly non-relativistic QM of many identical particles. It is exactly the same theory in a different language. The QFT formulation is called "second quantized" while the QM formulation is "first quantized". So for all non-relativistic quantum theories, the traditional textbook approach of starting with non-relativistic QM, is a great foundation. In both non-relativistic QFT and non-relativistic QM, the wave function ψ(x) has the meaning you stated in the OP, and does generalize as I described in post #2, and which Nugatory gave with even better mathematical accuracy in post #4.

    Even when one gets to relativistic QFT, one can approach via fields (eg. Srednicki http://web.physics.ucsb.edu/~mark/qft.html ) or via particles (eg. Weinberg https://www.amazon.com/The-Quantum-Theory-Fields-Volume/dp/0521670535 or Nair https://www.amazon.com/Quantum-Field-Theory-Perspective-Contemporary/dp/0387213864). The Hilbert space called Fock space in the non-rigourous QFT given in Srednicki, Weinberg, or Nair is formed by combining the spaces of quantum particles. Only in very strictly rigourous, exactly relativistic QFT is the particle approach less fundamental, because there the Hilbert space is not a Fock space.

    In all cases, the quantum particle does not have simultaneously well-defined position and momentum, ie. it does not have a classical trajectory. That is the important point about quantum particles. In non-relativistic quantum theory, it is well accepted that as a matter of interpretation, one can assign trajectories to quantum particles using some form of Bohmian Mechanics, but these trajectories are not governed by equations of motion that are the classical limit of the quantum equations.
     
    Last edited by a moderator: May 6, 2017
  25. Aug 25, 2014 #24

    Demystifier

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    I meant (1), of course.
     
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