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B Position of a particle immediately after a measurement

  1. Sep 15, 2016 #1

    Titan97

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    Say a particle was measured to be at a point C. Immediately after this, if i make another measurement, its given in griffith's book that the particle will still be found at C. Isn't this only possible if the particle was at rest?
     
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  3. Sep 15, 2016 #2

    A. Neumaier

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    is short for ''after time ##\Delta t## in the limit of ##\Delta t\to 0##''.
     
  4. Sep 15, 2016 #3

    Titan97

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    I am sorry. I don't get you :)
     
  5. Sep 15, 2016 #4

    Nugatory

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    We need to be clear about what is meant by "immediately". The shorter the time interval between the two measurements, the greater the probability that the second measurement will yield the same position as the first, and as that interval approaches zero the probability approaches certainty. That's what they mean by "immediately".
     
  6. Sep 15, 2016 #5

    Titan97

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    I need clarification on this: If the particle was found at C, does it mean there was a psi^2 maxima at/near C?
     
  7. Sep 15, 2016 #6

    Nugatory

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    The position operator does nor commu
    The particle is more likely to be found near that maximum so wherever you find it, that maximum is likely to be nearby. Once you find the particle to be at some particular location, the wave function collapses to a state in which the maximum is where you did find it.
     
  8. Sep 15, 2016 #7
    Ah I see. Before I thought measuring a quantum system twice, in immediate succession, will give the same result as measurement #1 did. Now I see it is more likely for this to happen, and there is always the probability of not getting the same result as the first measurement, provided the time interval between measurements isn't 0.

    I am reminded of the quantum zeno effect, but a quick google states it is around freezing the system's initial state.
     
  9. Sep 15, 2016 #8

    Nugatory

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    That's not quite right either; it depends on whether the observable in question commutes with the Hamiltonian. The measurement collapses the wave function to an eigenstate of the observable, and you'll see from the time-dependent Schrodinger equation that it will stay in that state if the observable commutes with the Hamiltonian but not otherwise.

    The position operator does not commute with the Hamiltonian, so we expect the position to evolve over time. (There are additional complications because the position operator has no physically realizable eigenstates, but that's a digression here).
     
    Last edited: Sep 16, 2016
  10. Sep 16, 2016 #9
    Thank you for the clarification!
     
  11. Sep 16, 2016 #10

    Titan97

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    Why does this happen? I mean why does the wave function collapse to an eigenstate after a measurement? What if it was never measured?
     
  12. Sep 16, 2016 #11

    PeroK

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    If you don't measure, then you don't know!
     
  13. Sep 16, 2016 #12

    Titan97

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    :DD
    what about this part of the question:

     
  14. Sep 16, 2016 #13

    PeroK

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    That's how observables work at the quantum level. Every observable is represented by an "operator". The only possible measurements are eigenvalues of that operator, and any measurement leaves the system in an eigenstate corresponding to the measured eigenvalue.

    Why? It's a postulate of QM. See for example:

    http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/qm2.html
     
  15. Sep 16, 2016 #14

    Titan97

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    Can you recommend a better book than griffiths? I keep getting more doubts. Like in the equation
    $$<p>=\int\psi *\frac{h}{2\pi i}\frac{\partial \psi}{\partial x}dx$$
    why did griffith choose
    $$\frac{h}{2\pi i}\frac{\partial}{\partial x}$$
    as the momentum operator?
     
  16. Sep 16, 2016 #15

    PeroK

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    You may find that that is a popular choice for the momentum operator. Try googling or take a look at:

    https://en.wikipedia.org/wiki/Momentum_operator

    Re text books, see previous threads, e.g.

    https://www.physicsforums.com/threads/introduction-to-qm-help.876886/#post-5508170

    The main alternatives to Griffiths seem to be Shankar, Sakurai and Ballentine. The consensus is, perhaps, that Griffiths is the most accessible, if not the best.
     
  17. Sep 16, 2016 #16

    Titan97

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    @PeroK is it wise to stick with Griffith considering my zero knowledge in linear algebra? Where can I learn linear algebra?
     
  18. Sep 16, 2016 #17

    PeroK

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    If you are actually trying to learn QM, you need linear algebra. There's no way otherwise. In a sense, QM is applied linear algebra. Griffiths goes perhaps as light as possible. There are other threads on here about linear algebra books. You need to go as far as Hilbert spaces at least.
     
  19. Sep 17, 2016 #18

    vanhees71

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    Well, I've the impression that Griffiths leads to a lot of confusion given the many questions by readers of this book here. I, however, cannot really judge it since I've only glanced over it yet. My favorite as an introductory textbook is Sakurai. Concerning interpretation, I recommend to read Ballentine.

    The (at least for me) convincing answer to the question, why the operators representing observables take the form they take in the usual representations (usually the position representation when the QM course uses the wave-mechanics approach which is pretty common for historical reasons) are symmetries. The fundamental physics nowadays is based on symmetry considerations. All starts with the principle structure of QT in terms of an observable algebra, realized as a representation on a Hilbert space. Then to fill this with concrete live for applications you have to start with the usual space-time models. To begin with, it should of course be Galilei-Newton spacetime although from a group-theoretical point of view it's a bit more complicated than Minkowski space. You meet the full case of complications, i.e., that the quantum realization of the Galilei group is both a covering group (substituting SU(2) as the representation of rotations in favor of the classical rotation group SO(3)) and a central extension (introducing mass as a central charge of the Galilei group; the mathematically possible case of 0 mass doesn't give a sensible physical model in non-relativstic QT) of the classical group.

    As very first aproach to the use of group theoretical arguments you can however boil down the problem to describe a particle in one-dimensional motion first, as is usually done in the hand-waving approach to wave mechanics. Roughly the argument goes as follows: You start somehow heuristically to introduce the wave function ##\psi(t,x)## as the probability amplitude and as a member of the Hilbert space of square-integrable functions of real numbers, ##\mathrm{L}^2(\mathbb{R})##. Symmetries are realized as unitary transformations on this Hilbert space (this is a bit imprecise but as a heuristic introduction it's enough to avoid confusion about subtle details which become important only later). The only symmetry of the real line as a model for space is its translation invariance (homogeneity), and you now may ask, what's the unitary transformation that describes translations, but that's very simple: Consider an observer A (Alice) who puts the origin of her coordinate at some place and another observer B (Bob) who puts the origin at the coordinate ##a## (in terms of Alice's reference frame). Then B uses ##x'=x-a## instead of Alice who uses ##x## as her coordinate for the location of the particle on the line. Now it's clear what happens: If the particle is in a (pure) state described by Alice by the wave function ##\psi(t,x)##, also for Bob it should be described by this wave function but in terms of his reference frame, ##\psi'(t,x')##. Which leads to
    $$\psi'(t,x')=\psi(t,x)=\psi(t,x'-a).$$
    This implies that the translation from A's to B's reference frame is described by the unitary operator
    $$\hat{T}(a) \psi(t,x)=\psi(t,x-a).$$
    That this is indeed a unitary transformation is trivial since for any two wave functions you have
    $$\langle \hat{T}(a) \psi_1|\hat{T}(a) \psi_2 \rangle=\int_{\mathbb{R}} \mathrm{d} x \psi_1^*(t,x-a) \psi_2(t,x-a) = \int_{\mathbb{R}} \mathrm{d} x \psi_1^*(t,x) \psi_2(t,x) = \langle \psi_1|\psi_2 \rangle.$$
    Now in classical mechanics momentum is the generating function of the canonical transformation in Hamilton mechanics that describes spatial translations, and so we define the momentum operator in quantum theory. Since we want to have self-adjoint rather than anti-self-adjoint operators we introduce and additional factor of ##\pm \mathrm{i}## (which sign you choose is convention, and we use simply the convention used all the time from Schrödinger's seminal paper on wave mechanics on).

    For an infinitesimal translation we have
    $$\hat{T}(\delta a) \psi(t,x)=\psi(t,x-\delta a)=\psi(t,x)-\delta a \partial_x \psi(t,x) + \mathcal{O}(\delta a^2)=:\psi(t,x) -\mathrm{i} \delta a \hat{p} \psi(t,x)+\mathcal{O}(\delta a^2).$$
    Now we have by comparison
    $$\hat{p}=-\mathrm{i} \partial_x.$$
    Now all this is in natural units, where ##\hbar=1##. If you want conventional units you have to introduce the appropriate power of ##\hbar## which by dimensional reasoning leads to
    $$\hat{p}=-\mathrm{i} \hbar \partial_x.$$
    This establishes why ##\hat{p}## should take the form it does from symmetry principles taken right from the classical theory (in terms of Hamilton mechanics).

    Now we also can state another important piece of mathematics. The whole formalism can be just based on purely algebraic arguments using the symmetries. The infinitesimal translations can be defined by the Lie algebra, i.e., the algebra of linear operators in Hilbert space with the composition as product. For the Lie algebra you only need the commutators, and the only non-trivial commutator of the algebra of a quantized particle along a line is that between position and momentum. Using the above result (again with ##\hbar=1## for convenience) we find
    $$\hat{x} \hat{p} \psi(t,x)=x (-\mathrm{i} \partial_x) \psi(t,x)$$
    and
    $$\hat{p} \hat{x} \psi(t,x)=(-\mathrm{i} \partial x) x \psi(t,x)=x(-\mathrm{i} \partial_x) \psi(t,x)-\mathrm{i} \psi(t,x),$$
    which by subtraction from the previous equation leads to
    $$(\hat{x} \hat{p}-\hat{p} \hat{x}) \psi(t,x)=[\hat{x},\hat{p}] \psi(t,x)=\mathrm{i} \psi(t,x) \; \Rightarrow \; [\hat{x},\hat{p}] =\mathrm{i} \hat{1}.$$
    You can build all the rules concerning operators of a particle moving along a line from this commutation relation!
     
  20. Sep 17, 2016 #19

    PeroK

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    That may be because Griffiths has a wider readership likely to be confused by QM. Or, at least, a wider readership amongst those completely new to QM. Ballentine's readers may, in general, be more mature in their knowledge of QM. I have a hunch that those confused by Griffiths would be just as confused by Ballentine!
     
  21. Sep 17, 2016 #20

    vanhees71

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    I'd not start with Ballentine, because it's too advanced. You get lost too early in a lot of debates of interpretation. As I said, I would start with Sakurai to learn the formalism first. I have the impression Griffiths take on the foundations of the formlism is too imprecise in an attempt to simplify the subject, but as Einstein said you should explain things as simple as possible but not simpler. Oversimplification is a barrier to learn a new subject as bad as the barrier errected by being too detailed an precise in the beginning. It's a really difficult task for a textbook writer to find the ballance between these contradicting poles!
     
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