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Position and Momentum

  1. Aug 24, 2005 #1
    What else? :smile:

    O.K., so there it is: I must be having a really stupid misconseption, so be gentle.

    X and P can never have rigorous (that is, without delta functions and infinity barriers) eigenstates, right?
    So, when I measure, for example, X, to what state does the system collapse?
     
  2. jcsd
  3. Aug 24, 2005 #2
    It collapses to a delta function centered at the measured position (in the case of an ideal measurement). You are, of course, leaving off the spin component of angular momentum.

    A delta function is normalizeable, it just doesn't really have continuous anything.
     
  4. Aug 24, 2005 #3

    George Jones

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    After a measurement on a system in state [itex]|\psi>[/itex], the state of the system is [itex]P|\psi>[/itex], where [itex]P[/itex] is an appropriate projection operator. This is true for observables that have non-degenerate spectra, degenerate spectra, continuous spectra, etc.

    No apparatus that measures an observable with a continuous is selective enough that a precise value is singled out.

    Consider a position measurement. The apparatus used to make the measurement will only be able to ascertain that the position lies within a certain interval, say [itex]x[/itex] is in the interval [itex]\left[a , c \right][/itex]. Then [itex]P = \int_{a}^{c} dx |x><x|[/itex] and after the measurement the state of the system is [itex]|\psi '> = \int_{a}^{c} dx |x><x|\psi>[/itex].

    Regards,
    George
     
  5. Aug 24, 2005 #4
    Thanks a lot!
     
  6. Aug 24, 2005 #5

    George Jones

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    Note that the states given in my previous post are normalizable, but not normalized. Usually, if [itex]|\psi>[/itex] is normalized and [itex]P[/itex] is a projector, [itex]P|\psi>[/itex] is not normalized. The normalized post-measurement state is [itex]P|\psi>/||P|\psi>||[/itex].

    Regards,
    George
     
  7. Aug 24, 2005 #6
    Suppose the state of the system is [itex]|\psi>[/itex]. Measurement of the position means to apply the operator X to the state [itex]|\psi>[/itex] to obtain the eigenvalue of position x, i.e. X[itex]|\psi>[/itex] = x[itex]|\psi>[/itex]. The system then falls into an eigenstate of position which we can write as [itex]|x>[/itex].

    On the other hand if you were to measure x-momentum you'd get something similar, i.e. measurement of x-momentum means to apply the operator Px to the state [itex]|\psi>[/itex] to obtain the eigenvalue of position px, i.e. Px[itex]|\psi>[/itex] = px[itex]|\psi>[/itex]. The system then falls into an eigenstate of mometum which we can write as [itex]|p_x>[/itex].

    None of this means that you can't simultaneously measure px and x exactly. What you can't do is start with the same system and expect to get the same eigenvalues regardless of what the other eigenvalue is. This implies there is no classical trajectory. This seems to be a huge misconception in QM but this is the correct interpretation.

    Pete
     
  8. Aug 25, 2005 #7

    Galileo

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    To expand on George's post. You are right in saying that although delta functions and plane waves are eigenstates of the position and momentum operators, they can not correspond to a physical state, since these are not properly normalizable. This is the case when measuring any observable with a continuous spectrum.

    Any measurement on a physical observable with a continuous spectrum has some uncertainty due to the limited selectivity of the device. For example, a position measurement with a slit has some uncertainty, since the slit must have some width. You can think of it as a filter with a certain bandwidth. (After the measurement you will keep only that part of the wavefunction in the selective interval, as explained in George's post.)

    To show that things can really go wrong if not done right, consider a particle in the ground state of an infinite potential square well of width a. We measure its position and we find the particle in the center: x=a/2. Immediately afterwards we measure its energy. What are the different result we can obtain?
    The following reasoning is wrong. After the position measurement the particle is in the eigenstate corresponding to the result found, so its wavefunction is proportional to [itex]\delta(x-a/2)[/itex]. If we measure the energy, the values that can be found are the usual [itex]E_n=(n\pi \hbar)^2/(2ma^2)[/itex], with probabilities proportional to:

    [tex]\left|\int_0^a \delta(x-a/2)\sqrt{\frac{2}{a}}\sin(\frac{n \pi x}{a})dx \right|^2=\frac{2}{a}\sin^2(\frac{n \pi a}{2})[/tex]

    which is 2/a if n is odd and 0 if it's even.
    So we find that all the probabilities of finding an energy with odd n to be equal. This is absurd, since the probabilities would sum up to infinity.

    We should take into account that, because of limited precision, the position x of the particle is: [itex]a/2-\epsilon/2\leq x \leq a/2 + \epsilon/2[/itex], where [itex]\epsilon[/itex] depends on the measurement device, but is never zero. If we take [itex]\epsilon[/itex] small enough we can say the wavefunction will be practically [itex]\sqrt{\epsilon}\delta^{(\epsilon)}(x-a/2)[/itex]. Where [itex]\delta^{(\epsilon)}(x-a/2)[/itex] is the null function except in the interval [itex][a/2 - \epsilon/2, a/2 + \epsilon/2][/itex], where it takes on the value [itex]1/\epsilon[/itex]. (It is normalized properly).
    Using this wavefunction you find that the probability P(E_n) of finding energy E_n is:

    [tex]P(E_n)=\left\{ \begin{array}{cc} \frac{8a}{\epsilon}(\frac{1}{n\pi})^2\sin^2(\frac{n\pi \epsilon}{2a}) & \mbox{if n is odd} \\
    0 & \mbox{if n is even} \end{array}\right.[/tex]

    If you plot this distribution you will find that it depends heavily on [itex]\epsilon[/itex]. The smaller [itex]\epsilon[/itex] the more the curve points towards the higher energies. It can be interpreted by saying that because the position is well known, it's momentum is very uncertain (could be very large), so kinetic energy is transferred to the particle.
    I think this close relation between prediction and the way you do your measurement is one the more bizarre aspects of QM.
     
  9. Aug 25, 2005 #8
    Galileo, thanks, interesting, I'd never noticed that infinity probability thing before. It could have saved me hours of arguing with certain people.
     
  10. Oct 31, 2005 #9
    Since the new state is very confined in position space, how does the time
    evolution operator make it spread again?
    After measuring position (for a free particle, particle in a box), how does [itex]|\psi '> = \int_{a}^{c} dx |x><x|\psi>[/itex] evolve in time?
    So [itex]|\psi '> = \int_{a}^{c} dx |x><x|\psi>[/itex] is one eigenstate of all the eigenstates that the state before measurement was superposed of? How does a superposition of states reemerge again (after measurement, when Schrödinger eq. reigns again?

    thanks
     
    Last edited: Oct 31, 2005
  11. Oct 31, 2005 #10

    jtbell

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    OK.

    OK.

    That's what I've been thinking for a while, too. However, the preceding statements above have made me have second thoughts. If you simultaneously measure both [itex]x[/itex] and [itex]p_x[/itex], then the system must at that point in time be in a simultaneous eigenstate of both of the corresponding operators. But there can be no such state because the position and momentum operators don't commute, right? Or does that restriction apply only to stationary states?

    I do agree that if you prepare many systems identically, and then simultaneously (or after equal elapsed times for each one), measure the momentum for half of them, and the position for the other half, you can make each individual measurement precisely; but the the two kinds of measurements (position and momentum) will be distributed randomly in a way that their variances satisfy the Heisenberg Uncertanty Principle.

    My doubts have to do with measuring the position and momentum simultaneously for a single system. Is it possible even to conceive a method for doing it?
     
  12. Oct 31, 2005 #11

    Physics Monkey

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    Suppose you could prepare a system with precise values of position and momentum, call the state [tex] | p_0\, x_0 \rangle [/tex]. Let's apply [tex] [X,P] [/tex] to this state and see what we get. We obtain
    [tex]
    i \hbar | p_0 \,x_0 \rangle = [X,P] | p_0 \,x_0 \rangle = (XP-PX) | p_0 \,x_0 \rangle = (x_0\, p_0 - p_0\, x_0) | p_0 \,x_0 \rangle = 0, [/tex]
    This is a contradiction, thus we cannot prepare a system with simultaneous values of position and momentum. If you prefer a less singular discussion, think about spin instead. The same arguement tells you that you can't prepare a system with simultaneous values of [tex] S_x [/tex] and [tex] S_y [/tex], for instance.
     
    Last edited: Oct 31, 2005
  13. Nov 1, 2005 #12
    What's wrong with my post number nine? I'm craving for answers from you knowledgeable people.
     
  14. Nov 1, 2005 #13
    All the eigenstates evolve according to Schrodinger eqn, and the state is given as a sum of the new eigenstates.
     
  15. Nov 1, 2005 #14

    jtbell

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    To take a free particle for example, the new state is a superposition of waves of the form [itex]A \exp [i(kx - \omega t)][/itex]. These waves have different phase velocties, so their phase differences change as time goes on, and the combined wave changes shape as a result.
     
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