4
$\begingroup$

First a simple example to illustrate the essence of the question: If we have a spin half particle in the $S_z+$ state and if one wants to choose from either measuring $S_z$ or measuring $S_y$, which are incompatible, one could execute one of two complementary actions: ''Orient the magnets in the z direction'' or '' Orient the magnets in the y direction''.

Now, the question: What are the corresponding complementary actions in the following case? :

At $\textbf{time t}$, we have a system in a state W such that the standard deviation in position observable x with respect to W is very small and is centered around a coordinate r, and the standard deviation of the incompatible momentum observable $P_x$ is very large(to the extent allowed by the uncertainty relation).

Now $\textbf{at time t}$ we want to choose either to measure x or $P_x$. This one seems a bit tricky to me because of the fact that most of the momentum measurements themselves involve position measurements along with time of flight measurements. Let me elaborate: If i want to execute the choice of measuring position, you might suggest: "Put the position measuring detectors on those spatial regions where there is non vanishing probability of detection at time t."

On the other hand, if I choose to measure momentum, you might say: "Put the position detectors along those directions of possible momentum at time and measure the time of flight ". It is this last suggestion I find a bit unconvincing. Why? Because when I do detect a particular value for the momentum when I execute this last suggestion, then I am really measuring the momentum at time $t+\Delta t$, where $\Delta t$ is the measured time of flight. I am not really measuring the momentum at time t as I did for position. Is there a way to resolve this dilemma? It would be very useful and more clear if the answer involves crisp experimental instruction/protocols.

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ Really nice question! Your dilemma is partly related to the fact that real measurements always involve at least a partial measurement of both position and momentum. Such measurements are possible: they are described by so-called positive-operator-valued measures. See Arthurs & Kelly's seminal article On the simultaneous measurement of a pair of conjugate observables and also Leonhardt's book Measuring the Quantum State of Light, where this question is discussed for light. This kind of measurements are routine in quantum-optics labs. $\endgroup$ – pglpm Jul 9 '20 at 8:08
  • 1
    $\begingroup$ Can I point out that it's not the momentum at $t+\Delta t$, rather the inferred average momentum for the entire span of $\Delta t$ after $t$ $\endgroup$ – Jim Aug 14 '20 at 17:22
0
$\begingroup$

What you propose isn't a momentum measurement. By measuring the position at two different times you have changed the state of your system twice, and you have not made a momentum measurement of the initial system you wanted to make a momentum measurement on.

Improve this answer
$\endgroup$
10
  • $\begingroup$ I agree. But the second position measurement could be considered a momentum measurement if combined with the time of flight measurement. $\endgroup$ – Varun Premkumar Jul 9 '20 at 4:07
  • $\begingroup$ In an ensemble of such setups, the momentum observation from the second position measurement gives a verification of the predicted standard deviation in momentum associated with the state at the end of first measurement(assuming that the state does not evolve significantly between the two measurements). $\endgroup$ – Varun Premkumar Jul 9 '20 at 4:13
  • $\begingroup$ This situation is in contrast with the verification of the predicted standard deviation in position with respect to the state at the end of first measurement: We can make the verification at the same time as the time at which the post measurement state of first measurement is considered. $\endgroup$ – Varun Premkumar Jul 9 '20 at 4:17
  • $\begingroup$ @ImmanuelVarun there is no time of flight measurement in quantum mechanical system dimensions ( where the heisenber uncertainty holds), only probability of finding a particle at (x,y,z,t) $\endgroup$ – anna v Jul 9 '20 at 5:27
  • 1
    $\begingroup$ @annav While you're technically correct, that's not a good argument here because even valid momentum measurements are probabilistic as well. The key point here is that the position measurements change the state of the system, so two position measurements cannot constitute a momentum measurement of the original system. $\endgroup$ – BioPhysicist Jul 9 '20 at 7:20
0
$\begingroup$

The point of TOF measurements is that the momentum doesn't change during the flight. You really do measure the momentum of the particle (in the sense that many measurements of the same setup will tell you the momentum distribution of the wavefunction) by measuring the position in that case, because we can deduce the momentum from the position.

However, in this case you would not prepare a momentum eigenstate, as the state of the particle after the measurement is a position eigenstate. To prepare a momentum eigenstate you really have to do an operationally different setup, which must not involve a position measurement.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.