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Regarding the free particle in QM, we are given that the general wave function is: $$\Psi(x,t) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}\phi(k)e^{i(kx-\frac{\hbar k^2}{2m}t)}dk.$$ The stationary states $$\Psi_{k}(x,t) = Ae^{i(kx-\frac{\hbar k^2}{2m}t)}$$ are not physically realizable since they are not square integrable. So in that sense a particle cannot have a definite momentum energy or momentum. That I understand. I just want to confirm what happens during measurement of say momentum or energy.

So we measure some definite value of momentum or energy which is an eigenvalue of the momentum or Hamiltonian (since the operators commute for a free particle). Then we would in principle collapse the wave function to some stationary state $\Psi_k$ but in this case we know that this is not possible (physically realizable). So do we measure a particular value of the momentum with some measurement uncertainty, with the uncertainty giving us the spread of values of the observable or do we measure a particular momentum and infer that there is a spread from above? Or do we not measure a particular value ever but rather a range of values for a given measurement?

Thanks.

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So we measure some definite value of momentum or energy which is an eigenvalue of the momentum or Hamiltonian (since the operators commute for a free particle).

The result of single measurement can be single value, but in case of quantities that have continuous domain we cannot say this is with certainty the actual value of the quantity. With any such measurement we always have uncertainty of the outcome that is greater than zero. This is unavoidable in practice, we do not have means to measure continuous variables with infinite accuracy.

Then we would in principle collapse the wave function to some stationary state $\Psi_k$ but in this case we know that this is not possible (physically realizable).

It is not important here whether such process is physically realizable; this depends on interpretation of the theory. There are interpretations that do not consider collapse as a result of measurement to be physical process at all, irrespective of whether the result is normalizable.

What is important here is that there is no normalizable function that would be eigenfunction of position operator (and there is no one that would be eigenfunction of the momentum operator). Therefore we cannot base our understanding of the theory on such fictive functions. Particle with definite position or momentum with zero uncertainty cannot be represented by normalized $\psi$ function.

So do we measure a particular value of the momentum with some measurement uncertainty, with the uncertainty giving us the spread of values of the observable

Yes, all measurements of position or momentum of particles have finite uncertainty, so the probability that the measured value equals the actual value that was sought is 0. When we look at particle tracks from bubble chamber, the track is thin but finite width, limiting the uncertainty of the particle coordinate to small but finite distance. In practice, I think microns at best.

Or do we not measure a particular value ever but rather a range of values for a given measurement?

When one particle is measured, one value plus uncertainty is usually recorded. If many particles are measured, then many values and uncertainties are recorded. In any case, no result is ever absolutely accurate, there is always some uncertainty.

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  • $\begingroup$ Those are just words. You may note that there's nothing about uncertainty of the coordinate measurement in $\hat{x}$. That's because it describes idealized arbitrarily precise measurement. If you want to describe actual imprecise measurements you have to use POVM. $\endgroup$
    – OON
    Sep 3 '16 at 7:35
  • $\begingroup$ How do you mean, "just words" - are you saying the answer is lacking in equations? I agree that there is nothing about uncertainty of results of measurement in $\hat{x}$, unless it is an Heisenberg operator, but I do not see why you bring it up. We do not have to talk about acronym high brow concepts to justify Born's formula for probability density. $\endgroup$ Sep 3 '16 at 10:23
  • $\begingroup$ @JánLalinský Thanks for your response. Using the derivation of position and momentum operator as described by Prahar in this other post of mine, firstly is this a satisfactory motivation for defining these observables? Also, is the QM theory and postulates motivated by this uncertainty in practical measurement (I'm referring to the unavoidable uncertainty coming from real world measurements) or does the uncertainty confirm the QM postulates and theory? $\endgroup$
    – Alex
    Sep 3 '16 at 12:42
  • $\begingroup$ In other words, are we trying to find a model which accommodates this uncertainty, so a model which gives a wavefunction which collapses to a wave function which is not normalizable because there are no definite measurements or does the theory precede this and is then the fact that the model accommodates the real world just a confirmation that the theory is correct? $\endgroup$
    – Alex
    Sep 3 '16 at 12:42
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$\hat{\cal{P}}(x)=|x\rangle\langle x|$ defines not a projection operator but what is called projection-valued measure i.e. it gives projection operators for some region $a<x<b$: $$\hat{\cal{P}}_{(a,b)}=\int\limits_a^b dx \hat{\cal{P}}(x)=\int\limits_a^b dx |x\rangle\langle x|$$ Acting on $|\psi\rangle$ it gives normalizable state.

This reflects that we can't talk about the probability of measuring the value $x$ but only probability density $p(x)$ and probability of $x$ being in the region $a<x<b$: $$P(a<x<b)=\int\limits_a^b dx\,p(x)$$ those are given by, $$p(x)=\langle\psi|\hat{\cal{P}}(x)|\psi\rangle=\langle\psi|x\rangle\langle x|\psi\rangle=|\psi(x)|^2$$ $$P(a,b)=\langle\psi|\hat{\cal{P}}_{(a,b)}|\psi\rangle=\int\limits_a^b dx\langle\psi|x\rangle\langle x|\psi\rangle=\int\limits_a^b dx|\psi(x)|^2$$

After the measurement that gives $a<x<b$ the state becomes normalizable $\hat{\cal{P}}_{(a,b)}|\psi\rangle$ or on the language of wavefunctions, $$\hat{\cal{P}}_{(a,b)}\psi(x)=\cases{0,&x<a\\\psi(x),&a<x<b\\0,&x>b}$$


UPDATE: I think it's useful to talk a bit about nature of collapse. If you would like to treat "collapse" as some sort of the objective change of the state you'll get into all sorts of nasty problems. "Collapse" of the wavefunction appears when we consider the conditional probability of some measurement of the initial state given some preceding measurement. It happens (for ideal projective measurements) that we can get this probability as probability of a single measurement of the collapsed state, $$P_\psi\Big(B(t_2)=\beta|A(t_1)=\alpha\Big)=\frac{P_\psi\Big(A(t_1)=\alpha,B(t_2)=\beta\Big)}{P_\psi\Big(A(t_1)=\alpha\Big)}=P_\chi\Big(B(t_2)=\beta\Big)$$ $$|\chi\rangle=\frac{\hat{\cal{P}}_{A(t_2)=\alpha}|\psi\rangle}{\sqrt{\langle\psi|\hat{\cal{P}}_{A(t_1)=\alpha}|\psi\rangle}}$$ For projective measurement of the continous variables you can still find conditional probability densities but this "collapse" idea is not very useful. There are three ways to still apply it, all of which requires you to talk about some error margin,

  1. Remain in the realm of the idealized measurement and restrict yourselves only to discussion of the conditions $a<x<b$ (see above)

  2. Discretize the variable, which may be good mathematically but is not very close to the actual measurements

  3. Realistically your measurements are not ideal and have some fundamental imprecision. That however means that different results for values of $x$ are not mutually exclusive. That means that you are no longer describing your measurement with idealized $x$ with its orthogonal projectors but instead use some POVM that depends on the measurement device you use. Then you can apply collapse idea to the singular value of $x$ but this is no longer $\hat{x}$ in the textbook sense.

As I said you can perfectly live without going into all that if you ask only right questions. Think less about "collapse" and more about what you measure in the experiment.

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