Can we measure the momentum of a particle in a specific position?

[amjad-sh]

Hello all;
Sorry if my questions are stupid or weird, but I'm still a beginner in quantum mechanics. So please be patient with me!
I'm reading shankar text of quantum mechanics,and I reached the part related to momentum operator and the momentum wave function in position basis.
He derived that \langle x | p \rangle=\frac{1}{√2πħ} e^{ipx} which means that it is a plane wave.
so \langle x | p \rangle can't be normalized and we can't associate with them a sensible probability distribution as the probability density will be uniform in all space.
But prof shankar then said:"But since the plane waves are eigenfunctions of P does it means that well-defined states of momentum do not exist ? yes,in strict sence".

my first question is:1)does he mean by saying"well-defined states of momentum do not exist" that we can't measure the momentum of a particle in a specific position?I know that we can't but what he means by this?
2)does this mean that |p> does not belong to Hilbert space?

Now prof shankar continued:"however there do exist states that are normalizable to unity(i.e correspond to proper vectors)and come arbitrarily close to having a precise momentum".

3) does this mean that for proper states we can detect momentum which is very close to precise momentum?
while for improper states it is impossible to detect momentum?

then prof shankar said:"thus a particle coming out from an accelerator with some advertised momentum ,say 500 Gev/c,is in a proper normalizable state(since it is located in our laboratory)and not in a plane wave state correponding to | p=500Gev/c \rangle.
4) I didn't get this at all, if someone can simplify it to me.

Thanks in advance!

 

[DrDu]

Yes, 2) means that |p> does not belong to the Hilbert space.
Concerning 3, we can't prepare improper states, in the first place. Hence the question whether we could measure their momentum makes no sense.

 

[amjad-sh]

Concerning 3, we can't prepare improper states, in the first place. Hence the question whether we could measure their momentum makes no sense.

why we can't prepare them?

 

[DrDu]

Because they aren't in Hilbert space.

 

[amjad-sh]

Because they aren't in Hilbert space.

why can't we measure accurate momentum for proper states,just close to accurate?

 

[DrDu]

I think this can already be understood in the case of classical light. Can you prepare a truly monochromatic lightwave? No, because a monochomatic light wave would have infinite size. The light produced by any source will be localized and each monochromatic wave will only contribute infinitesimally to a localized wavepacket.

 

[kith]

then prof shankar said:"thus a particle coming out from an accelerator with some advertised momentum ,say 500 Gev/c,is in a proper normalizable state(since it is located in our laboratory)and not in a plane wave state correponding to | p=500Gev/c \rangle

Having the particle located in the laboratory gives you a maximum position uncertainty (maybe something like 10\,m). Using the Heisenberg uncertainty relation, you can calculate a minimum uncertainty for the momentum. Although this is extremely small, it isn't zero. So you don't have a plane wave with a sharp wavelength but something like a Gaussian distribution of momenta which is centered very narrowly at p = 500\,GeV.

 

[amjad-sh]

Thanks for the replies, but I still didn't get what "states of well-defined momentum do not exits" really means??

 

[kith]

Have you understood DrDu's post #6?

If yes, then the basic logic is simple: if a certain state can't be prepared even in principle, then it doesn't describe a possible physical situation and therefor is unphysical. Related with this is a more technical reason. It is a central postulate of QM, that the possible states of a quantum system form the mathematical structure of a Hilbert space. The elements of this space need to be square-integrable which means that a certain integral over all of space needs to be finite (this is related to the probabilistic part of QM). This isn't the case for momentum eigenstates because they extend over all of space. So only approximations of momentum eigenstates may describe actual physical situations. However, it is often convenient to use the unphysical, idealized states in calculations.

If no, you need to be more precise about what you don't get.

 

[amjad-sh]

I think I got the idea much more, thanks to all (DrDu and kith)

 

[f95toli]

I have a follow up question.
What happens if we instead talk about generalized momentum? In an electronic system this would be charge, which (presumably) is well defined and you can prepare eigenstates (for the sake of argument we could consider a simple LC-oscillator).

It is not clear to me if the same applies in this case.

 

[DrDu]

What do you mean with generalized momentum? Charge can take on only discrete values in contrast to momentum.

 

[PeroK]

I have a follow up question.
What happens if we instead talk about generalized momentum? In an electronic system this would be charge, which (presumably) is well defined and you can prepare eigenstates (for the sake of argument we could consider a simple LC-oscillator).

It is not clear to me if the same applies in this case.

Angular momentum could be an example of generalised momentum, where physically realisable eigenstates exist.

 

[f95toli]

What do you mean with generalized momentum? Charge can take on only discrete values in contrast to momentum.

In e.g. solid-state qubits we refer to charge and phase as generalized momentum and position, respectively.
This turn out to be pretty natural (e.g. the "mass" of en electronic circuit is given by its capacitance) and there is also a charge-phase uncertainty relation. The momentum-position / charge-phase correspondence turns out to be "exact" at least in most practical situations, and I was wondering if this was actually an example of when this was not true.

A good introduction can be found in Devoret's review on "Quantum Fluctuations in Electrical Circuits" (should be freely available online)

 

[vanhees71]

Again, it is important to distinguish between the principle quantum-theoretical restriction to determine an observable precisely (i.e., to prepare a system in such a way that an observable, e.g., momentum, takes a precise value) and the possibility to measure this quantity. You can measure the momentum of a particle as precisely as you like (of course, as a continuous variable there will be always a finite resolution of your measurement device, but this you can in principle make as small as you want, given you have the resources to do so).

The Heisenberg uncertainty relation for position and momentum $\Delta x \Delta p \geq \hbar/2$ in its usual textbook meaning tells you that there is no state, which makes position and momentum of a particle both sharply defined. Of course, you can make $\Delta p$ as small as you like (that's what the accelerator physicists try to achieve when constructing an accelerator like the LHC), but at the cost of the precision of its position. Indeed, in the usual accelrators the particles come in bunches which are pretty wide along the beam axis with a pretty well determined momentum.

You can measure the momentum of each particle of course as precisely as you want, supposed you have the corresponding detector resolution.

 

[DrDu]

In e.g. solid-state qubits we refer to charge and phase as generalized momentum and position, respectively.

Ok, but phase is not even a hermitian operator, so the relation seems a bit far fetched.
The point of whether a variable can be measured with vanishing dispersion or not does not depend on the variable being momentum like but on it having a continuous spectrum. If the spectrum is continuous the corresponding eigenstates don't lie in Hilbert space (though they may be elements of some extended space, like a rigged Hilbert space).

 

[f95toli]

The point of whether a variable can be measured with vanishing dispersion or not does not depend on the variable being momentum like but on it having a continuous spectrum. If the spectrum is continuous the corresponding eigenstates don't lie in Hilbert space

I understand that. However, we can design circuits where either the charge (i.e. generalized momentum) or phase (generalized position) are quantized (depending on which energy scale dominates). Hence, coming back to the OP it would seems that for some types of generalized momentum we can talk about well-defined states.

As far as I remember one can solve the problem of there not being a "proper" Hermitian phase operator by introducing two operators (something like \hat{\sin \phi} and \hat{\cos \phi} ). However, for most practical situation one can to a very good approximation introduce a phase operator. You can probably find the details in Zagoskin's book on Quantum Engineering. Note also that the first experimental demonstration of macroscopic quantum tunneling was actually tunneling of phase in a Josephson junction. and
phase qubits is currently one of the leading contenders to build a (simple) quantum computer with error correction (surface code). this is what is used by e.g. Google. .

 

[DrDu]

I understand that. However, we can design circuits where either the charge (i.e. generalized momentum) or phase (generalized position) are quantized (depending on which energy scale dominates). Hence, coming back to the OP it would seems that for some types of generalized momentum we can talk about well-defined states.

Of course. Although for me PeroK's example of angular momentum corresponds more to my understanding of a generalized momentum. I.e., it is the generator of some symmetry which is not so obvious with charge.

 

[vanhees71]

Well, charge itself is the generator of a symmetry. E.g., in electrodynamics it's the phase invariance of the Lagrangian (and thus the action) of the electron-positron field, i.e., the invariance under the global transformation
$$\psi \rightarrow \exp(\mathrm{i} \alpha) \psi, \quad \bar{\psi} \rightarrow \bar{\psi} \exp(-\ii \alpha) \psi.$$
The Noether current is the vector current
$$j^{\mu}=\bar{\psi} \gamma^{\mu} \psi.$$
Thus it obeys the continuity equation and thus
$$Q(t)=\int \mathrm{d}^3 \vec{x} j^0(t,\vec{x})=\text{const}$$
and it's the generator for the symmetry transformation (in the quantized version of this game).

Gauging this symmetry together with the principle of minimal coupling leads directly to QED.

 

[DrDu]

Yes, I should have said space time symmetry. The pair angular momentum - angle is quite analogous to the pair charge - phase.
The charge operator is more complicated, in my opinion, because there is a superselection rule for charge.

 

[vanhees71]

True, but there's also a superlection rule for angular momentum. There must not be any superposition of half-integer with integer angular-momentum states, because otherwise rotation symmetry would not be realized.