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Topic: Is it possible to calculate the momentum of an electron precisely?  (Read 4291 times)

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Offline AdiDex

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In a national level, undergraduate examination (TIFR-2018) following question was asked -

In my view answer should be (D). My explanation - To calculate an electron's momentum precisely, it must have a totally underminable position! Since our apparatus have some definite size, you have to confine it in some definite space.
ΔxΔp≥ℏ2
Since we want to make Δp to approach zero, Δx must approach to infinite, Which is not possible. So we can't measure the electron's precise momentum. So option (B) shouldn't be the answer.

Moreover, As far as I know, Heisenberg's uncertainty principle restricts us from measuring two quantities(which doesn't commute each other) simultaneously. We can measure them subsequently(one after another). So option (D) should be the answer.

Offline Corribus

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Re: Is it possible to calculate the momentum of an electron precisely?
« Reply #1 on: February 06, 2018, 10:34:01 AM »
I don't like the way this problem is written. Full of double negatives. Also, answer B doesn't distinguish between linear and angular momentum. But anyway, assuming they mean linear momentum: in principle you can measure the (linear) momentum of any particle exactly as long as it's not spatially confined. As soon as you put a confinement on the position, then the momentum becomes uncertain. There are, for example, a lot of applications that rely on the de Broglie relation between electron momentum and wavelength; e.g., electron diffraction.

If the problem allows for the inclusion of electron angular momentum (it's not clear), then you can definitely measure that precisely, because changes in discrete electron angular momentum states is the basis of many forms of spectroscopy. Although here there's always still a broadening due to lifetime uncertainty (since the electron stays in a state for a finite amount of time, the energy of the state has an uncertainty imposed upon it). This translates into a finite linewidth for a spectroscopic transition regardless of how precise the spectrometer is.

Answer D: this seems to be a restatement of the observer effect - that is, measuring the system disturbs the system in some way, which impacts subsequent measurements. Historically, I believe that the uncertainty principle was based on the idea of the observer effect until it was discovered that the uncertainty was a fundamental principle of quantum nature that is independent of the actual process of measurement. But, I know the two principles are often conflated and many people still regard the observer effect and uncertainty principle as one and the same. Even though they aren't. How rigorous is the question being with regard to what the "uncertainty principle" refers to?
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline AdiDex

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Re: Is it possible to calculate the momentum of an electron precisely?
« Reply #2 on: February 06, 2018, 01:11:50 PM »
Quote
I don't like the way this problem is written. Full of double negatives.

It's their job to confuse the students ;D , it's the one of the toughest national exam for the undergraduate students.Look at their papers
http://univ.tifr.res.in/gs2018/Prev_QP/Prev_QP.htm
If possible,Please give your comment on the level of their questions.You must have seen many question papers in your career. I am curious about the nature and level of the foreign universities. Is it similar? or Even harder ?


in principle you can measure the (linear) momentum of any particle exactly as long as it's not spatially confined. As soon as you put a confinement on the position.
Q.1 What is meaning of "spatially confined"?

Q.2 How is this possible to measure an electron's momentum without confining it? You have to confine it in the instrument. I can understand, If the size of the apparatus is big enough we assume that it's infinitely bigger as compared to the size for which quantum effects are predominant. But still it's an assumption, in reality, it has some finite confinement.

Historically, I believe that the uncertainty principle was based on the idea of the observer effect until it was discovered that the uncertainty was a fundamental principle of quantum nature that is independent of the actual process of measurement.

Q.3 Can you please shed some light on the historical aspect of observer effect? And how people used to think that it's a consequence of Heisenberg's uncertainty principle.

As per my understanding, It's just an independent quantum phenomenon due to which you can get any value for the measurement from its eigenvalues randomly. It's totally random on an individual level, one of the best hypothesis to explain this is many worlds interpretation.

How can somebody get confused between Heisenberg's uncertainty principle and observer's effect?
« Last Edit: February 06, 2018, 01:22:03 PM by AdiDex »

Offline Corribus

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Re: Is it possible to calculate the momentum of an electron precisely?
« Reply #3 on: February 07, 2018, 09:38:27 AM »
Honestly your first question is a hard question to answer. I haven’t been in academia in a long time for one thing, and for another “difficulty level” of an exam is something that’s hard to objectively quantify.

Q.1 What is meaning of "spatially confined"?

A.1. Meaning that there is a potential boundary that restricts where a particle may be physically located in space. A free particle can be anywhere without restriction as long as it has sufficient potential energy. A classic example of a confined particle is a particle-in-a-box. In this case the particle is defined as being confined to a box because the potential energy at the box walls is infinite. Therefore, it is impossible for the particle to be located beyond the box walls. It is this confinement that results in quantization of many observables, including the particle’s energy.  Essentially, confinement is the basis of much of quantum mechanics. Note that the confinement to a potential boundary is not absolute in most real systems due to the phenomenon of tunneling. This is because the potential boundary isn’t infinite.

Q.2 How is this possible to measure an electron's momentum without confining it? You have to confine it in the instrument. I can understand, If the size of the apparatus is big enough we assume that it's infinitely bigger as compared to the size for which quantum effects are predominant. But still it's an assumption, in reality, it has some finite confinement.

A2. I guess if you want to be pedantic, every particle must be ultimately be confined if we accept that the universe is not infinite in size and the particle must be in the universe. We are all in a box!

I mean, let’s consider a (hypothetical) free particle, which is unbound and can be anywhere (for the sake of argument, let’s assume the particle can truly be anywhere, universe boundaries be damned). The Heisenberg principle does still hold.

[tex] \Delta x \Delta p \geq \frac {\hbar}{2}[/tex]

Since ΔX for an unbound (not confined) particle approaches infinity, then ΔP may approach zero and still satisfy the uncertainty principle. You can be annoying and say, well, the particle HAS to be in the universe, therefore it is confined, therefore the uncertainty in momentum can’t, strictly speaking, ever equal zero. I’ll grant you that, but what’s the practical point at which the uncertainty is basically zero?  If you are feeling industrious, you can start with a particle-in-a-box model and look at how the variances of position and momentum behave as the box length is increased to infinity. If you are REALLY industrious you can put in actual box lengths and make a guess as what the practical limit is for when a particle essentially becomes "unbound" and the uncertainty in momentum and energy level differences become practically meaningless. I would guess, for an electron, it is somewhere above 1 nm but far below 1 micron. This is, probably not coincidentally, where the "nano" properties of small particles begin to become significant. E.g., quantum dots become "quantumish" at around 8-9 nm or so. This is when electron confinement starts to manifest itself.

Q.3 Can you please shed some light on the historical aspect of observer effect? And how people used to think that it's a consequence of Heisenberg's uncertainty principle.

A. 3. You can find some answers to this by sleuthing around the internet. I guess the confusion lies in the fact that both give rise to a measurement uncertainty at quantum space scales, but whereas one is a practical effect of the act of measurement, the other is a fundamental limitation of measurement independent of the actual act of measuring. Apparently, Heisenberg himself got it wrong and mistook the one for the other, because some of his original thought experiments referred to the observer effect and not the principle that would eventually bear his name. Other common quantum mechanical thought experiments, such as Schrodinger’s Cat, allude to uncertainty in measurement at quantum scales, but refer to observer effects and not the Uncertainty Principle. I imagine this is at the root of a lot of confusion, particularly those who have a laymen’s familiarity with the bizarreness of the quantum mechanics through popular literature and whatnot but haven’t taken any actual rigorous coursework. It wouldn’t be a stretch to think that test writers can make errors along these lines. I guess that’s the challenge of writing tests on subjects that still to this day aren’t fully understood. Didn’t Richard Feynman say something to the effect that anyone who says they understand quantum mechanics doesn’t understand quantum mechanics?

Maybe you find these links interesting:

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

http://factmyth.com/factoids/observing-a-phenomenon-affects-its-outcome/
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline AdiDex

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Re: Is it possible to calculate the momentum of an electron precisely?
« Reply #4 on: February 07, 2018, 01:00:20 PM »
Thank you for your explanations :) .
Quote
I haven’t been in academia in a long time
Just out of curiosity what do you do nowadays?

Quote
Didn’t Richard Feynman say something to the effect that anyone who says they understand quantum mechanics doesn’t understand quantum mechanics?
I remember one of his lectures while addressing normal crowd where he said, "Well, graduate students also don't understand this topic! Why don't they understand it? Because their teacher also don't understand it ;D "

Offline Enthalpy

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Re: Is it possible to calculate the momentum of an electron precisely?
« Reply #5 on: February 08, 2018, 01:30:34 PM »
What is getting increasingly unclear to me: how a measurement should be different from any interaction, and why the outcome should be unique.

I mean, if some experiment tells some day that human observers have magic powers, I'll accept it. To my knowledge, it has not been observed, despite experiments were made. The apparatus just makes some interaction and displays a result.

And then, the present trend is not to say "the interaction has collapsed the wave function" or "has let the particle choose its state", but rather to say "all outcomes coexist after the interaction, as a consequence of the multiple possibilities before the interaction". Even "wave function collapse" looks out-fashioned; never mind, it was an ad hoc hypothesis anyway, not deduced from the rest of QM.

So when the formulation of the detection of a photon at one single place was, with the collapse hypothesis:
"The absorption forced the photon to be at one place, and then it can't be elsewhere"
without collapse it could look like:
"In the situations where [or in the universes where] we observe the photon at one place, we don't observe it elsewhere" but all the outcomes still coexist.
And then, the measurement affects the observed particle only by the physical means used. If all observations coexist, and all states of the observed particle, they can still interfere in the future. In some situations where the many states are perturbed little enough to keep the phase coherence, a future interference can make stable hence observable fringes.

Once again, the double slit experiment leads to incomplete conclusions. If we think instead of interferences of atoms:

An atom (rather a jet of atoms) can absorb a first photon in a first interference location, it flies further to a second location where it can absorb a second photon coherent with the first photon. We observe interference fringes because in the time between both possible absorptions, the atom is on both states, having interacted with a photon and not having interacted.

Or did I get something wrongly? To my knowledge (or ignorance), these interferences can only result from single atoms, since the jet isn't coherent enough to make a stable inference between several atoms.

And then, since the state of the atom between both places is a superposition, so is the state of the first photon: both absorbed and not, simultaneously.

I see no excellent reason to treat other observations=interactions differently from this absorption of a photon by an atom in the atom interference. That is, the destruction of a photon does not collapse the future possibilities of the story. It happens and not, here and elsewhere.

One way I could be horribly wrong is that the light used for atom interferences has many coherent photons, and maybe that the same photon can irradiate the atom at both locations.

Anyway, where should the inventory of possibilities shrink, collapse, call it as you want? At a measurement, and then what does a measurement do differently from other interactions that don't collapse the wave function? Or is it only our conscience that is unable to grasp the superposition of states it lives in?

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