Understanding the Measurement and Interaction of Electrons: A Beginner's Guide

[UncertaintyAjay]

And on many many websites I've red that because the wavelength of macroscopic objects is small that they are almost exactly where we see them.

Link please.

The hyperphysics page does not say what I have quoted you as saying. It says that you cannot see evidence of their wave nature. That is different to saying that "because the wavelength of macroscopic objects is small that they are almost exactly where we see them."

Also the deBroglie equation is an inverse relationship. Momentum is inversely proportional to wavelength. SO if your momentum is large wavelength is smaller not bigger.

I think you should forget about that stuff about wavelength determining the position of a particle. Look up the uncertainty principle on hyper physics. That might help you out.

 

[durant35]

http://pigeonsnest.co.uk/stuff/macroscopic-quantum-phenomena.html

The conventional wisdom has it that quantum effects, such as wave/particle duality, are only noticeable when dealing with atomic-sized or smaller objects. To express it in simple terms, all particles have an associated wavelength, which roughly equates to the distance around the notional position of the object where its quantum behaviour may be observed. The wavelength gets shorter as either the mass or the energy - which are basically the same thing expressed in different ways - gets larger. For a not particularly energetic electron, the wavelengths are of atomic size, which is how we get stable atomic structures and all the wonderful phenomena of chemical bonding, and tunnel diodes and LEDs and other cool dang. For a more massive particle like a proton, its wavelength is of nuclear size, and we get all the nuclear phenomena which are a bit like chemical ones only smaller and more energetic.

 

[UncertaintyAjay]

which roughly equates to the distance around the notional position of the object where its quantum behaviour may be observed

^This,
is not the same as this:

I have imagined wavelength as a boundary which contains all the possible positions of the object.

 

[durant35]

What's the difference?

 

[UncertaintyAjay]

The article does not say wavelength contains all the possible positions of an object. That's the difference. ( Sorry if I'm sounding a bit brusque. Not intentional.)

 

[durant35]

It equates to the distance around the object, so it implies a boundary where we can observe the quantum behavior. And that quantum behavior implies a range of positions where an object can be found, depending on what we measure.

 

[UncertaintyAjay]

And that quantum behavior implies a range of positions where an object can be found, depending on what we measure.

How? What's your logic?
This is moving into realms that I know not of. All i know is what I said about the uncertainty principle.

 

[durant35]

I am clearly referring to macroscopic objects and the 'emergence' of the classical world from the underlying quantum world, it is stated that macro objects have very little wave nature and that's why they almost have a fixed position with little uncertainty. Some even go as far as to say that uncertainty principle doesn't matter for macro objects.

I hope a mentor will see this and analyze it so that we have a clarified picture about the localization of macro objects.

 

[bhobba]

I hope a mentor will see this and analyze it so that we have a clarified picture about the localization of macro objects.

You are over complicating it. A better view is they are wave packets:
https://en.wikipedia.org/wiki/Wave_packet

Interaction with the environment prevents it from spreading.

QM is silent on what that wave is, it tells tells us the position of the object if you were to measure it. The thing is, for macroscopic objects, the width of the packet is way below what we can measure so for all intents an purposes is actually at that location.

Thanks
Bill

 

[durant35]

You are over complicating it. A better view is they are wave packets:
https://en.wikipedia.org/wiki/Wave_packet

Interaction with the environment prevents it from spreading.

QM is silent on what that wave is, it tells tells us the position of the object if you were to measure it. The thing is, for macroscopic objects, the width of the packet is way below what we can measure so for all intents an purposes is actually at that location.

Thanks
Bill

Does that mean that the uncertainty in momentum is high because even thought we know that the mass is great the velocity is hard to measure because of the motion of all the individual components of the macro object.

 

[bhobba]

Does that mean that the uncertainty in momentum is high because even thought we know that the mass is great the velocity is hard to measure because of the motion of all the individual components of the macro object.

For wave packets the uncertainty in both momentum and position are about the same. For macro objects both are way below our ability to detect.

Added Later:
As correctly pointed out below they are not necessarily the same - merely below our ability to detect.

Thanks
Bill

 

[vanhees71]

That thing about a small wavelength meaning that a particle is localised is wrong. Hence the confusion.

I cannot stress this enough- the uncertainty principle does not talk about the actual values of momentum and position but the error in your measurement of them.

That's also not entirely correct, and this statement lead to a lot of confusion for myself when I learned quantum theory. The reason is that quantum theory doesn't tell too much about measurements. Classical theoretical physics doesn't tell much about measurements either although, of course, the entire edifice of physics rests on the possibility to quantitatively measure observables on objects which measurement procedures are the true definitions of these quantities.

Nevertheless, what QT describes is how to describe the properties ("states") of objects (some very careful people say in the context of QT only probabilistic properties of objects, i.e., they describe only ensembles of objects). Thus the uncertainty relation says that for any possible state of a particle the standard deviations of the components of the position vector and that of the components of momentum obeys the Heisenberg uncertainty relation, ##\Delta x_j \Delta p_k \geq \hbar/2##. This tells you that as more precise the position of a particle is determined the less precise the momentum of this particle is determined.

You can always measure position or momentum with any precision you like, but still repeating these measurements very often on an ensemble of particles always prepared in the same state, won't give you smaller fluctuations than the uncertainty relation allows. Note that you always measure either a position-vector component or a momentum-vector component on each particle, but measuring these quantities on equally prepared particles, you can estimate the standard deviations of both observables, and they will always obey the Heisenberg uncertainty relation.

The other question about the socalled measurement disturbance is much more complicated, and it depends very much on the precise definition of measurement procedures whether some accuracy-disturbance relation exists and which precise form it takes. There was a lot of debate about this in the community. If needed, I can search for some papers about the subject.

 

[vanhees71]

For wave packets the uncertainty in both momentum and position are about the same. For macro objects both are way below our ability to detect.

Thanks
Bill

This I don't understand. You cannot compare a momentum with a position uncertainty. So it doesn't make sense to state they are the same. You can easily construct wave packets with any given position or momentum uncertainty. A nice example for a QM 1 exercise, which can be exactly solved analytically, including the full time evolution are Gaussian wave packets for the free particle (exhausting the uncertainty relation, i.e., making ##\Delta x \Delta p=\hbar/2##) or the harmonic oscillator (which are certain unitary transformations of its ground state, called coherent states). It's very illuminating to solve these initial-value problems of the Schrödinger equation!

 

[bhobba]

This I don't understand.

No wonder you don't understand - what I said was wrong. However for macro objects both uncertainties are way below our ability to detect.

Thanks
Bill

 

[drvrm]

For descriptions in a micro world i.e. electron one has to do the basic quantum mechanics in the same manner as we do describe the classical world of particles in classical or Newtonian mechanics.
For example the rules or norms or behavior of quantum particle will be different- we denote a classical one with a point in 3 dim. space and study the time development of its position using equations of motion(Newtonian framework) but in QM the electron can be described by a wave and its position can be determined by method of "measurement" in new new mechanics.
one can represent a particle by a wave function say psi which can be function of its position /momentum or any physical attribute of its state- the wave function ideally spans the whole space but practically has its modulus squared representing the position probability of finding the particle.
one should look up the discussions in an intr. book on new mechanics and proceed step by step say quantum mechanics by powell and craseman or feynmanns lectures on physics ( QM-vol iii) which is available online-one can discuss an area of physics when you traverse the concepts rather than jumping to conclusions.

 

[durant35]

No wonder you don't understand - what I said was wrong. However for macro objects both uncertainties are way below our ability to detect.

Thanks
Bill

So for macroscopic objects the uncertainites in both position and momentum are very small and that's why the classical world 'emerges' from the underlying quantum microscopic world? So the width of the packet basically represents where can we find the macroscopic object?

How isolated do macroscopic objects need to be to exhibit quantum behavior so that their locations spread?

 

[bhobba]

So for macroscopic objects the uncertainites in both position and momentum are very small and that's why the classical world 'emerges' from the underlying quantum microscopic world? So the width of the packet basically represents where can we find the macroscopic object?

Basically

How isolated do macroscopic objects need to be to exhibit quantum behavior so that their locations spread?

Very eg they need to be nearly at absolute zero and even then its difficult.

Thanks
Bill

 

[durant35]

Basically
Very eg they need to be nearly at absolute zero and even then its difficult.

Thanks
Bill

Okay, thanks Bill. Just an off-question, do molecules in everyday interacting objects also have a small width of the wave packet so that they are quite well localized

 

[bhobba]

Okay, thanks Bill. Just an off-question, do molecules in everyday interacting objects also have a small width of the wave packet so that they are quite well localized

Sure. But for exactly what's going on you need to chat to a solid state physicist - which I am not - but some that post here are.

Thanks
Bill

 

[vanhees71]

So for macroscopic objects the uncertainites in both position and momentum are very small and that's why the classical world 'emerges' from the underlying quantum microscopic world? So the width of the packet basically represents where can we find the macroscopic object?

How isolated do macroscopic objects need to be to exhibit quantum behavior so that their locations spread?

If you say, something is "small" you've to say, compared to what. The uncertainties of position and momentum (or the position in phase space), which obey the Heisenberg uncertainty relation ##\Delta x \Delta p_x \geq \hbar/2##, are usually very small compared to the necessary resolution of the phase-space position on a macroscopic scale. This means that very many different quantum states cannot be distinguished on a macroscopic scale. Also usually it is hard to isolate a macroscopic system sufficiently from the environment, so that you have always a mixture of many quantum states due to this perturbance of the system by interactions with the environment, which leads to decoherence and thus classical behavior.

On the other hand there are astonishing examples for the quantum behavior of macroscopic objects. E.g.,

http://physicsworld.com/cws/article/news/2011/dec/02/diamonds-entangled-at-room-temperature

 

[durant35]

If you say, something is "small" you've to say, compared to what. The uncertainties of position and momentum (or the position in phase space), which obey the Heisenberg uncertainty relation ##\Delta x \Delta p_x \geq \hbar/2##, are usually very small compared to the necessary resolution of the phase-space position on a macroscopic scale. This means that very many different quantum states cannot be distinguished on a macroscopic scale. Also usually it is hard to isolate a macroscopic system sufficiently from the environment, so that you have always a mixture of many quantum states due to this perturbance of the system by interactions with the environment, which leads to decoherence and thus classical behavior.

On the other hand there are astonishing examples for the quantum behavior of macroscopic objects. E.g.,

http://physicsworld.com/cws/article/news/2011/dec/02/diamonds-entangled-at-room-temperature

Believe it or not, I saw this article while googleing some stuff. How is this possible, I mean how didn't the decoherence kick in?

 

[vanhees71]

That's a good question. I'm not 100% sure, but the reason must be that the energy gap between the used phonon mode to the next excited state is very large and thus that even at room temperature the probability for transitions is very low. Perhaps you find the detailed answer in the Science article:

https://people.phys.ethz.ch/~reimk/Media/Science-2011-Lee-1253-6.pdf

 

[durant35]

That's a good question. I'm not 100% sure, but the reason must be that the energy gap between the used phonon mode to the next excited state is very large and thus that even at room temperature the probability for transitions is very low. Perhaps you find the detailed answer in the Science article:

https://people.phys.ethz.ch/~reimk/Media/Science-2011-Lee-1253-6.pdf

I'm sorry, but I don't understand it. I know only a little bit about entaglement and as far as I know it occurs when the wavefunctions overlap so that the system acts as one. How could the wavefunctions overlap spatially at that kind of temperature, what did actually happen? It is mentioned that each of the diamons was simultaneously in the state of 'vibrating and non vibrating' which is even more confusing, could you please clarify it a bit?

 

[vanhees71]

Ok, I'll try. I only have to find the time to read the paper! Please be patient with me ;-)).

 

[durant35]

Ok, I'll try. I only have to find the time to read the paper! Please be patient with me ;-)).

No problem, thank you.
:)

 

[durant35]

Do superpositions of macroscopic objects occur naturally in the world or do the experimenters have to induce them like with the diamonds in question?

 

[bhobba]

Do superpositions of macroscopic objects occur naturally in the world or do the experimenters have to induce them like with the diamonds in question?

The technical meaning of superposition is rather mathematical and makes questions like the above not well posed so can't be answered.

A better query would be do quantum effects occur in the everyday world around us. Yes - but they usually are not obvious. For example transistors work because of things called holes which are in fact quasi particles and depend entirely on QM. There are others like the strange behaviour of liquid helium. How common is this strange stuff - maybe more common than we generally think - but its seems the exception rather than the rule - mostly classical physics is good enough

That said even the simple phenomena of light traveling though glass is rather complex and dependant on advanced QM. Do a post about it and and someone into solid state physics may explain what's really going on - its quite interesting.

Thanks
Bill

 

[durant35]

Thanks Bill. What I meant was quantum effects on the macro scale like the 'vibrating-non vibrating' state from the diamonds experiment. I know that many things like transistors work on the miscroscopic qm background. But can a state like the one mentioned occur without experiments?

 

[bhobba]

But can a state like the one mentioned occur without experiments?

To the best of my knowledge - no.

Thanks
Bill

 

[durant35]

To the best of my knowledge - no.

Thanks
Bill

Ok. Do you know how did they manage to achieve the entaglement in conditions that aren't cold and isolated?

 

[zonde]

That's a good question. I'm not 100% sure, but the reason must be that the energy gap between the used phonon mode to the next excited state is very large and thus that even at room temperature the probability for transitions is very low. Perhaps you find the detailed answer in the Science article:

https://people.phys.ethz.ch/~reimk/Media/Science-2011-Lee-1253-6.pdf

It seems that phonons used in the experiment are of much higher frequency than that of thermal vibrations. And these phonons are relatively stable because of specific structure of crystal ("bulk vibration consisting of two counter-oscillating sublattices within the diamond structure.")

On the first glance the experiment seems to show rather bizarre effect - phonon shared by two distant diamonds. But on the second glance it seems that experiment is consistent with explanation that there are two polarization entangled phonons in crystals.

 

[bhobba]

Ok. Do you know how did they manage to achieve the entaglement in conditions that aren't cold and isolated?

No.

Thanks
Bill

 

[durant35]

It seems that phonons used in the experiment are of much higher frequency than that of thermal vibrations. And these phonons are relatively stable because of specific structure of crystal ("bulk vibration consisting of two counter-oscillating sublattices within the diamond structure.")

On the first glance the experiment seems to show rather bizarre effect - phonon shared by two distant diamonds. But on the second glance it seems that experiment is consistent with explanation that there are two polarization entangled phonons in crystals.

But how did they manage to cross paths of the two diamonds, I've red that electron paths had something to do about it.

 

[zonde]

But how did they manage to cross paths of the two diamonds, I've red that electron paths had something to do about it.

They didn't of course.
First they shine a pump laser pulse on both diamonds. A photon splits in phonon and redder photon than the rest of pump photons. They collect these redder photons from both crystals and analyze them together using two polarization beam splitters and half wave plate. After short time they shine second pulse on diamonds and phonon combines with one pump photon and creates bluer photon. These are collected and analyzed together in separate channel using two PBSes and wave plates.

 

[durant35]

But is that a vibration of a diamond per se or a state where the phonon is spread as a wave in both diamonds?