How do physicists measure electron position if the wavefunction is everywhere?

[mfb]

I was wondering can an random atomic decay in normal circumstances cause a superposition on a macro level.

No. Not even in not normal circumstances.

 

[bhobba]

I didn't mean to get off topic, I was wondering can an random atomic decay in normal circumstances cause a superposition on a macro level.

OK rather than answer the question directly can you define what you mean by superposition?

Thanks
Bill

 

[durant35]

A mixture of states like in the case of Schrödinger's cat. But mfb already answered the question so thanks to both of you sincerely for the patience.

 

[bhobba]

A mixture of states like in the case of Schrödinger's cat. But mfb already answered the question so thanks to both of you sincerely for the patience.

No worries.

Just a technical aside. A superposition is NOT a mixture.

Thanks
Bill

 

[durant35]

No.
The uncertainty principle does not talk about the actual value of position or momentum. It merely says this:
The position and momentum of something cannot be simultaneously measured with arbitrarily high accuracy.
It is a statement about the accuracy with which it is possible to measure momentum and position. If you measure one to a high accuracy, your measurement of the other must necessarily be more inaccurate. The relationship between the error in measurement of momentum ( Δp) and error in position ( Δx) are related to each other by:
ΔpΔx≥h/2π ( equation 1)
where h is Planck's constant ( 6.63 * 10^-34). Let's say your measurement of the electron's position is fairly accurate and the error is tiny. Then the error in position is necessarily larger than h/(2π*Δx). I.e:
Δp≥h/(2π*Δx) ( equation 2).
So if your error in measurement of position is small, you can see from equation 2 that error in measurement of position is large.

So increased momentum does not result in more certainty in position.

It is also important to note that the uncertainty principle does not talk about the accuracy of your apparatus. I could be using the most accurate apparatus ever and this law would still apply. Try as hard as you want. The uncertainty principle is inescapable. The reason for this is because the very act of measurement interferes with the system and changes a state.

For example if you wanted to measure the position of an electron, you could do it by having a light source and observing a flash as the electron goes by. But the interaction between the electron and the photon will result in a change in the momentum of the electron. So the act of measuring changes the system. Only, in macroscopic systems this phenomenon is so small it can easily be neglected.

A quick scroll through the posts and this gave me a bit of a worry and confusion.

If we increase the momentum, the wavelength becomes smaller, right?
And the small wavelength implies better localization and a smaller region where we can find the object.

Does UncertaintyAjay just mean that the value of the momentum doesn't have any effect on the position/momentum standard deviation in the equation?

 

[UncertaintyAjay]

Yes.

 

[durant35]

Yes.

I'm still confusing myself because I can't conceptualize this so let me try to explain you.

I have imagined wavelength as a boundary which contains all the possible positions of the object. Now the problem is that we can't calculate the precise wavelength because of the velocity uncertainty. So in idealized conditions let's take let say 45 for the momentum which gives us a particular wavelength. Due to uncertainty, it can be a bigger and a smaller number than 45 and if it is bigger than the wavelength also varies and becomed bigger so the boundary gets bigger, which would imply that greater uncertainty in velocity gets greater uncertainty in position which makes no sense. Please correct me because this is confusing me.

 

[UncertaintyAjay]

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

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

Not true.

Due to uncertainty, it can be a bigger and a smaller number than 45

I suppose by 'it' you mean momentum?

if it is bigger than the wavelength also varies and becomed bigger

How?

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.

Here is what I am going to do.Say momentum is y units. Say the error in this measurement is Δp. Say you simultaneously measure the position of the particle and get that it is some distance z from your origin. Then your measurement of position will have an error Δx that must be greater than h/(4π*Δp). That is what the uncertainty principle states. If you see, there is absolutely no mention of z ( the actual position) or y ( the actual momentum) in the uncertainty relation( highlighted in bold just above). There are only the errors in your measurement of the two properties. I don't know where you got the wavelength thing from but I don't think it is true. A mentor or someone better versed in QM than me could tell you more about that.

 

[durant35]

Check out the Broglie wavelength for anything on an observable scale. A dust particle or something. And for a baseball...

I inferred it from this. 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. And on some websites I've red that wavelength represents the boundary for a big object. I hope the mentor will clarify this.

 

[durant35]

"If you explore the wavelength values for ordinary macroscopic objects like baseballs, you will find that their DeBroglie wavelengths are ridiculously small. Comparison of the power of ten for the wavelength will show what the wavelengths of ordinary objects are much smaller than a nucleus. The implication is that for ordinary objects, you will never see any evidence of their wave nature, and they can be considered to be particles for all practical purposes."

Quote from the hyperphysics webpage.

 

[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