Can someone explain HUP with some mathematical rigor?

[SeventhSigma]

I was watching one of Susskind's videos and I had some questions:

Briefly summarizing from the video:

E = hf (Planck's constant times frequency)
E = cp (speed of light times momentum)
or p = E/c = hf/c

And we know that if we have wavelength lambda, then if t = 1/f and vel = dist/time, then c = lambda*f

So f = c/lambda, or p = E/c = hf/c = h/lambda (de Broglie's equation -- the smaller the wavelength, the larger the momentum)

So say we take a photograph of something -- if we want it to not be fuzzy, we need lambda < deltaX if you want an image with precision deltaX.

If we want to measure something with sufficiently high precision, we need to use a short-wavelength, high-momentum entity. So we shoot a high-momentum photon to get a decent position, but then we're going to knock our object away at a random direction of uncertain magnitude. Immediately after we measure the position, the momentum has been changed, so there is a tradeoff between position and momentum.


My questions:

1. Why does lambda need to be smaller than DeltaX? What does this really mean? I understand that, for instance, if I want to make a picture in Photoshop of sufficient resolution, I can't do it if my pixels are too large (better the resolution, the more pixels I need. I can't make a drawing of a face with a 2x2 grid). But how does this analogy play into determining the "resolution" of an object with wavelength?

2. How exactly do we measure position? I know we measure it by firing light at something and then having that something bounce the light back at us so we can interpret it, but why does high momentum mean we "know" more precisely where it bounced back from?

3. Is there no other way to determine an object's position instead of having objects hit off one another?

4. How is HUP, at this rate, not a measurement problem? I always hear how QM has "intrinsic randomness" and how the uncertainties in HUP are also intrinsic and that this isn't a measurement problem in itself, but to me this sure sounds like it -- akin to searching for an object in a dark room (bumping into something and therefore changing its position). How can I make the leap to understanding how HUP is an "intrinsic" problem?

 

[SpectraCat]

Check out: http://en.wikipedia.org/wiki/Uncertainty_principle#Mathematical_derivations

that shows why the HUP is intrinsic to quantum systems, rather than a feature of measurement.

The whole page is actually pretty good for understanding the different forms and interpretations that the HUP can be expressed in.

 

[SeventhSigma]

I must admit, that math didn't make too much sense to me. It implies that position and momentum are "Hermitian operators" -- why?

I still do not understand why we say HUP is intrinsic.

 

[SpectraCat]

The best qualitative explanation of what it means for the HUP to be intrinsic to quantum systems was given to me by ZapperZ. You can search around to get it in his own words, but I think the following is an accurate paraphrasing of his thought experiment:

Suppose you pass a particle through an arbitrarily narrow slit with width dx ... at the moment it passes through the slit, you have made a very precise measurement of it's position, which means that the x-component of momentum of the particle at that moment has a very large uncertainty. Now suppose that you measure the x-component of momentum of the same particle at some later time .. this can be done with arbitrary precision, for example by simply allowing the particle to impact an imaging detector, and measuring the x-displacement from the position of the center of the slit. So you can *measure* both the position and the momentum of a particle with arbitrary precision ... the HUP places no restrictions on this.

Where the HUP comes in is if you try to repeat the experiment .. you will almost certainly find a very different value for the x-component of the momentum for the next particle that passes through the slit. Furthermore, if you repeat the experiment many times, you will build up an image that represents the momentum distribution that is predicted by the HUP.

So, you can say that, once the the uncertainty of the position has been restricted to a very small value, the distributions of possible momentum values becomes very broad. This places a restriction on the precision with which you can make predictions about the momentum measurements *before* they are made .. you can only say that they will fall somewhere within the width of the momentum distribution determined by the HUP. However, each individual momentum measurement can be made to arbitrary precision.

 

[SpectraCat]

I must admit, that math didn't make too much sense to me. It implies that position and momentum are "Hermitian operators" -- why?

I still do not understand why we say HUP is intrinsic.

Hermetian operators are those which have real eigenvalues (this may not be a helpful distinction). Since when we make observations of physical observables, we always measure real numbers (as opposed to complex numbers), there is a postulate of quantum mechanics that says all physical observables correspond to Hermetian operators.

 

[Matterwave]

I've always like the simple analogy of a wave on a string. If you think of a particle as a wave-packet, then the displacement of the particle is simply given by where the wave packet "is", and the momentum of the particle is given by the wavelength of the wave packet. If we have a wave-packet which was very localized (small uncertainty in x), then it's very hard to measure the wavelength (momentum) of such a wave-packet (the wave-packet must necessarily contain many different waves of different wavelengths). If, on the other hand, you had a very nice plain wave wave-packet over many many oscillations, then you can get a very good measure of the wavelength (momentum), but it's now hard to tell where the wave packet "is" as it is spread out over space. There's an intrinsic trade off because the narrower you make your wave packet, the more frequencies you need to build it.

 

[SeventhSigma]

Okay, so we say that momentum and position are real numbers -- that's all fine and well and intuitive. But why do we have to say that one necessarily implies less precision in the other no matter what?

Matterwave: I'm not understanding the analogy -- what do you mean by "wave-packet" specifically, and why does small uncertainty of x mean you need many different waves of different wavelengths?

 

[jtbell]

We assume that momentum p corresponds (via de Broglie) to wavelength

[tex]\lambda = \frac{h}{p}[/tex]

and wavenumber

[tex]k = \frac{2\pi}{\lambda} = \frac{2\pi p}{h} = \frac{p}{\hbar}[/tex]

Further, we assume that a superposition of waves with different wavelengths (momenta) represents a particle that can have any of the momenta, with a probability determined by the square of the amplitude of each wave in the superposition.

Now add together several waves with momenta that span a range whose standard deviation is [itex]\Delta p[/itex]. The superposition forms a "packet" whose spatial distribution has a "width" (standard deviation) of [itex]\Delta x[/itex].

The product [itex]\Delta x \Delta p[/itex] is constant for a given shape of packet. The constant is different for different shapes. See this post for an example, with an attached picture:

https://www.physicsforums.com/showthread.php?p=2972106#post2972106

From Fourier analysis it is possible to prove that the smallest possible value for the constant is [itex]\hbar / 2[/itex] and that it occurs for a certain shape packet called "Gaussian."

 

[SeventhSigma]

What is the actual formula for determining probability? I mean we could have infinitely many waves making up a superposition, no?

EDIT: Reading the rest, one moment

 

[SeventhSigma]

Okay, so given that we have wavelength corresponding to momentum by definition, we can take a bunch of waves with different momenta/wavelengths and add them together to get wave packet superpositions.

So when y2 has longer wavelengths/wavenumbers than y1, we see that y2 consequently has less variance than y1 when we're talking about the outskirts of the wave packet.

Alright I think I am starting to better understand -- but I am still not quite getting it. I can imagine that if we were to add in even more longer-wavelength waves to the packet, we'd eliminate the variance. What would that resulting graph start to look like? A vertical line indicating a precise x value?

And so if we instead use a single cosine wave to represent a packet, we'd know its momentum perfectly (just h/lambda of the cosine wave) but it would have no localization (it'd span negative to positive infinity across x) and therefore we'd have no idea where to determine its position?

Am I understanding this correctly?

 

[jtbell]

What is the actual formula for determining probability? I mean we could have infinitely many waves making up a superposition, no?

For a discrete set of waves (a finite number, or a "countably infinite" number of them), the probability of the momentum corresponding to one wave is the "modulus squared" of the amplitude [itex]P = |a|^2 = a^*a[/itex]. In general, the amplitude (coefficient) can be complex. For a continuous set of waves ("uncountably infinite" number corresponding to a continuous range of momenta) you interpret this as a probability distribution.

So when y2 has longer wavelengths/wavenumbers than y1, we see that y2 consequently has less variance than y1 when we're talking about the outskirts of the wave packet.

Careful! what counts is not the magnitude of the wavenumbers (k's) themselves, but rather the "width" of their range. In my example, y1 is the sum of waves with k = 9, 9.2, 9.4, 9.6, 9.8, 10.0, 10.2, 10.4, 10.6, 10.8 and 11.0. y2 is the sum of waves with k = 8. 8.4, 8.8, 9.2, 9.6, 10.0, 10.4, 10.8, 11.2, 11.6 and 12.0. The average k is the same (10.0) for both of them, but the range of k's is twice as large for y2 (+/- 2.0) as for y1 (+/- 1.0).

Physically, these would represent two particles with the same expectation value for momentum p, but with uncertainty [itex]\Delta p[/itex] twice as large for y2 as for y1.

I can imagine that if we were to add in even more longer-wavelength waves to the packet, we'd eliminate the variance.

As you add more waves with larger and smaller values of k, keeping the average the same, the packet gets narrower in terms of [itex]\Delta x[/itex]. In order to make the packet "infinitesimally narrow" you'd have to add waves spanning the infinite range of k values: -infinity to +infinity.

And so if we instead use a single cosine wave to represent a packet, we'd know its momentum perfectly (just h/lambda of the cosine wave) but it would have no localization (it'd span negative to positive infinity across x) and therefore we'd have no idea where to determine its position?

Exactly.

 

[SeventhSigma]

Yes, sorry, that's what I meant by variance (and thusly standard deviation). y1's k values and y2's k values have the same EV but the y1's set has less variance (smaller SD). Consequently, we see fewer "large wobbles/fluctuations" in the y2 graph as we move out from the center (it starts to peter off into lower-probability realms).

I guess my next question is this: So I understand how position and momentum would have tradeoffs if we describe them via a wavefunction. Do we absolutely have to use wavefunctions to describe these attributes/operators? For instance, I don't need to use a wavefunction to describe macroscopic objects. Is there something fundamental about the quantum world that demands we use wavefunctions? Is there an obstacle that prevents us from leveraging another method?

 

[f95toli]

For instance, I don't need to use a wavefunction to describe macroscopic objects.

That depends on how accurate your description needs to be. It is true that we can usually (but not always) get away with neglecting the UP (and other quantum effects) when dealing with macroscopic objects, but that is (mainly*) because the effects are negligible. There is no set scale at which quantum mechanics stops being valid; what we consider to be "macroscopic" laws of nature (Newton's laws etc) are just good approximations.
Hence, the HUP IS valid for macroscopic objects (and in some cases we can even observe the effects).

 

[SeventhSigma]

Sure, I understand that the effects are negligible for macroscopic events (and therefore we could also argue that strange phenomena like suddenly materializing in a different position is possible but extraordinarily unlikely, as you'd need every single particle of your body to take on an extremely improbable value of its distribution function at the same time). That's what I meant by "don't need" -- we don't need it for practical purposes.

But my question is why we need wavefunctions to describe things on a quantum level to begin with to achieve that type of precision. This, to me, sounds like a measurement problem of sorts.

 

[mquirce]

For energy, the uncetainty is expresed by dE = h /2pi *t. i think that dE is smoll or large,it is eneergy which formula is E = h * f where f is frequence that has ollways an integer number, hence a precise one. Isn't this a dilema about what we need to throw out of both?
Something new in this post is the concept of paket of waves. Does this give a paket frequency les than one ? That is f1 +f2 +...finf.= F < 1 ?
If my post you think is senseless do not respond.

 

[f95toli]

But my question is why we need wavefunctions to describe things on a quantum level to begin with to achieve that type of precision. This, to me, sounds like a measurement problem of sorts.

Because it turns out that wavefunctions describe reality; meaning we don't have a choice.

 

[SeventhSigma]

But my question is how wavefunctions are the fundamental entities for describing reality. I understand that we can model reality by leveraging wavefunctions, but why do we have no choice? I always hear about how the wave/particle duality doesn't really exist. Is light a particle? "Well, not exactly." Is it a wave? "Well, not exactly." And so we describe phenomena by simply measuring and predicting certain values/attributes that we DO have access to instead of trying to turn everything into a picture.

But at the end of all this, it still feels like a measurement problem.

 

[Matterwave]

Sure, I understand that the effects are negligible for macroscopic events (and therefore we could also argue that strange phenomena like suddenly materializing in a different position is possible but extraordinarily unlikely, as you'd need every single particle of your body to take on an extremely improbable value of its distribution function at the same time). That's what I meant by "don't need" -- we don't need it for practical purposes.

But my question is why we need wavefunctions to describe things on a quantum level to begin with to achieve that type of precision. This, to me, sounds like a measurement problem of sorts.

The use of wave-functions may not be the ONLY way to describe quantum phenomena, but there have been proofs that say that this probabilistic nature of the Universe is very fundamental. In order to get away from this probabilistic description, one would have to at the very least violate locality (that nothing travels faster than the speed of light, as postulated by SR and confirmed by numerous experiments).

 

[SeventhSigma]

Why would we violate locality without probability? Is there a basic example?

 

[Matterwave]

The problem has to deal with Entanglement. If you do a search for "Bell test" or "EPR paradox" you should be able to get some background on this. There are numerous posts, even on this forum itself, discussing this.

The basic idea is, if you have 2 entangled particles, then the quantum description of things leads to one correlation probability between the two particles, while a local realistic (i.e. local+fundamentally non-probabilistic) leads to another correlation between the two particles. Tests have been done, and the statistics show that the quantum description gives the correct correlation. Other descriptions which can give the correct correlation are either probabilistic or non-local. No local realistic theory can produce the correct correlation, this has been proved.

 

[SeventhSigma]

jtbell: In your example of cosine waves, wouldn't I technically get many possible position values? Even though the local "range" between -3 and 3 shows decreasing uncertainty as a result of the amplitudes fading out from the edges, wouldn't this exact same "bump" happen elsewhere along the x-axis and infinitely so?

For instance, for y2, replacing x = 0 there gives you a local amplitude of 11 (and then it starts to peter off as you go towards x=3 or x=-3), but then you get another "burst" where the amplitude approaches 11 again at x=-15.7 or so.

In other words, how does adding waves together tell us where the particle is when waves have no defined range standalone? And when we add such waves together, we can have multiple high-probability clusters that could designate position?

 

[BruceW]

but then you get another "burst" where the amplitude approaches 11 again at x=-15.7 or so.

This is true when you only use a finite number of waves, but when we are using all possible wavenumbers, the wavefunction can become localised.

 

[jtbell]

jtbell: In your example of cosine waves, wouldn't I technically get many possible position values? Even though the local "range" between -3 and 3 shows decreasing uncertainty as a result of the amplitudes fading out from the edges, wouldn't this exact same "bump" happen elsewhere along the x-axis and infinitely so?

Yes, because in this example we're adding a finite number of waves. Try this exercise if you have a spreadsheet program handy, or write a little computer program if you know a programming language: Repeat the sum for y1, but "interpolate" more waves in between, so that y1 uses k-values of 9.0, 9.1, 9.2, 9.3, ..., 10.8, 10.9, 11.0. This sum contains about twice as many waves, but in the same range [itex]\Delta k[/itex]. The "bumps" that you describe above will be twice as far apart as before, I think.

Adding more and more waves in the same range [itex]\Delta k[/itex] makes the "bumps" get further and further apart. Adding an infinite number of waves, spaced infinitesimally close together in k, gives only a single "bump" at the origin. To do this you have to use an integral, not a summation.

Added: in this case the integral is pretty easy to do if you've had basic calculus:

[tex]\psi(x) = \int^{11}_9 {\cos(kx)dk}[/tex]