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What is it about position and momentum that forbids knowing both quantities at once?

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Roo2
#19
Jul26-11, 12:07 AM
P: 41
Quote Quote by CJames View Post
Is there something wrong with the link Fredrik posted earlier?

http://www.kevinaylward.co.uk/qm/bal...ation_1970.pdf

Page 365

I don't mean this to be sarcastic if it comes across that way.
Thanks! I was skimming this thread at work and didn't notice the link. That's exactly what I was looking for. I was actually amused to find this thread; I had a discussion with my room mate a couple weeks ago about HUP. Both of us are college undergrads, and our experience with quantum mechanics is limited to one semester of physical chemistry apiece (at different universities). I mentioned that I had been taught that HUP applied to sets of measurements of p and r and didn't apply to the position and momentum of a single particle. He replied that since the particle is itself a wave, and a wave has no defined absolute position, then HUP applies to a single particle. I don't have the physics background to respond to that, so I let it be. That paper is clearing up the picture for me. Forgive me for asking again for references, but I don't have the physics education to know off the bat... have there been any revisions to the interpretation of the statistical model in regard to a single particle measurement in the 40 years since this paper was published?
fuesiker
#20
Jul28-11, 07:24 AM
P: 49
The most intuitive way, imho, to view the uncertainty principle is to recognize that the wavefunction of some system in momentum space is the Fourier transform of the wavefunction of that same system in position space. You can arrive at this by honoring the fact that the momentum operator is the propagator of translation, and that is how you can relate both wavefunctions to each other to arrive at the equation showing they are related by the Fourier transform.

Now that we are convinced of this, we can look at the properties of the Fourier transform. If a function K is the Fourier transform of a function F, then the sharper F is, the broader is K.

The uncertainty principle has nothing to do with observation. It does not arise due to observation. It is the nature of things that there are non-commuting observables (time and energy, momentum and position) that cannot be determined simultaneously to arbitrary precision. We find this unintuitive because in our universe, Planck's constant is way too small for quantum effects to have become intuitive to our senses, just like special relativisitic effects are also counter-intuitive to our senses since c is so big in our world. Imagine if we lived in a world where c is 30 mph ;)
Dickfore
#21
Jul28-11, 07:48 AM
P: 3,014
@OP:

The canonical commutation relation:
[tex]
[\hat{x}_{i}, \hat{p}_{k}] = i \, \hbar \, \delta_{i k}
[/tex]

EDIT:
How to derive these relations? You need to identify momentum as the generator of translations in space (according to the correspondence principle and the fact that the physical quantity conserved due to homogeneity of space is linear momentum) and it acts on states that are eigenstates of position according to:
[tex]
\hat{T}(\mathbf{a}) \, \left|\mathbf{x}\right\rangle = \left|\mathbf{x} + \mathbf{a}\right\rangle, \; \hat{T}(\mathbf{a}) = \exp{\left(-\frac{i}{\hbar} \, \mathbf{a} \cdot \hat{\mathbf{p}}\right)}
[/tex]
ZealScience
#22
Jul28-11, 08:00 AM
P: 351
I heard that these states of momentum and position are orthogonal in Hilbert space. Probably mathematics thinks that it is intuitive, but not us. Because in experiments that can be seen with naked eyes have no violation to classical mechanics which is described by definite phases...
Dickfore
#23
Jul28-11, 08:05 AM
P: 3,014
[tex]
\begin{array}{l}
\hat{x}_{i} \, \hat{T}(\mathbf{a}) \, \left|\mathbf{x}\right\rangle = (x_{i} + a_{i}) \, \left|\mathbf{x} + \mathbf{a}\right\rangle \\

\hat{T}(\mathbf{a}) \, \hat{x}_{i} \, \left|\mathbf{x}\right\rangle = x_{i} \, \left|\mathbf{x} + \mathbf{a}\right\rangle
\end{array} \Rightarrow \left[\hat{x}_{i}, \hat{T}(\mathbf{a})\right] = a_{i} \, \hat{T}(\mathbf{a})
[/tex]

Expanding to linear power in the translation vector [itex]\mathbf{a}[/itex], we get:
[tex]
-\frac{i}{\hbar} \, a_{k} \, \left[\hat{x}_{i}, \hat{p}_{k}\right] = a_{i} = \delta_{i k} \, a_{k}
[/tex]
Since this has to hold for arbitrary components [itex]a_{k}[/itex], we must have the commutation relations to hold.

After you have a commutation relation between two operators, you can apply the same math that you mentioned in the OP (Cauchy-Schwartz ineqality) to derive an Uncertainty relation.
Dickfore
#24
Jul28-11, 08:08 AM
P: 3,014
Quote Quote by ZealScience View Post
I heard that these states of momentum and position are orthogonal in Hilbert space.
They are not. For example:
[tex]
\langle x | p \rangle = \frac{1}{(2\pi \hbar)^{1/2}} \, \exp{\left(\frac{i}{\hbar} \, p \, x\right)} \neq 0
[/tex]
ZealScience
#25
Jul28-11, 08:12 AM
P: 351
Quote Quote by Dickfore View Post
They are not. For example:
[tex]
\langle x | p \rangle = \frac{1}{(2\pi \hbar)^{1/2}} \, \exp{\left(\frac{i}{\hbar} \, p \, x\right)} \neq 0
[/tex]
Can you specify the eigenvectors of position and momentum? I just started learning it and I am not familiar with it.
Dickfore
#26
Jul28-11, 08:21 AM
P: 3,014
Quote Quote by ZealScience View Post
Can you specify the eigenvectors of position and momentum? I just started learning it and I am not familiar with it.
FFFFFFFFFFFFF

I pressed the page back button on my keyboard and my whole response is gone. I hate this glitch on PF.
ZealScience
#27
Jul28-11, 08:39 AM
P: 351
Quote Quote by Dickfore View Post
FFFFFFFFFFFFF

I pressed the page back button on my keyboard and my whole response is gone. I hate this glitch on PF.
Accidents can happen... it is not deterministic~~~
Fredrik
#28
Jul28-11, 08:46 AM
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Quote Quote by Dickfore View Post
FFFFFFFFFFFFF

I pressed the page back button on my keyboard and my whole response is gone. I hate this glitch on PF.
I have found that you can usually get it back if you press the page forward button and answer yes to the question if you want to send the information again.
Dickfore
#29
Jul28-11, 09:04 AM
P: 3,014
Quote Quote by Fredrik View Post
I have found that you can usually get it back if you press the page forward button and answer yes to the question if you want to send the information again.
For me the text gets erased.
atyy
#30
Jul28-11, 09:19 AM
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Quote Quote by Fredrik View Post
Good post nkadambi, but I must point out an inaccuracy in what you said. I didn't really understand this myself until recently. It is possible to measure position and momentum simultaneously. In fact, we often measure the momentum by measuring the position and interpreting the result as a momentum measurement. (Check out figure 3 in this pdf).
If I remember correctly, after measuring position, the particle collapses to an eigenstate of position, then spreads out due to time evolution. After measuring momentum, it collapses to an eigenstate of momentum, then remains in the same state since it is an eigenstate of the free particle Hamiltonian. What state does it collapse to after position and momentum are simultaneously measured?
Fredrik
#31
Jul28-11, 09:25 AM
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Quote Quote by atyy View Post
If I remember correctly, after measuring position, the particle collapses to an eigenstate of position, then spreads out due to time evolution. After measuring momentum, it collapses to an eigenstate of momentum, then remains in the same state since it is an eigenstate of the free particle Hamiltonian. What state does it collapse to after position and momentum are simultaneously measured?
In this example, the particle is absorbed by the detector, so no new state is prepared. If it had been the kind of detector that lets the particle pass through it, the new state would have been one with a sharply defined position (the wavefunction will be close to zero far from the detector).

Quote Quote by atyy View Post
After measuring momentum, it collapses to an eigenstate of momentum
This is what I was taught as well, but I wonder if it really makes sense. Even if we disregard all the difficulties associated with the fact that an eigenfunction of [itex]-i\,\nabla[/itex] isn't square-integrable, and interpret the statement as "after a momentum measurement, the wavefunction will have a sharply defined momentum", it still looks false to me. I don't know much about how things are measured, but it seems to me that momentum is always measured by measuring the position instead. It's possible that in most cases, the position measurement is inexact enough to give us something like exp(-x2) as the new wavefunction, so that both the wavefunction and its Fourier transform have a single peak, with widths of the same order of magnitude in units such that [itex]\hbar=1[/itex].
atyy
#32
Jul28-11, 10:11 AM
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Quote Quote by Fredrik View Post
This is what I was taught as well, but I wonder if it really makes sense. Even if we disregard all the difficulties associated with the fact that an eigenfunction of [itex]-i\,\nabla[/itex] isn't square-integrable, and interpret the statement as "after a momentum measurement, the wavefunction will have a sharply defined momentum", it still looks false to me. I don't know much about how things are measured, but it seems to me that momentum is always measured by measuring the position instead. It's possible that in most cases, the position measurement is inexact enough to give us something like exp(-x2) as the new wavefunction, so that both the wavefunction and its Fourier transform have a single peak, with widths of the same order of magnitude in units such that [itex]\hbar=1[/itex].
I haven't really thought this through myself, and am just going to ask questions as they come to mind (but OP please stop me if this is hijacking!). How about in the relativistic case, say like Compton scattering? Isn't momentum there measured without measuring position? OTOH, there is no "sensible" position operator in relativistic QFT, so maybe that's different?
Fredrik
#33
Jul28-11, 10:16 AM
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Quote Quote by ZealScience View Post
Can you specify the eigenvectors of position and momentum? I just started learning it and I am not familiar with it.
Neither of those operators have eigenvectors in the semi-inner product space of square-integrable functions from [itex]\mathbb R[/itex] into [itex]\mathbb C[/itex]. If we view this space as a subspace of the vector space of all functions from [itex]\mathbb R[/itex] into [itex]\mathbb C[/itex], then the functions [itex]u_p[/itex] defined by [itex]u_p(x)=e^{ipx}[/itex] for all x are eigenfunctions (with eigenvalue p) of the momentum operator. Note that they are not square-integrable. The position operator doesn't have any eigenfunctions in this space either, but there's a trick. The "functions" [itex]v_x[/itex] such that [itex]v_x(y)=\delta(y-x)[/itex] for all y, where [itex]\delta[/itex] is the Dirac delta, can be thought of as "eigenfunctions" (with "eigenvalue" x) of the position operator, even though they aren't really functions.
Fredrik
#34
Jul28-11, 10:31 AM
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Quote Quote by atyy View Post
How about in the relativistic case, say like Compton scattering? Isn't momentum there measured without measuring position?
I don't think so. The way I see it, a measuring device is just a device that produces a signal (that can be approximately described as classical) that informs us that an interaction has taken place. The location of the device (or the location of the relevant component of it) can always be interpreted as the result of a position measurement. So it's not possible to measure anything without measuring the position of the particle(s) that participated in the interaction that produced the signal.

Hm, I suppose it's possible that those particles aren't the same ones as the one we're trying to measure the momentum of.
Dickfore
#35
Jul28-11, 11:19 AM
P: 3,014
Regarding measurement and falling back into the eigenstate of the observable corresponding to the observed eigenvalue, I have doubts

If you read section 7 in "Quantum Mechanics vol. 3" (the wave function and measurements) by Landau, Lifgarbagez, they discuss in the 12th paragraph the possibility that the wave function in which the electron drops after the measurement is not necessarily an eigenfunction of the observable being measured.
fuesiker
#36
Jul28-11, 01:49 PM
P: 49
In principle, once you measure a quantity on a certain wavefunction, this wavefunction will collapse onto an eigenstate of that observable quantity, say O. Then, it depends on the dynamics of your system. In general, there is time evolution going on propagated by some Hamiltonian H, and then it depends on whether or not H and O commute. If they do, then your wavefunction will remain an eigenstate of O acquiring merely a phase shift. If O and H do not commute, then this wavefunction will no longer remain an eigenstate of O.


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