what is it about position and momentum that forbids knowing both quantities at once?by jeebs Tags: forbids, knowing, momentum, position, quantities 

#19
Jul2611, 12:07 AM

P: 41





#20
Jul2811, 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 noncommuting 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 counterintuitive to our senses since c is so big in our world. Imagine if we lived in a world where c is 30 mph ;) 



#21
Jul2811, 07:48 AM

P: 3,015

@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] 



#22
Jul2811, 08:00 AM

P: 343

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...




#23
Jul2811, 08:05 AM

P: 3,015

[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 (CauchySchwartz ineqality) to derive an Uncertainty relation. 



#24
Jul2811, 08:08 AM

P: 3,015

[tex] \langle x  p \rangle = \frac{1}{(2\pi \hbar)^{1/2}} \, \exp{\left(\frac{i}{\hbar} \, p \, x\right)} \neq 0 [/tex] 



#25
Jul2811, 08:12 AM

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#26
Jul2811, 08:21 AM

P: 3,015

I pressed the page back button on my keyboard and my whole response is gone. I hate this glitch on PF. 



#27
Jul2811, 08:39 AM

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#28
Jul2811, 08:46 AM

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#29
Jul2811, 09:04 AM

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#30
Jul2811, 09:19 AM

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#31
Jul2811, 09:25 AM

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#32
Jul2811, 10:11 AM

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#33
Jul2811, 10:16 AM

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#34
Jul2811, 10:31 AM

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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. 



#35
Jul2811, 11:19 AM

P: 3,015

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. 



#36
Jul2811, 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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