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sokrates
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Is it possible to derive Heisenberg's Uncertainty principle just using Schrodinger Equation?
If yes, is there any source that you can refer me to?
If yes, is there any source that you can refer me to?
If you leave this discussion considering that it is "added on top", then I feel you may have missed some of the point here.sokrates said:This verifies my initial conjecture then.
Indeed, it is added on top.
StatusX - it is very clarifying thank you
sokrates said:[tex](i \hbar \frac{\partial}{\partial t}+qV) \Psi = \frac{1}{2m}(\sigma \cdot (\overline{p}+q\overline{A}))^2 \Psi[/tex]
is what I consider the Schrödinger equation ( Derived from Dirac equation to "naturally" include spin.)
A= vector potential
V= scalar potential
sigma = Pauli spin matrices in vector form
So how do I exactly derive the uncertainty principle ==== starting from this equation ====
Please show it mathematically.
No, that is incorrect.sokrates said:An equation by itself is not meaningless.
Schrödinger equation predicted the observed spectra of Hydrogen atom (firstly) and many others...
Just by itself. No commutators, no Hilbert space.
I'm not arguing against StatusX's statements, but based on your statements you seem to be taking away only partial information, and/or wrong interpretation, from this discussion. I'm trying to help, but I seem to be failing to communicate clearly.sokrates said:Secondly, I asked whether Schrödinger equation was sufficient to derive Heisenberg's Uncertainy Principle ('by itself') or not and StatusX was right on the spot in his/her answer.
This is essentially the answer given to you previously by several people. Since you already understand steps 1 and 2: Notice #1 is written in the position basis. How do you know that #1 is an eigen state of the momentum operator? How do you know that in #2, a Fourier transform relates the momentum and position basis?sokrates said:This definitely looks promising. Thank you very much for taking the time to outline it.
But it surpassed my knowledge at the moment, I'll probe further if I run into difficulties.
JustinLevy said:No, that is incorrect.
Look at what you wrote and claimed was the Schrodinger equation above. You cannot even use that to calculate anything until you can specify the operators in some basis. Once you've taken momentum to be [tex]i\hbar\nabla[/tex] in the position basis, you have already used some of the essential pieces that lead to the HUP.
JustinLevy said:I'm not arguing against StatusX's statements, but based on your statements you seem to be taking away only partial information, and/or wrong interpretation, from this discussion. I'm trying to help, but I seem to be failing to communicate clearly.
JustinLevy said:This is essentially the answer given to you previously by several people. Since you already understand steps 1 and 2: Notice #1 is written in the position basis. How do you know that #1 is an eigen state of the momentum operator? How do you know that in #2, a Fourier transform relates the momentum and position basis?
jensa said:The OP seems to get a little irritated so it is probably not a good idea to post a reply but I must say that I agree with what JustinLevy said. The Schrödinger equation as a differential equation alone does not have any physical meaning. Physics only comes in when you attach meaning to the wave-function you are solving it for. So in order to calculate something of value you must first specify: How do I obtain physical observables from this wavefunction
It is in this step the HUP pops up. If you have not made any statements about how to obtain observables from your wavefunction you obviosly cannot make any statement about observables, regardless if you have an equation for the wavefunction.I think the point JustinLevy makes is very relevant to your original question and not just some attempt at opening a philosophical debate.
vanesch said:Just my 2 cents. I agree fully with what StatusX wrote, in that the HUP is part of the kinematical description, while the SE is what defines the dynamics. However, I suppose people have different opinions on this, depending on how they got heuristically to a suggestion of what a wavefunction is, what the SE is and so on: there are different ways to suggest how to do so, and different textbooks have different approaches.
But in general, you first set up your Hilbert space by defining the spectra of a complete set of commuting observables (for instance, X, Y and Z with spectra (x,y,z) all real 3-tuples for a featureless point particle). You then introduce their conjugate observables by *requiring* the HUP: [x,px] = i hbar etc... which turns out to define their eigenstates.
And from there on, you try to guess a sensible hamiltonian (by analogy with a classical system, or by inspiration, or whatever), which fixes the time evolution of your states through the SE.
However, as others suggested here, several textbooks take a more suggestive approach by first suggesting that a particle in uniform motion must correspond to something like a plane wave, and then try to deduce how one could get something that is "position" and "momentum" from that wave, and how we could then set up a first order equation that takes these suggestions, and gives the original plane wave as a solution.
jensa said:I would like to add that HUP does not really need to be *required* with this approach. One of the postulates of QM is that observables are represented by Hermitian operators which then define the eigenstates. The observables quite naturally do not need to commute with one another and if you define the momentum operator as a generator of translations, i.e. an operator which takes position eigenstate [tex]|x\rangle[/itex] to [itex]|x+dx\rangle[/itex], the position and momentum operators will naturally satisfy the HUP.
In a similar way the Hamiltonian can be thought of as the generator of time-translations which, if you follow this road, automatically leads you to the Schrödinger equation (with a as of yet undetermined Hamiltonian). So with this approach the Schrödinger equation really does not take a fundamental role.
I am not challanging that form of the Schoridinger eq. The point is, you cannot derive the measurements you claim from that equation alone. For that equation contained several operators, and to put it into a useful form you would need to write what those operators are in some basis, which you need additional information/equations to do. ... as soon as you have written the momentum operator as [tex]i\hbar\nabla[/tex], you have incorporated details intimately related to your question.sokrates said:I am not claiming that Schrodinger Equation. You can find that formula in textbooks if you don't recognize it.
Historically, Schrodinger didn't even know what the "wave" he was solving for was. Several papers later, he was incorrectly suggesting it could be the electron density. It was Max Born that soon figured out that the appropriate choice was to consider [tex]\Psi^*\Psi[/tex] a probability density.sokrates said:When Schrodinger invented his formula, as I tried to explain to you, there was no established "quantum-mechanical formulation" yet. Without operators, bases (all this spectacular terminology you persistently bring up) the prediction was evident. It's a simple equation that has a strong correlation with reality.
This is not a philosophical issue.sokrates said:If you want to drag this debate onto a philosophical frame, I am not the right person.
sokrates said:Is it possible to derive Heisenberg's Uncertainty principle just using Schrodinger Equation?
If yes, is there any source that you can refer me to?
JustinLevy said:---
... as soon as you have written the momentum operator as [tex]i\hbar\nabla[/tex], you have incorporated details intimately related to your question.
---
It is not philosophical fodder, but actually the crux of the issue in responding to your question.
JustinLevy said:Historically, Schrodinger didn't even know what the "wave" he was solving for was. Several papers later, he was incorrectly suggesting it could be the electron density. It was Max Born that soon figured out that the appropriate choice was to consider [tex]\Psi^*\Psi[/tex] a probability density.
What was the best answer? My answer is that the HUP would hold even if the Schrödinger equation didn't. QM says that states are represented by wavefunctions*, and observables by self-adjoint operators. That alone is enough to derive the UP for non-commuting observables. The HUP (which is specifically about position and momentum) is a consequence of the above and the interpretation of x as position and i(d/dx) as momentum.sokrates said:I started the thread with a simple question. I think the best answer has been presented.
Equations means nothing if you don't give specific meanings to the symbols present in there.sokrates said:The truth is out there. And this is our way of 'simulating' it. Schrodinger equation, as I have stressed a number of times, includes a lot of information by itself. And personally for me, it's enough to leave it at that, even to simulate and understand extremely sophisticated devices - like spin-torque transistors.
Fredrik said:What was the best answer? My answer is that the HUP would hold even if the Schrödinger equation didn't. QM says that states are represented by wavefunctions*, and observables by self-adjoint operators. That alone is enough to derive the UP for non-commuting observables. The HUP (which is specifically about position and momentum) is a consequence of the above and the interpretation of x as position and i(d/dx) as momentum.
So the HUP holds in our theory even if we stop there and don't add the assumption that the time evolution of the states is described by the Schrödinger equation. Of course, if we don't, we won't be able to predict the results of many experiments.
quZz said:Fredrik, that is quite right... sokrates should realize that Schroedinger equation describes evolution in time, while uncertainty principles are consequences of definitions of corresponding operators. Any solution of Schroedinger equation with Hamiltonian constructed with those operators satisfy uncertainty principles.
lightarrow said:Equations means nothing if you don't give specific meanings to the symbols present in there.