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Schrödinger equation 
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#1
Dec1312, 10:51 PM

P: 21

Does the Schrödinger equation completely neglect the uncertainty principle? If so, wouldn't this imply that our intensity distribution has its own probability distribution?



#2
Dec1412, 07:27 AM

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The Schrödinger equation predicts the wave function with certainty; but from this wave function the uncertainties of observables can be derived exactly



#3
Dec1412, 07:43 AM

Mentor
P: 11,602

The momentumspace wave function ##\Phi(p,t)## is basically the Fourier transform of the positionspace wave function ##\Psi(x,t)##. The uncertainty principle comes from the properties of Fourier transforms. Any pair of functions that are related by Fourier transforms has a similar uncertainty principle.



#4
Dec1412, 08:43 AM

P: 21

Schrödinger equation
I understand what both of you are saying and I appreciate the replies.
In the Schrödinger equation we input values for energy/mass assuming we know with 100% certainty what these values for energy/mass are. Due to the input of these values is where my question holds its regards. 


#5
Dec1412, 10:21 AM

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P: 5,364

The input is a wave function, the output is a wave function at a later time. This predicts with certainty that a system will be in a state A' at time t' > t provided that it was in state A at time t; A is specified by a wave function or a state vector A>. In case of the timeindep. SE the input is not energy, the input is nothing! The outputs are a) the allowed energy eigenvalues and b) the corresponding eigenfunctions. The SE does not tell you in which state the system is, in only tells you what the allowed state are 


#6
Dec1412, 02:28 PM

P: 21




#7
Dec1412, 03:56 PM

P: 1,042




#8
Dec1412, 04:12 PM

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HW Helper
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#9
Dec1412, 06:52 PM

P: 21

If all direct observables have some uncertainty, won't this mess up our intensity distribution even more than the fouriers already do?



#10
Dec1412, 09:02 PM

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Thanks
P: 3,432

In the timedependent Schrodinger equation [itex]H\Psi=E\Psi[/itex] the Hamiltonian is written as if all of its inputs were exactly known. For example, if we're dealing with two charged particles, there will be a [itex]\frac{1}{r1r2}[/itex] term somewhere in it, where r1 and r2 are the positions of the two particles. You should read that as saying not that the two particles are at those exact positions, but rather that if they were in those positions that would be the exact distance between them. The uncertainty principle doesn't stop us from talking about how things would be if we knew exactly where a particle was, it just forbids us from knowing exactly where it is. Once I have the Hamiltonian written down, I solve Schrodinger's equation; and as tom.stoer said in #2, the uncertainty principle is inherent in the ψ that comes out. 


#11
Dec1412, 10:02 PM

P: 21

Ahh yes, talking in this fashion resolves my concerns.
Thank you nug 


#12
Dec1512, 02:15 AM

P: 1,020

the wave function when described in terms of a fourier integral,gives directly relationship to uncertainty principle.you can see something about it here on page 51
http://books.google.co.in/books?id=7...packet&f=false 


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