Schrödinger Eq: Neglect Uncertainty Principle? Intensity Probability?

In summary, the Schrödinger equation predicts the wave function with certainty, but from this wave function the uncertainties of observables can be derived exactly. The momentum-space wave function is related to the position-space wave function through Fourier transforms, which results in a similar uncertainty principle. The Hamiltonian used in the Schrödinger equation takes into account the fundamental observables, such as position and momentum, but does not determine their exact values due to the uncertainty principle. This is inherent in the wave function that is output from the equation. Additionally, the Fourier transform of the wave function provides a direct relationship to the uncertainty principle.
  • #1
Mr-T
21
0
Does the Schrödinger equation completely neglect the uncertainty principle? If so, wouldn't this imply that our intensity distribution has its own probability distribution?
 
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  • #2
The Schrödinger equation predicts the wave function with certainty; but from this wave function the uncertainties of observables can be derived exactly
 
  • #3
The momentum-space wave function ##\Phi(p,t)## is basically the Fourier transform of the position-space 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
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
Mr-T said:
In the Schrödinger equation we input values for energy/mass assuming we know with 100% certainty what these values for energy/mass are.
No

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 time-indep. 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
tom.stoer said:
In case of the time-indep. SE the input is not energy, the input is nothing!

If you are not inputting any information into the T-I SE then how do you know what particle it is talking about?!
 
  • #7
Mr-T said:
If you are not inputting any information into the T-I SE then how do you know what particle it is talking about?!
Do you mean you specify a potential, then solve the SE equation for a given potential? Or you plug in the values of the eigenvalues?
 
  • #8
Mr-T said:
If you are not inputting any information into the T-I SE then how do you know what particle it is talking about?!

The remark by Tom is an overstatement, an exaggeration. The input is the specific form of the Hamiltonian in terms of fundamental observables such as position, momentum, spin.
 
  • #9
If all direct observables have some uncertainty, won't this mess up our intensity distribution even more than the fouriers already do?
 
  • #10
Mr-T said:
If all direct observables have some uncertainty, won't this mess up our intensity distribution even more than the fouriers already do?

OK, I think I see where you were going with your original question...

In the time-dependent 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}{r1-r2}[/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
Ahh yes, talking in this fashion resolves my concerns.

Thank you nug
 

1. What is the Schrödinger equation?

The Schrödinger equation is a fundamental equation in quantum mechanics that describes how the wave function of a physical system changes over time.

2. What is the Uncertainty Principle?

The Uncertainty Principle, also known as the Heisenberg Uncertainty Principle, states that the more precisely the position of a particle is known, the less precisely its momentum can be known, and vice versa.

3. How does the Schrödinger equation neglect the Uncertainty Principle?

The Schrödinger equation does not neglect the Uncertainty Principle. In fact, the Uncertainty Principle is a fundamental part of the equation and cannot be neglected.

4. What is intensity probability in relation to the Schrödinger equation?

Intensity probability refers to the likelihood of finding a particle in a particular location at a given time, as described by the wave function in the Schrödinger equation.

5. How is the Schrödinger equation used in modern science?

The Schrödinger equation is used to understand and predict the behavior of particles at the quantum level. It is a crucial tool in fields such as quantum mechanics, chemistry, and materials science.

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