Heisenberg Uncertainty Principle & Wave Function

In summary, the conversation discusses the application of Heisenberg's uncertainty principle to a top-hat wave function and the estimation of minimum kinetic energy in a 1,2,3-D box. The first link shows how this can be done, while the second link deals with a particle in an infinite square well potential. However, there may be issues with using the infinite box potential to demonstrate the uncertainty relation as it does not have a proper momentum operator. The Hamiltonian in the second link is self-adjoint, making it a more accurate approach.
  • #1
Sum Guy
21
1
Considering how Heisenberg's uncertainty principle is applied to a top-hat wave function:

This hyperphysics page shows how you can go about estimating the minimum kinetic energy of a particle in a 1,2,3-D box: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/uncer2.html

You can also investigate a particle in a box via the following treatment: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/schr.html#c2

I'm having difficulty reconciling these two things? The first I would think of as applying the uncertainty principle to a top hat function, yet in the second link (the same scenario I think) we clearly have sinusoidal (i.e. non top hat) wavefunctions. How can you bring these two things together? Can you in general apply the uncertainty principle to a top hat function?
 
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  • #2
I'm not sure I understand how you are interpreting those web pages.

The first one is a square potential, not a "top hat". The secon one is a1D infinite square well. These represent the POTENTIAL profile. BOTH will result in "sinusoidal" wavefunctions.

Zz.
 
  • #3
To use the infinite box potential to "demonstrate" the uncertainty relation between position and momentum is quite common in the introductory QT textbook literature, but it's nevertheless one of the sins in physics didactics one should avoid. The reason is that for this problem no proper momentum operator exists. So it doesn't make sense to talk about a momentum probability distribution in this case either. The reason is pretty formal concerning the self-adjointness of operators in the Hilbert space ##L^2([-L/2,L/2])## with "rigid boundary conditions" ##\psi(-L/2)=\psi(L/2)=0##.

The second link is however correct since the Hamiltonian
$$\hat{H}=-\frac{\hbar^2}{2m} \partial_x^2$$
is indeed self-adjoint in this space and thus energy is properly defined for the problem.
 
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1. What is the Heisenberg Uncertainty Principle?

The Heisenberg Uncertainty Principle, also known as the Uncertainty Principle, is a fundamental concept in quantum mechanics that states that it is impossible to know both the exact position and momentum of a particle at the same time. This means that the more precisely we know the position of a particle, the less we can know about its momentum, and vice versa.

2. What is the significance of the Heisenberg Uncertainty Principle?

The Heisenberg Uncertainty Principle has significant implications for our understanding of the behavior of particles at the quantum level. It means that there is a fundamental limit to the precision with which we can measure certain properties of particles. It also highlights the probabilistic nature of quantum mechanics, as the exact position and momentum of a particle cannot be simultaneously known.

3. How does the Heisenberg Uncertainty Principle relate to the wave function?

The wave function is a mathematical representation of a particle's state in quantum mechanics. The Heisenberg Uncertainty Principle can be derived from the properties of the wave function, specifically its position and momentum components. The uncertainty principle essentially limits the accuracy with which we can determine the values of these components.

4. Can the Heisenberg Uncertainty Principle be violated?

No, the Heisenberg Uncertainty Principle is a fundamental principle of quantum mechanics and cannot be violated. It is a consequence of the wave-particle duality of particles at the quantum level and has been confirmed through numerous experiments and observations.

5. How does the Heisenberg Uncertainty Principle impact our everyday lives?

While the Heisenberg Uncertainty Principle may seem to only apply to particles at the quantum level, its effects can be seen in everyday life. For example, the principle explains why we cannot simultaneously know the exact position and speed of a car, as measuring one will inherently affect the other. It also has applications in technologies such as MRI machines and semiconductor devices.

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