Solving 1D Infinite Well: Momentum Problem & Heisenberg Uncertainty

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In summary, p_{av} in the one-dimensional infinite well is equal to 0. (p^2)_{av} can be calculated using the equation (\frac{\hbar\pi^2n^2}{L})^2. The uncertainty in momentum, \Delta p, is equal to \frac{\hbar\pi n}{L}. When compared to the Heisenberg uncertainty relationship, \Delta p \Delta x = \frac{\hbar}{2}\sqrt{2n^2\pi^2 -1}, it appears that the uncertainty increases with n. However, further calculations are needed to determine \Delta x.
  • #1
Hyperreality
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(a) In the infinite one-dimensional well, what is [tex]p_{av}[/tex]?

(b) What is [tex](p^2)_{av}[/tex]?

(c) What is [tex]\Delta p = \sqrt{(p^2)_av - (p_av)^2}[/tex]?

(d) Compute [tex]\Delta p \Delta x[/tex], and compare with the Heisenberg uncertainty relationship.

Here's my working:

(a) [tex]p_{av}=0[/tex].

I'm not so sure about this bit
(b)[tex](\frac{p^2}{2m})_{av} = E_{n} = \frac{\hbar^2\pi^2n^2}{2mL^2}[/tex].
There fore [tex](p^2)_{av}=(\frac{\hbar\pi^2n^2}{L})^2[/tex]

(c)Therefore,
[tex]\Delta p = \frac{\hbar\pi n}{L}[/tex].

(d)[tex]\Delta p\Delta x = \frac{\hbar}{2}\sqrt{2n^2\pi^2 -1}[/tex]

Part (d) it seems the most suspicious, that is, the uncertainty increases with n^2. Have I done anything wrong?
 
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  • #2
Actually it increases with "n"...You have square root form a "n^{2}"...It looks okay...Though you didn't show the calculations leading to [itex] \Delta x [/itex]...

Daniel.
 
  • #3


Your working for parts (a) and (b) is correct. For part (c), you seem to have forgotten to square the term inside the square root, so it should be \Delta p = \frac{\hbar\pi n}{L}. For part (d), you are correct that the uncertainty increases with n^2. This is a fundamental aspect of the Heisenberg uncertainty principle, which states that the product of uncertainties in position and momentum must be greater than or equal to a certain value (in this case, \frac{\hbar}{2}). So as n increases, the uncertainty in momentum also increases, in order to satisfy this relationship. This shows the inherent uncertainty in the position and momentum of a particle in the infinite one-dimensional well.
 

1. What is the 1D infinite well problem?

The 1D infinite well problem is a theoretical concept in quantum mechanics that involves a particle confined to a one-dimensional space with infinite potential barriers on both ends. This problem is often used to illustrate the principles of quantum mechanics, such as the quantization of energy levels and the wave-particle duality of matter.

2. How is momentum involved in solving the 1D infinite well problem?

In solving the 1D infinite well problem, momentum is used to determine the allowed energy levels of the particle. The momentum of the particle is quantized, meaning it can only take on specific discrete values, based on the size of the well and the mass of the particle.

3. What is the Heisenberg uncertainty principle?

The Heisenberg uncertainty principle is a fundamental principle in quantum mechanics that states that it is impossible to simultaneously know the exact position and momentum of a particle. This means that the more precisely we know one of these properties, the less precisely we can know the other.

4. How does the Heisenberg uncertainty principle relate to the 1D infinite well problem?

In the 1D infinite well problem, the Heisenberg uncertainty principle is important because it places a limit on how precisely we can know the momentum of the particle, given that its position is confined to a specific region. This uncertainty in momentum leads to the quantization of energy levels in the well.

5. What are some real-world applications of the 1D infinite well problem and the Heisenberg uncertainty principle?

The 1D infinite well problem and the Heisenberg uncertainty principle have many practical applications, such as in the development of semiconductors and electronic devices. They also play a crucial role in understanding the behavior of subatomic particles and can be used to explain phenomena such as the tunneling effect and the stability of atoms.

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