Particle-in-a-box and the uncertainty principle

In summary: P = 2 Pi hbar / L dekta x = .385 L so delta x * delta p = .385 Pi * hbarThis is consistent with the principle.cheers,PatrickIn summary, the possible energies for a particle-in-a-box are given by E_n = (n^2h^2)/(8mL^2) and the corresponding momentum is given by p_n = (nh)/(2L). However, the uncertainty principle states that the product of the uncertainties in position and momentum must be greater than or equal to h/2pi. This seems to be violated in the case of a particle confined within a box, but
  • #1
broegger
257
0
For a particle-in-a-box it can be shown that the possible energies are given by

[tex] E_n = \frac{n^2h^2}{8mL^2} [/tex]

where L is the length of the box. The corresponding momentum are given by:

[tex] p_n = \frac{nh}{2L} [/tex]

I don't think it's a problem that the energy has a definite value ([tex] \Delta E = 0 [/tex]) since it is a stationary state ([tex] \Delta t = \infty [/tex]).

But how is it possible for the momentum to be definite ([tex] \Delta p = 0 [/tex]) and, at the same time, the particle to be confined within the box ([tex] \Delta x < \infty [/tex]). Doesn't this violate the uncertainty principle [tex]

\Delta x \Delta p_x \geq \frac{h}{2\pi} [/tex].
 
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  • #2
Do the calculation and find out...it just takes a few integrals to find the answer.
 
  • #3
By my understanding it does. We have [tex] \Delta x \approx L/2 [/tex] and [tex] \Delta p_x = 0 [/tex], and thus:

[tex] \Delta x \Delta p_x = (L/2) \cdot 0 = 0 \leq \frac{h}{2\pi} [/tex]

A violation! What is wrong with this??
 
  • #4
What you write as p_x is in fact |p_x|.
Indeed, given the energy, the momentum can be to the left or to the right, so there is an uncertainty on p_x (even if there isn't on |p_x|).

cheers,
Patrick.
 
  • #5
Oh, you're right.. So we have [tex] \Delta p_x \approx \frac{h}{2L} [/tex]?
 
  • #6
broegger said:
Oh, you're right.. So we have [tex] \Delta p_x \approx \frac{h}{2L} [/tex]?

Without plunging into the calculation, I'd guess so. If (for n = 1), we can say that |p| = Sqrt[ 2 m E_1] = h/2L. This is of course a bit naive, because the particle is not really free, it undergoes an interaction at the walls. But it is "most of the time" free. So we'd think then that the corresponding momentum density function can be represented by two dirac pulses, one at -h/2L and one at h/2L. The average momentum being 0, this means that the standard deviation is h/2L for p_x.
But as I said, we've forgotten the role of the potential energy here. Indeed, the position representation of such a momentum double dirac function would be a sine or cosine, with period 2L. Well, this is exactly the ground state !
It is sin(x / (pi L) ) ! But wait... it is this function within the interval 0-L, but the wave function is 0 outside the interval, while the Fourier transform of our 2 dirac peaks would indicate sin(x/(pi L) ) function everywhere. So we have multiplied in the position space with a square window function (1 in the interval 0-L, 0 outside). This translates into a convolution with a sin(p)/p function in the momentum representation, which will smoothen out the dirac peaks and will probably change a bit the value of Delta-p_x.

At least, the above is my guess, I never worked it out myself.

cheers,
Patrick.
 
  • #7
Integral calculation

Might I suggest you redo your calculations? The uncertainty is p is desinqrt as
sqrt ( <p^2> - <p>^2)

where <p^2> = (nh/(4L))^2 and ,p. = 0 so the uncertainty in the momentum is not zero in the eigenstates of the particle-in-a-box.

achemist
 
  • #8
Integration typos

Hi, sorry for the typos. If you do the calculation for n=1 you will get
delta P = Pi hbar / L
delta x = .181 L
so indeed we get delta x * delta p = .181 Pi * hbar
 

1. What is a particle-in-a-box?

A particle-in-a-box is a theoretical concept used in quantum mechanics to describe the behavior of a particle confined to a one-dimensional region. It is often used as a simplified model to understand the behavior of electrons in atoms or molecules.

2. How does the particle-in-a-box model relate to the uncertainty principle?

The particle-in-a-box model demonstrates the uncertainty principle by showing that the more precisely we know the position of a particle, the less precisely we can know its momentum, and vice versa. This is because the particle is confined to a specific region, limiting its possible states and creating uncertainty in its position and momentum.

3. Can the particle-in-a-box model be applied to real-world scenarios?

While the particle-in-a-box model is a useful tool for understanding the uncertainty principle, it is a simplified and idealized representation. In reality, particles are not confined to a specific region and can have a range of possible states. However, the concept of uncertainty still applies to all particles, and the model can be used to make predictions and calculations that align with experimental results.

4. How does the size of the box affect the uncertainty principle?

The size of the box in the particle-in-a-box model directly affects the uncertainty principle. A smaller box means a more confined particle, leading to a higher uncertainty in its position and a lower uncertainty in its momentum. Conversely, a larger box means a less confined particle, resulting in a lower uncertainty in its position and a higher uncertainty in its momentum.

5. Are there any limitations to the particle-in-a-box model?

Yes, the particle-in-a-box model is limited in its applicability to real-world scenarios, as it only considers one-dimensional confinement and does not account for external factors such as interactions with other particles. It is also a simplified representation and cannot fully capture the complexities of quantum mechanics. However, it remains a useful tool for conceptualizing and understanding the uncertainty principle.

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