Derivation Problem: Particle in a Box

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    Derivation
In summary, the analytical expression of DeltaX for a particle in a box is:\Deltax\Deltap = h/2pi\sqrt{(n\pi)^{2} - 6} / \sqrt{12}
  • #1
nadeemo
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Homework Statement



The analytical expression of [tex]\Delta[/tex]x[tex]\Delta[/tex]p for a particle in a box is:

[tex]\Delta[/tex]x[tex]\Delta[/tex]p = h/2pi[tex]\sqrt{(n\pi)^{2} - 6}[/tex] / [tex]\sqrt{12}[/tex]
for any quantum number, n

Homework Equations



([tex]\Delta[/tex]x)[tex]^{2}[/tex] = <x[tex]^{2}[/tex]> - <x>[tex]^{2}[/tex]

and ([tex]\Delta[/tex]p)[tex]^{2}[/tex] = <p[tex]^{2}[/tex]> - <p>[tex]^{2}[/tex]

[tex]\Psi[/tex] = [tex]\sqrt{2/L}[/tex] sin(nxpi/L)

The Attempt at a Solution



so i tried to find (delta x)^2 and multiplied it with (delta p)^2 and rooted it, but i still have "L" left over in my derivation which doesn't work out...
 
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  • #2
Can you show the calculations for the uncertainties in x and p? What you're doing is correct from what you've described.
 
  • #3
<x^2> = integral of (x^2)(psi^2) = 2/ L integral (x^2)sin^2(nxpi/L) dx

integrate by parts from zero to L andd
<x^2> = L^2 - L + 1

<p^2> = integral (conjugate psi)(momentum operator)^2(psi)dx
<p^2> = -h/Lpi integral (conjugate psi)(d^2 psi/ dx^2)dx
after taking derivative and the integral
<p^2> = [ (h bar)(n)(pi)/(L) ] ^2

<p>^2 i found to be zero
and <x>^2 i found to be L^2 / 4

so ( delta x)^2 = L^2 - L + 1 - L^2 / 4 = 3L^2 /4 - L + 1
and (delta p)^2 = [(h bar)(n)(pi)/(L)] ^2

(delta x)(delta p) = sqrt [(3L^2/4 - L + 1)[(hbar)(n)(pi)/(L)]^2]

you can see that L is un removable...it doesn't really work out for me..
 
  • #4
psi(x) must be zero at the ends, so your -L+1 is wrong.
Check the dimensions. <x*2> has to ~L^2.
 
  • #5
<x^2> = 2/L integral (x^2)sin^2 (nxpi/L) dx
= 2/L[x^2(x/2 - (L / 4npi)sin (2nxpi/L)) - integral x sin^2 (nxpi/L)] from 0 to L
= 2/L[x^2(x/2 - (L / 4npi)sin (2nxpi/L)) - x(x/2 - (L/4npi)sin (2nxpi/L)) - integral of sin^2 (nxpi/L)] from 0 to L
=2/L[x^2(x/2 - (L / 4npi)sin (2nxpi/L)) - x(x/2 - (L/4npi)sin (2nxpi/L)) - (x/2 - (L/4npi)sin (2nxpi/L] from 0 to L
= 2/L [ L^3 /2 - L^2/2 + L /2]
= L^2 - L + 1

i don't think i did anythign wrong in my integration...well clearly something is wrong

but i found this
6a1042615e1f7e4a192736255848f2f1.png


its from -a/2 to a/2..so how do i make that form 0 to L?
its the same right?
 
Last edited:
  • #6
so this pic basically solves my problem
its just that...
i wouldn't know how to integrate to get that in the first place
and
<x>^2 = L^2 / 4...and i can't incorporate that..
because well sqrt( <x^2> * (delta p)^2) = the answer that they want
but i need (delta x)^2(delta p)^2

(delta x)^2 = <x^2> - <x> ^2...and if i do that its no longer correct...
 
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  • #7
Things to note: [tex]\langle x \rangle = \langle p \rangle = 0[/tex] -- can you show this?

Further, you should be able to integrate [tex]\int x^n sin(x)\,dx[/tex] as an indefinite integral. Can you do so for n=1 or n=2?
 
  • #8
<x> doesn't equal <p>
in my textbook they show that,
<x> = L/2 and <p> = 0..
and that integral doesn't really relate because x^n sin x...when i have (x^n)(sin^2 x)
 
  • #9
True -- my bad when posting too quickly. But you should still be able to do those integrals by parts...
 
  • #10
6a1042615e1f7e4a192736255848f2f1.png


from my integration by parts for <x^2> i get (L^2)/3 when accoring to that equation
i should be gettin L^2[(n^2)(pi^2) - 6] / [12(n^2)(pi^2)]

a = L, but you hafta multiply it by 2/L, since the wavefunction is sqrt(2/L) sin(nxpi/L)
 

Related to Derivation Problem: Particle in a Box

What is the Derivation Problem: Particle in a Box?

The Derivation Problem: Particle in a Box is a mathematical model used in quantum mechanics to describe the behavior of a particle confined to a one-dimensional box. It involves solving the Schrödinger equation for a particle that is free to move within a certain region, but is constrained by impenetrable walls at the boundaries.

Why is the Particle in a Box model important?

The Particle in a Box model is important because it provides insight into the behavior of particles at the quantum level. It allows scientists to study the properties of particles in confined spaces and understand how their energy levels and wave functions are affected by the boundaries of the box.

What are the key assumptions made in the Particle in a Box model?

The key assumptions of the Particle in a Box model are that the walls of the box are infinitely high and impenetrable, and that the particle is free to move within the box but cannot escape it. Additionally, the model assumes that the particle has no interactions with other particles or external forces.

How is the Particle in a Box model solved?

The Particle in a Box model is solved using the Schrödinger equation, which is a differential equation that describes the time evolution of a quantum system. The equation is solved using boundary conditions at the walls of the box, and the resulting solutions give information about the energy levels and wave functions of the particle.

What are some real-world applications of the Particle in a Box model?

The Particle in a Box model has applications in various fields, including nanotechnology, materials science, and quantum computing. It is used to understand the behavior of electrons in semiconductor devices, the properties of atoms and molecules in confined spaces, and the behavior of quantum bits in quantum computers.

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