Rigid Box Particle and Heisenberg Principle

[knowLittle]

Homework Statement

For the ground state of a particle in a rigid box, we
have seen that the momentum has a definite magni-
tude (h/2π)k but is equally likely to be in either direction.
This means that the uncertainty in p is Δp≈ (h/2π)k. The
uncertainty in position is Δx≈ a/2. Verify that these
uncertainties are consistent with the Heisenberg
uncertainty principle.

Homework Equations

Heisenberg Relations
Δp= h/2pi Δk ...(1)
ΔxΔk ≥ 1/2
ΔxΔp≥ (h/2pi)/2

The Attempt at a Solution

(a/2)Δp≥h/4pi

Δp≥h/(2pi*a )

p=(pi/a)h-bar

I am not sure, if I am in the right direction. I feel that I am going in circles. Any help?

 

[TSny]

You're given that Δp ≈ hk/(2π). Use what you know about the ground state of a particle in a box to express k in terms of a.

Then see what you get for ΔpΔx.

 

[knowLittle]

k=π/a

a≈(h/2)(1/Δp) ,a=2Δx
2Δx=(h/2)(1/Δp)

Δp*Δx=(h/4)

I still don't see it. Help, please.

 

[TSny]

Yes, k = π/a. I think what you did is ok.

I would have written it out a little differently:

You were given in the statement of the problem that you could assume Δp≈hk/(2π). So, when you substitute for k you get

Δp ≈ h(π/a)(1/(2π)) = h/(2a).

You were also given that Δx≈a/2.

So, putting it together, ΔpΔx ≈ h/(2a) (a/2) = h/4. This is the same as your result.

The question asks you to state whether or not this is consistent with the uncertainty principle. What do you think?

 

[knowLittle]

Thank you, Freunde.

It agrees with Heisenberg equation.
$\Delta p \Delta x \geq \dfrac {\hbar}{2} = 5.272 \times 10^{-35} m^{2} \times \dfrac {kg}{s}$
$\Delta p \Delta x= h/4 = 16.5 \times 10^{-35} \times m^{2} \times \dfrac{kg}{s}$

 

[TSny]

Good!

 

[m1rohit]

I Thought this will be helpful

I have attached two pdf files.problem 4.2 in the pdf named uncertainty will give you the actual uncertainty relation for your problem and the second pdf file will show you that how this uncertainty results in zero point energy.

 

[knowLittle]

I have attached two pdf files.problem 4.2 in the pdf named uncertainty will give you the actual uncertainty relation for your problem and the second pdf file will show you that how this uncertainty results in zero point energy.

Thank you for the information.
Could you cite your sources?
Also, can anyone else corroborate that the sources are right?

 

[m1rohit]

oh it is taken from a excellent book named concept and application of quantum mechanics by Nouredine Zettili.this book is easily available you can check it yourself