# Uncertainty product

Sushmita

## Homework Statement

For the ground state of a particle moving freely in a one-dimensional box 0≤x≤L with rigid reflecting end points, the uncertainity product (Δx)(Δp) is
(A) h/2
(B) h√2
(C) >h/2
(D) h/√3

## Homework Equations

The uncertainity principle says that -
(Δx)(Δp) ≥ ħ/2
Ground state energy in a one dimensional box of size L is E = π2ħ2/(2mL2)

## The Attempt at a Solution

The particle will be localised in the resion between 0 and L. So the positional uncertainity is L
(Δx) =L

Particle is in the ground state
E= π2ħ2/(2mL2)
p2/(2m) = π2ħ2/(2mL2)
From here I have calculated the momentum ( p= πħ/L) of the particle but I cannot find the uncertainity in position. Can you give me a direction.

Homework Helper
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From here I have calculated the momentum ( p= πħ/L) of the particle but I cannot find the uncertainity in position. Can you give me a direction.
You really need to find Δp. How would you do that?
As for the uncertainty in position, you just stated it is L.
The particle will be localised in the resion between 0 and L. So the positional uncertainity is L

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Homework Helper
I think you mean the uncertainty in momentum. I"m a little rusty in QM but I believe what you just calculated is the expectation value of p22, <p2>. Momentum is not a single number (which would violate the uncertainty principle).

I believe the average value of p is 0, so the variance in p (the square of Δp) is <p2> - <p>2 = <p2>. So the expression you gave for p, the square root of the variance, is actually the momentum uncertainty you're looking for.

Sushmita
I think you mean the uncertainty in momentum. I"m a little rusty in QM but I believe what you just calculated is the expectation value of p22, <p2>. Momentum is not a single number (which would violate the uncertainty principle).

I believe the average value of p is 0, so the variance in p (the square of Δp) is <p2> - <p>2 = <p2>. So the expression you gave for p, the square root of the variance, is actually the momentum uncertainty you're looking for.

Oh yeah. I was totally missing this out. I tried the whole problem by calculating Δp from √(<p2> - <p>2)
I am getting uncertainity in momentum too now which is Δp = πħ/L.
So now (Δx)(Δp) = L × πħ/L = πħ = π× (h/2π) = h/2.
But this is not the answer.