# Uncertainty of position in an infinite potential well

The ground state energy of a particle trapped in an infinite potential well of width a is given by (ħ2π2)/2ma2. So the momentum is given by (2mE)1/2 = ħπ/a. Since this is a precise value, doesn't that mean that we know momentum with 100% certainty? And if that is the case shouldn't the uncertainty in position be infinite?

Simon Bridge
Homework Helper
Your relation for the momentum is incorrect.
You can find the momentum in QM by applying the momentum operator - just like you find the energy by applying the energy operator. So what happes when you apply the momentum operator to the ground-state energy eigenfunction for a particle in a box?

Note: The momentum wavefunction is the fourier transform of the position wavefunction.

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Ananthan9470
stevendaryl
Staff Emeritus
The ground state energy of a particle trapped in an infinite potential well of width a is given by (ħ2π2)/2ma2. So the momentum is given by (2mE)1/2 = ħπ/a. Since this is a precise value, doesn't that mean that we know momentum with 100% certainty? And if that is the case shouldn't the uncertainty in position be infinite?

But if all you know is $E$, then the momentum is $\pm \sqrt{2mE}$. So the expectation value of $p$ is 0 (it's just as likely to be pointing in one direction as another). A rough estimate of the uncertainty in a quantity is the standard deviation, which would tell us:

$\delta p = \sqrt{\langle p^2\rangle - \langle p \rangle^2}$ where $\langle X \rangle$ means the expectation value, or average value. So for this problem:

$\langle p \rangle = 0$
$\langle p^2 \rangle = 2mE$

So
$\delta p = \sqrt{2mE} = \frac{\hbar \pi}{a}$

So the uncertainty of $p$ is not zero.

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Ananthan9470 and Simon Bridge
jtbell
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Ananthan9470