Uncertainty of position in an infinite potential well

  • #1
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?
 

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  • #2
Simon Bridge
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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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  • #3
stevendaryl
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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 [itex]E[/itex], then the momentum is [itex]\pm \sqrt{2mE}[/itex]. So the expectation value of [itex]p[/itex] 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:

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

[itex]\langle p \rangle = 0[/itex]
[itex]\langle p^2 \rangle = 2mE[/itex]

So
[itex]\delta p = \sqrt{2mE} = \frac{\hbar \pi}{a} [/itex]

So the uncertainty of [itex]p[/itex] is not zero.
 
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jtbell
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