Show that the uncertainty in the position of a ground state harmonic oscillator is 1/22/mk1/4.

Short Answer

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The uncertainty in position in the ground state of harmonic oscillator is 122mk1/4.

Step by step solution

01

Uncertainty

The uncertainty in the position of harmonic oscillator is the deflection from the actual position of the oscillator.

02

Determination of uncertainty in the position of harmonic oscillator

The relation between uncertainty in position and momentum is given as:

ΔxΔp=2

The angular frequency of the harmonic oscillator is given as:

ω=km

The energy of a quantum harmonic oscillator is given as:

E=Δp22m+12mω2Δx2E=12m2Δx2+12mω2Δx2E=28mΔx2+12mω2Δx2

Differentiate the above expression with respect to position.

dEdx=ddx28mΔx2+12mω2Δx2dEdx=-228mΔx3+12mkm22ΔxdEdx=-24mΔx3+kΔx

The uncertainty in position in the ground state will be zero so substitute dEdx=0in the above expression.

0=-24mΔx3+kΔx24mΔx3=kΔxΔx4=24mkΔx=122mk1/4

Therefore, the uncertainty in position in the ground state of harmonic oscillator is 122mk1/4.

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