Uncertainty of energy in a quantum harmonic oscillator

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SUMMARY

The uncertainty of the kinetic energy of a quantum harmonic oscillator in the ground state can be calculated using the equations provided. Specifically, the average kinetic energy is given by \(\left\langle E_{kin} \right\rangle = \frac{\left\langle p^2_x \right\rangle}{2m}\) and the uncertainty is determined by \(\Delta E_{kin}=\sqrt{\left\langle E^2_{kin} \right\rangle - \left\langle E_{kin} \right\rangle^2}\). The relationship \(\left\langle E^2_{kin} \right\rangle = \frac{\left\langle p^4_x \right\rangle}{4m^2}\) is confirmed as valid, allowing for the calculation of the uncertainty in kinetic energy.

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  • Knowledge of operators and expectation values
  • Familiarity with the Heisenberg uncertainty principle
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bobred
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Homework Statement


Find the uncertainty of the kinetic energy of a quantum harmonic oscillator in the ground state, using

[itex]\left\langle p^2_x \right\rangle = \displaystyle\frac{\hbar^2}{2a^2}[/itex] and
[itex]\left\langle p^4_x \right\rangle = \displaystyle\frac{3\hbar^2}{4a^2}[/itex]


Homework Equations


[itex]\Delta E_{kin}=\sqrt{\left\langle E^2_{kin} \right\rangle - \left\langle E_{kin} \right\rangle^2}[/itex]

[itex]\left\langle E_{kin} \right\rangle = \displaystyle\frac{\left\langle p^2_x \right\rangle}{2m}[/itex]

The Attempt at a Solution


With [itex]\left\langle E_{kin} \right\rangle^2[/itex] I have no problem with but am I valid in saying

[itex]\left\langle E^2_{kin} \right\rangle = \displaystyle\frac{\left\langle p^4_x \right\rangle}{4m^2}[/itex]?
 
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bobred said:
am I valid in saying

[itex]\left\langle E^2_{kin} \right\rangle = \displaystyle\frac{\left\langle p^4_x \right\rangle}{4m^2}[/itex]?
Yes.
 

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