Ground State Energy of quantum oscillator

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PsychonautQQ
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My textbook says the ground state energy of the QSHO is given by 1/2*h_bar*w and that this is the minimum energy consistent with the uncertainty principle. However I am having trouble deriving this myself... ΔEΔt ≥ h_bar / 2.. so then ΔE/Δfrequency ≥ h_bar / 2...

ΔE*2*pi / w ≥ h_bar / 2
ΔE ≥ h_bar*w / 4*pi

what am I doing wrong?
 
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You aren't doing anything wrong per say, at least not as far as I can tell. ##\Delta t## is the characteristic time scale of the system and certainly since we have a characteristic frequency scale ##\omega##, the characteristic time scale should be given by ##\Delta t = \frac{2\pi}{\omega}##.

I'm not entirely sure why you chose to start with the energy-time uncertainty relation; if you take a look at its derivation, particularly the assumptions about the evolution of expectation values of operators through a characteristic time period of the system, you would see that it's much more subtle a relation than the usual position-momentum uncertainty relation, but even that withstanding, the text is referring to ##\langle E \rangle ##, not ##\Delta E##, when it talks about the minimum energy consistent with the uncertainty principle. In other words ##\langle E \rangle \geq \frac{1}{2}\hbar \omega## is the desired result.

So start instead with ##\Delta x \Delta p \geq \frac{\hbar}{2}##. Write down the expectation value for the total energy using the Hamiltonian for the QSHO and use the definition of variance and the position-momentum uncertainty relation to get an inequality for the energy expectation value involving only ##\Delta p##. Then minimize the result.
 
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