# Ground State Energy of quantum oscillator

1. Apr 5, 2014

### PsychonautQQ

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?

2. Apr 5, 2014

### WannabeNewton

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.

Last edited: Apr 5, 2014