Recent content by Sciencenerd3

Thanks so much. i cannot believe I wasted so much time on something so simple and trivial like forgetting order of operations. I'm new to the forum, is there any specific way I am supposed to give thanks or credit?

Could it be that I am acting on both wave functions multiplied together with the momentum operator instead of just the second wave function in the expectation equation?

\psi_0=(\frac{\alpha}{\pi})^\frac{1}{4}e^{-\frac{\alpha x^2}{2}} So using the expectation equation: <p^2>=∫\psi_0^*\hat{p}\hat{p}\psi_0dx <p^2>=∫\psi_0^*\frac{\hbar^2}{i^2}\frac{∂^2}{∂x^2}\psi_0dx <p^2>=∫\frac{\hbar^2}{i^2}\frac{∂^2}{∂x^2}(\frac{ \alpha}{\pi})^\frac{1}{2}e^{-\alpha...

Yes, the wavefunction is normalized because it is the ground state harmonic oscillator wavefunction. However, when I try to use the expectation value equation: ⟨A⟩=∫ℝdxψ∗(x)Aˆψ(x). I get <P2>=h2(bar)/i2[-2a+2a]=0. However, I know that this value is not 0. I am supposed to use the...

This seems as though it would work but I have to solve it in a specific way. My answer would be close if too many things were not cancelling but I cannot seem to work it out fully. After I integrate after taking the double derivative and keeping in mind to used the product rule, I am getting zero.

Ground State Wave Equation: ψ0=(a/∏)(1/4)e(-ax2/2) Prove the Heisenberg Uncertainty principle ≥h(bar)/2 by way of expectation values. First I found <x>=0 because it was an odd function then I found <Px>=0 because it was an odd function Then <x2>=∫(a/∏)(1/2)x2e(-ax2)/2dx=1/2a by way of...