Recent content by Unredeemed

I've got it! Thanks very much.

It's my university notes - we could well have to use ladder operators to solve this. In the past we've occasionally had questions that can only be covered with material from the next chapter. What in specific should I be looking for with regard to ladder operators?

Not yet. But they do come in the next chapter of the notes?

We're then asked to show by considering <\psi\mid (P\pm im\omega X)^{k}\psi> for k=1,2 and using orthogonality properties of eigenvectors that: \mathbb{E}_{\psi}(P)=0=\mathbb{E}_{\psi}(X) and \mathbb{E}_{\psi}(P^{2})=m^{2}\omega^{2}\mathbb{E}_{\psi}(X^{2})=mE I've shown that (P-im\omega...

Homework Statement The vector \psi =\psi_{n} is a normalized eigenvector for the energy level E=E_{n}=(n+\frac{1}{2})\hbar\omega of the harmonic oscillator with Hamiltonian H=\frac{P^{2}}{2m}+\frac{1}{2}m\omega^{2}X^{2}. Show that...

Would you mind going into a little more depth about how you'd modify the Picard's theorem proof here? I'm a little confused as the version I've seen is really long.

Can you take the matrix out of the integral even though its entries are polynomials in t?

Homework Statement Prove that the convolution of e^{-\left|x\right|} is (1-x)e^{x} for x<0 and (1+x)e^{-x} for x>0 Homework Equations The Attempt at a Solution I plug through the integral in the standard way and take the limits as x tends to positive and negative infinity etc...

I feel like using the distance formula is over complicating things. After all, you haven't been asked to find the ACTUAL distance, just the point which is closest. You could do that by minimising the distance, but I think it's easier to think about what this would look like on the graph. So...

I'd have used the fact that the point closest to (-2,4) will lie on the line that passes through (-2,4) and is perpendicular to the parabola at the point where they cross.

Well ## \frac{14}{13} - \frac{(\sqrt[3]{14})^{3}}{13}=\frac{14}{13}-\frac{14}{13} = 0 ## So that's the root?

So is my interval ## [-\frac{\sqrt{39}}{3},\frac{\sqrt{39}}{3}] ## ? Because that doesn't contain a root of the equation? Sorry, I'm very confused.

Sorry, I don't understand what you mean?

But if we want ## |g'(x)| < γ ## And ## |g'(x)| = \frac{-3x^{2}}{13} ## So ## |\frac{-3x^{2}}{13}| < γ ## And then ## |x^{2}| < \frac{13γ}{3} ## Implies ## |x| < \sqrt{\frac{13γ}{3}} ## Then ## |x| < \frac{\sqrt{39}}{3} ## since ## γ < 1 ## No?

I've found that abs(g'(x)) < \frac{\sqrt{39}}{3} But, this value is less than \sqrt[3]{14} which is obviously the root. So is there no interval?