Recent content by hoch449

For part c) the integral is performed under all space. When you write out the integral, the integrand is [sin(\frac{x}{3})]^*cos(\frac{x}{3}) with a constant on the outside of the integral. Now for this to be 0. X would have to be those values that I wrote in my previous post. Does this seem...

Homework Statement a) Show that the functions f=sin(ax) and g=cos(ax) are eigenfunctions of the operator \hat{A}=\frac{d^2}{dx^2}. b) What are their corresponding eigenvalues? c)For what values of a are these two eigenfunctions orthogonal? d) For a=\frac{1}{3} construct a linear...

Ah that totally makes sense! Non-degenerate eigenvalues means different ie. (not the same) eigenvalues does it not?

Homework Statement Prove that if two linear operators A and B commute and have non-degenerate eigenvalues then the two operators have common eigenfunctions. Homework Equations [A,B]= AB - BA= 0 Af=af Bg=cg,\ let\ g=(f+1) --> B(f+1)=c(f+1)\ where\ a\neq c The Attempt at a...

Solve the following differential: \frac{d^2}{d\theta^2} + cot\theta\frac{dS}{d\theta} - \frac{m^2}{sin^2\theta}S(\theta) + \frac{cS(\theta)}{\hbar}=0 The first step is: "For convenience we change the independent variable, by making the substitution w=cos\theta" So my question is...

What does this mean?? (Intro QM) I am on the angular momentum unit in my introductory quantum mechanics course and we had to examine the following two commutators: 1) [Lx,Ly] it ends up equating to i\hbar Lz [Lx,Ly]= i\hbar Lz [Ly,Lz]= i\hbar Lx [Lz,Lx]= i\hbar Ly Tt was concluded...

I don't believe there are any specific parentheses. The exact question is to find the following commutator [Lx,Ly] where: Lx= (y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y}) Ly= (z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z})

Homework Statement Simplify the following two expressions: y\frac{\partial}{\partial z}z\frac{\partial}{\partial x} z\frac{\partial}{\partial y}x\frac{\partial}{\partial z} The Attempt at a Solution for the first one: y\frac{\partial}{\partial z}z\frac{\partial}{\partial x}...

The Right Hand Side of the equation gives me the difficulty. I am sure I am making an elementary mistake. \frac{\partial\theta}{\partial y}: \frac{d}{dy}cos\theta=\frac{d}{dy}(\frac{z}{r}) \frac{d}{d\theta}(cos\theta)\frac{d\theta}{dy}=\frac{d}{dr}r^-1\frac{dr}{dy}...

An earlier part to this question was to find \frac{\partial r}{\partial y} and I solved it correctly. Here is how I did it. r^2= x^2 + y^2 + z^2 \frac{d}{dy}r^2= \frac{d}{dy}y^2 \frac{d}{dr}r^2\frac{dr}{dy}=2y 2r\frac{dr}{dy}=2y so therefore \frac{\partial r}{\partial y}=...

Homework Statement Find \frac{\partial\theta}{\partial y} z=rcos\theta x=rsin\theta\cos\phi y=rsin\theta\sin\phi r^2=x^2 + y^2 + z^2 The Attempt at a Solution We know cos\theta=\frac{z}{r}=\frac{z}{\sqrt{x^2 + y^2 + z^2}} So implicit differentiation says to differentiate...

Perfect it all makes sense now! You guys are great thanks!

Homework Statement I am supposed to find if the following commutes: [Lx,Ly] Homework Equations Lx= -i\hbar[y(\partial/\partialz) - z(\partial/\partialy)] Ly= -i\hbar[z(\partial/\partialx) - x(\partial/\partialz)] where [Lx,Ly]=LxLy-LyLx If it commutes then [Lx,Ly]=0...