# Recent content by perishingtardi

1. ### Raising to half power = PRINCIPAL square root?

This may seem like a very elementary question...but here goes anyway. When a positive number is raised to the power 1/2, I have always assumed that this is defined as the PRINCIPAL (positive) square root, e.g. 7^{1/2} = \sqrt{7},. That is, it does not include both the positive and negative...
2. ### Differentiation of inverse trig

Say we want to differentiate \arcsin x. To do this we put y=\arcsin x. Then x=\sin y \implies \frac{dx}{dy}= \cos y. Then we use the relation \sin^2 y + \cos^2 y = 1 \implies \cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}. Therefore \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}. My question is that...
3. ### Absolute value in separable ODEs?

Suppose I have a variable separable ODE, e.g., \frac{dy}{dx} = 3y. We all know that the solution is y=Ae^{3x} where A is a constant. My question is as follows. To actually find this solution we rearrange the equation and integrate to get \int \frac{dy}{y} = 3 \int dx, which gives \ln...
4. ### Modified Bessel function with imaginary index is purely real?

I think I know what the problem is. The first equality is wrong, i.e., \left( K_{i \beta}(x) \right)^* is not simply K_{-i\beta}(x).
5. ### Modified Bessel function with imaginary index is purely real?

I'm trying to decide if the modified Bessel function K_{i \beta}(x) is purely real when \beta and x are purely real. I think that is ought to be. My reasoning is the following: \left (K_{i \beta}(x)\right)^* = K_{-i \beta}(x) = \frac{\pi}{2} \frac{I_{i \beta}(x) - I_{-i \beta}(x)}{\sin(-i...
6. ### 2nd order PDE using integration by parts

its zero?? how does that help though?
7. ### 2nd order PDE using integration by parts

Homework Statement Find the general solution of the equation (\zeta - \eta)^2 \frac{\partial^2 u(\zeta,\eta)}{\partial\zeta \, \partial\eta}=0, where ##\zeta## and ##\eta## are independent variables. Homework Equations The Attempt at a Solution I set ##X = \partial u/\partial\eta## so that...
8. ### Difficult integral

Yeah I did that substitution myself to make it easier... now how do I show that the integral is equal to pi??
9. ### Difficult integral

I've found through the transformation u=pa/\hbar that it is equivalent to showing \int_{-\infty}^{\infty} \frac{\sin^2 u}{u^2}\,du = \pi, if that helps anyone.
10. ### Difficult integral

Mod note: Moved from the math technical sections. I need to show that \int_{-\infty}^\infty \frac{\sin^2 (pa/\hbar)}{p^2} \, dp = \frac{\pi a}{\hbar}. I haven't got a clue how to integrate this function! Any help would be much appreciated thanks.
11. ### Finding momentum distribution for particle in square well

Homework Statement A particle of mass ##m## is trapped between two walls in an infinite square well with potential energy V(x) = \left\{ \begin{array}{cc} +\infty & (x < -a), \\ 0 & (-a \leq x \leq a), \\ +\infty & (x > a).\end{array} \right. Suppose the wavefuntion of the particle at time...
12. ### Parameterization for method of characteristics

Consider the PDE xu_x + yu_y = 4u. Suppose that we want to find the solution that satisfies u=1 on the circle x^2 + y^2 = 1 using the method of characteristics. I have read that the boundary condition can be parameterized as x=\sigma, \qquad y=(1-\sigma^2)^{1/2}, \qquad u=1. My...
13. ### Show that inner product is zero.

Thanks for that - I think we both posted at the same time! My question is related to quantum mechanics. I think physicists use the words Hermitian and self-adjoint interchangeably. I know that pure mathematicians will distinguish them but it's not important for my area. Thanks!
14. ### Show that inner product is zero.

Actually I think I got it: a_i \langle u_i |B|u_j \rangle - \langle u_i |B| u_j \rangle a_j = 0 using the Hermiticity of A for the first term, and then since a_i \neq a_j we get \langle u_i |B| u_j \rangle = 0
15. ### Show that inner product is zero.

Let A be a Hermitian operator with n eigenkets: A|u_i\rangle = a_i |u_i\rangle for i=1,2,...,n. Suppose B is an operator that commutes with A. How could I show that \langle u_i | B | u_j \rangle = 0 \qquad (a_i \neq a_j)? I have tried the following but not sure how to proceed: AB -...