Recent content by Wong

hi kakarukeys. To prove it, first note that the set represented by a^{2}-b^{2}-b^{2}-c^{2}=1, a>1 is in fact a subset of A^2-B^2-C^2-D^2>0, A>0. Then what we need to prove becomes Prove Aa+Bb+Cc+Dd>0, where (A, B, C, D) and (a, b, c, d) both satisfies, h^2-i^2-j^2-k^2>0, h>0...

Let me try to give some motivation about the inequaility. Consider the equivalent problem in 2D. That is, a^{2}-b^{2}>0, a>1 A^2-B^2=1, A>1 Prove or disprove aA+bB>0 In 2D, the region represented by the first equation is like a "quadrant" rotated by 45 degrees. (Try to plot it.) The set...

In the second case, f may tend to infinity at a or at b. So f being continuous on [a, b] is required.

First try to think about what you want to prove. That is, S^{-1}AS=D, where D is a diagonal matrix. This is equivalent to proving AS=DS, where D is diagonal. Now each column of S is an eigenvector of A. So A acting on S should produce something quite simple. (Try to think of what is the defining...

To get you started, Av=\lambda v U^{-1}Av=\lambda U^{-1}v Now how may you manipulate the last equation to get U^{-1}AU?

Unitarity of U is not required. To prove your assertion, try to start with the equation Av=\lambda v. How may you manipulate this equation to obtain U^{-1}AU?

a) U is a unitary matrix <=> U*U = I, where "*" denotes conjugate transpose <=> \sum_{j} u_{ji}^{*}u_{jk} = \delta_{ik} <=> u_{i}^{*}u_{k}=\delta_{ik}, where u_{i} is the ith column of U. The last relation implies orthogonality of columns of U. b)This one needs a little thought. If u is an...

The modified drunken walk with a wall at the origin is in fact equivalent to the "absolute" random walk, {|N|}, where N is the random displacement at the nth step.

In fact, "putting a barrier" at zero is equivalent to asserting that the probability of going to the right equals one when the person is at the origin and 1/2 otherwise. With a little bit of work one is able to show that the probaility that the person is at k after n steps is...

A conductor contains many free charges. One should not think that the *excess* charges are all free charges that are present. There are also free electrons. It is because of this that in *static* cases there can be no electric field inside a conductor.

True...what a silly mistake...I'm talking about static field. But the main point is that the state is not even metastable.

Just want to put in my two cents... There can be *no* field inside a conductor by definition. The central (positive) test charge will be neutralised by the free electrons in the conductors, giving zero charge at the centre. This state is not even metastable.

hi, emob2p. I think I would start from the time independent schroedinger equation, H\phi=E\phi, where H is the Hamiltonian, \phi the wavefunction. Normally, H=p^2/(2m)+V(x), where p is the momentum operator. First multiply both sides with \phi^{*}, then note that p is hermitian (and the...

Hi all, I started on my physics program this semester. But things begin to puzzle me much, especially quantum mechanics. I hope that you wuold take the time to answer a few of my puzzles. I used Sakurai's book in my quantum mechanics course. It is said that in the position space, the...

In fact what you said is a general theorem regarding harmonic functions in any dimensions. By definition a harmonic function is a function f satisfying \sum_{i} {\partial_{i}^{2} f}= 0 . To prove the assertion in two dimension, you may like to recall that in a source free region, the...