Homework Statement
The radial portion of the wavefunction of a particle contained in a spherically symmetric infinite well potential (ie U= infinite outside a certain radius "a" and 0 inside the radius) is given by the spherical Bessel functions.
J_{l}(kr) =...
Homework Statement
You are given in a earlier stage of this problem that the wavefunction is separable, ie.)
\Psi(x,y) = X(x)Y(y)
The problem asks you to solve for the wavefunction of a particle trapped in a 2D infinite square well using Parity. ie.) solve
\Psi(-x,-y) = \Psi(x,y) and...
So if my initial equations were correct then I can't figure out why I'm having problems actually computing my transmission and reflection coefficients. Using my 2 boundary conditions I get
A = \frac{C}{2}(1-\frac{k_{2}}{k_{1}})
B =\frac{C}{2}(1+\frac{k_{2}}{k_{1}}).
Solving for Probability...
I've actually always had trouble understanding why they didn't all just vanish, thanks for the clarification there. But without the finite condition that leaves me with only 2 boundary conditions (continuity of \psi and its derivative at the boundary). I won't be able to eliminate 3 unknowns...
Well here's what I know about the topic. The spatial portion of the time independent schrodinger equation in 1D satisfies the differential equation
\frac{-h(bar)}{2m}\frac{d^{2}\psi}{dx^{2}} + U(x)\psi = E\psi.
When U(x) is constant then the general solution for this equation is
\psi...
Well I guess I could add a De^{ik_{2}x} term to \psi_{2} but wouldn't that be representative of a particle going over the barrier and then heading back towards the right (I'm thinking in terms of incident, transmitted, and reflected)?
My prof wants us to eliminate all but one constant term...
Umm well before the step down the particle is under the influence of a nonzero potential. I thought the general form for this type of situation (in 1D) was the one I had listed for the first wavefunction. After the step the potential is zero so I just used the equation of a free particle...
Homework Statement
Write out the spatial portion of the wavefunction for a particle coming from the right with energy (E) greater than the step potential (U) and passing over the step.Homework Equations
The Attempt at a Solution
I just want to know if this is the right set of equations for this...
Thanks for the clarification on the superscript problem. I honestly have no idea what force would be acting on the bullet in the horizontal direction, its been a long time since I've done any classical mechanics and I always hated it (still do).
As to the differentiation part, If I have z(\rho)...
Homework Statement
A gun can fire shells in any direction with the same speed v0. Ignoring air resistance and using cylindrical polar coordinates with the gun at the origin and z measure vertically up, show that the gun can hit any object inside the surface
z = \frac{v_{0}^{2}}{2g} -...
I know what you are thinking, here's another person who's too lazy to do the research himself about grades and grad school, but I honestly have tried to find information.
I am doing a joint honors in physics and mathematics with a minor in Computer Science with hopes of pursuing a doctorate in...
Oh yeah it should be x = 5cos(theta)t, although it didn't change my answer in the long run. I guess that first answer was just luck then. Thanks for all the help!
Thanks.
There is a second part to this problem that asks at what angle the boat should be launched in order to arrive directly across from the starting position. Here's what I tried.
V(t) = <5cos(theta), u(5t) - 5sin(theta)>
Does it still take him 8 sec to reach the other side or will it be...
Ok here's my attempt:
0 <= t <= 8 where t is time
x is the horizontal distance across the river so it my be expressed as x = 5t
the vertical velocity is therefore u(5t) = 3sin((pi)t/8).
Integrating over dt I get a result of 48/pi, is that what you were getting at?
I'm afraid I don't see how that is true, if the river is flowing perpendicular to the boat then the boat will be moving along a hypotenuse, so if the river is flowing fast enough then couldn't it take longer than 40s for the boat to reach the other side? And without knowing the displacement I...