Thanks, Just checked, its available from the university library, I will pick it up and start ready. Will this be accessible to me? or will I need further QM exposure?
I have just finished my 2nd year of undergraduate study, which had the first pass of QM in it. I want to do a PhD in Quantum information/computing. What can I learn in my own time to get me closer to understanding QM information/computing? Currently I only know what we covered in lectures, and...
Wow, I was being an absolute idiot, and inputting A back in incorrectly, for some reason i was just putting the value of A back into
<E_0|\phi> and <E_1|\phi> expecting to get one.
Thankyou lol
1. Homework Statement
a particle is in a linear superposition of two states with energy E_0 \ and\ E_1
|\phi> = A|E_0> + \frac{A}{(3-\epsilon)^{1/2}}|E_1>
where:
A \ > \ 0, \ 0\ <\ \epsilon \ <\ 3
What is the value of A expressed as a function of epsilon
2. Homework Equations
P(E_0) \...
objects are certain colours due to the fact that they REFLECT that colour of light. such that there phonons are opaque to that frequency.
a photon is absorbed by an atom causing it to be excited and an electron moves to a higher energy state. photons will only be absorbed by atoms if they have...
yes W is always the hypotenuse, so if u want to use the angle alpha then yeah Wx would be W cos alpha. this is exactly the same as W sin theta.
always resolve the triangle like this. when its in the form shown in the book its harder to visualise as it makes it seem that Wx is the hypotenuse...
W is always the hypotenuse due to it being the greatest value.
imagine completing the vector triangle between W and Wx. the value at W would be theta from the original triangle
you are trying to use constant velocity equation v=s/t instead of a constant acceleration equation
here are the variables s=displacement, t=time, u=initial velocity v=final velocity a=acceleration
here are the constant acceleration equations
s=ut+0.5at^2
v=u+at
s=(v+u/2)*t
v^2=u^2+2as...
Maui,
Let me show you with calculation as you are completely refusing to even think about what everyone is saying.
the radius of the moon = 1737km
if we assume the moon is a flat disc the diameter is 1737x10^3*2 = 3.474x10^6
divide the number by a millisecond (1x10^-3)
we get 3.474x10^9...
I believe it is due to the degeneracy pressure not allowing electrons movement. I just had a thought would the white dwarf act as a solid as such where it has energy bands instead of energy levels? is this why it glows white?
there are no particles, light is an electromagnetic wave, a wave consisting of an electric field and a magnetic field sloshing back and forth so to speak. if you are talking about "particles" being photons, photons have particle type characteristics and are discrete packets of energy not...