Um
1+1 =2
2+1 =3
3+1 =4
4+1 =5
5+1 =6=0
0+1 =1
So 1 generates Z6
0+5=5
5+5=10=4
4+5=9=3
3+5=8=2
2+5=7=1
1+5=6=0
So does 5!
Wait... aren't 5 and 6 relatively prime? Hmmmm...what to do about those other groups...?
Psi* d3(Psi) = d(Psi*d2(Psi))-d(Psi*)d2(Psi)
The first term will vanish since Psi at infinity vanishes. Play around with the second term. Sorry, I'm i a bit of a rush. I'll look at the problem later this evening.
To get a better feel for the equation, note that it is quadratic in both variables. By completing the square you will be able to rewrite this equation as one that describes an ellipse. The center of this ellipse is the top of the hill.
For part (a) you are on the right track. Squaring both sides we get the equation for a circle with center (-a,-b) in the plane. Take the parametrization of a circle at the origin and perform horizontal and vertical shifts to it to get the parametrization of this particular circle.
The last...
This is the brute force method, at will take some time.
Make a group table. Recall that for a group, each element in the group must appear only once in each row and in each column. If you see any element appearing more than once then you know the structure you have is not that of a...
Use the binomial approximation, and learn to love it:
(1+b)^-2 ~= 1 - 2b + 3b^2 for b<<1
Start with your first expression...forget combining fractions. Factor an x^2 out of all the denominators to get terms like the one I just wrote. Apply the approximation and collect like terms. You'll...
First you have to choose a sequence of functions to work with whose limit is the dirac delta. For instance the sequence
y(x,n) = n^2 x + n for -1/n < x < 0
-n^2 x + n for 0 < x < 1/n
0 otherwise
Then delta(x) = lim(y(x,n),n->infinity)
Use this sequence to prove...
Use the uncertainty principle to find out the spatial resolution of a particle with this momentum.
(hands wave around)
x ~= h/(4*pi*p)
Is this x larger than the diameter of the nucleus?
Well...
The alpha particle is nonrelativistic since KE/mc^2 = 10^-3, so you can use the classical formula for momentum p = sqrt(2mKE).
Use this momentum to find the deBroglie wavelength.
Here is a nice derivation of how an we get the exponentiated operator to generate translations:
Consider the differentiable function f at x. Translating f to the right means changing the argument to get f(x-a). Now take the displacement a and cut it up in N pieces, each piece having length...