Yes, for a mass moving in 1D, the Schrodinger equation gives
Hψn(x) + Vψn(x) = Eψn(x)
H = Hamiltonian
So if I solve the equation for a 1D box I would get something like
ψn(x) = Asin(kx)+Bsin(kx)
Do I then solve for k,A, and B?
Homework Statement
Consider a particle which is confined in a one-dimensional box of size L, so that the position space wave function ψ(x) has to vanish at x = 0 and x = L. The energy operator is H = p2/2m + V (x), where the potential is V (x) = 0 for 0 < x < L, and V (x) = ∞ otherwise.
Find...
sorry, do you mean a circle centered at the origin? if that's what you mean then
x2 + y2 = R2where x(t) = Rcos(Ωt) and y(t) = Rsin(Ωt) in polar coordinates
then
##\vec {r} = x_0\vec {i} + R\cos (Ωt)\vec {i} + R\sin (Ωt)\vec {j} = x_0\vec{i} + x(t)\vec{i} + y(t)\vec{j}##
or is the last step...
Homework Statement
The position of a participle in a fixed inertial frame of reference is given by the vector
r = i(x0 + Rcos(Ωt)) +j(Rsin(Ωt))where x0, R and Ω are constants.
a) Show that the particle moves in a circle with constant speed
Homework Equations
F = mv2/r
The Attempt at a...
Homework Statement
A cannon that is capable of firing a shell at speed v0 is mounted on a vertical tower of height h that overlooks a level plain below.
Show that the elevation angle α at which the cannon must be set to achieve maximum range is given by the expression
csc2(α) = 2(1+gh/V02)...
Okay, so I multiply by m to get the potential energy, but I don't understand what you mean by adding a constant. So my first term in my expansion should just be 0 somehow?
Homework Statement
Show that the variation of gravity with height can be accounted for approximately by the following potential function
V = mgz(1+z/re)
in which re is the radius of the Earth. find the force given by the above potential function.
Homework Equations
V = GM/r
The Attempt at a...
Right... so in this case, I can say that λ = b' since it asks me to label the common eigenstates as |a',b'>
So for an eigenstate of A to be an eigenstate of B, they must have |a'> on the RHS of the equation
Edit: I was confused because I thought it told me to show that
A|a'> = a'|a'> = B|b'>...
Should I say that Since A|a'> = B|a'> then A and B must share a common eigenbasis? So the eigenstates of A must be the eigenstates of B?
edit: and since they only differ by a constant λ then they must be non-degenerate?