Thanks @Baluncore, so I thought about this as well and it makes sense for high frequency signals where the skin depth is much smaller than the radius of the wire.. but from what I understand, Rosa's derivation applies to DC currents, which is what makes it less intuitive to me
I know that the whole topic of inductance in a straight wire is complicated (and has led to some heated discussions in this forum :smile:). I followed Rosa's derivation and can see that it leads to an inverse relation of the inductance to the wire radius, and from what could understand, the...
I was teaching the basics of quantum states and was showing the students that an arbitrary state in a quantum two-level system could be written as ##|\psi\rangle = C_1 |+\rangle + C_2 |-\rangle = R_1 |+\rangle + R_2 e^{i \alpha} |-\rangle##, with {##C_i##} complex and {##R_i##} real.
Then...
I'm using the general Hubbard model (H = U \sum n_{i,\uparrow} n_{i,\downarrow} - t \sum (c^{\dagger}_{i,\sigma} c_{i+1,\sigma} + c^{\dagger}_{i+1,\sigma} c_{i,\sigma})) to solve for eigenstates of simple quantum dot configurations.
For the case of a double dot with two electrons in singlet...
I've calculated the eigenstates of the Hubbard Hamiltonian for two fermions.
The ground state is (U2 - (U2 + 16t2)1/2)/2
For U = infty, I get 0.
For U >> t, I should get the exchange energy J = -4t2/U
How do I get from the ground state equation to J?