Recent content by hokhani

In the ferromagnetic materials not only the atoms have magnetic dipoles but also the dipoles are aligned well in different domains. However, what is the differnce berween atomic structure of a soft ferromagnet like iron and a hard ferromagnet like a bar magnet? My first guess is that the atomic...

In the youtube lecture “electron interaction and the Hubbard model” at the time 2:23:00, we have the following self-consistent equation with energy appearing at both sides: $$(\hat P \hat H_0 \hat P+\hat P \hat H_1 \hat Q (E-\hat Q \hat H_0 \hat Q)^{-1} \hat Q \hat H_1 \hat P) |\phi...

Thank you so much for your attention. Could you please explain how did you obtain ##c^{\dagger}_{\uparrow} d_{\uparrow} |1100 \rangle =- |0110 \rangle##? In other words, my question is about the appearance of the minus sign. EDIT: Your answer is quite to the point and I am lookig for the reason...

Of course. In the following virtual lecture on youtube at the minute 52 (the blue text): The Hamiltonian is written in the subspace ##Q=2, S_z=0## and I am only interested in the difference between the arrays ##H_{1,3}## and ##H_{2,3}## so I neglected the other terms of the Hamiltonian in my...

In a system with two orbitals ##c## and ##d## (each with two spin degrees of freedom), consider the Hamiltonian ##H=V(d^{\dagger}_{\uparrow} c_{\uparrow} + c^{\dagger}_{\uparrow}d_{\uparrow}+d^{\dagger}_{\downarrow} c_{\downarrow} + c^{\dagger}_{\downarrow}d_{\downarrow})##. Also suppose that...

Thanks, I got it. Like the parity in one-particle system, provided that there is no degeneracy, the eigenstates must have even or odd symmetry as ##\frac{1}{\sqrt(2)}(|1,1,0,0>\pm|0,0,1,1>)##. Otherwise, there is no demand for symmetry of the eigenstates.

Sorry if I didn't convey myself. To explain in a better way; for a left-right symmetric system, can we have the eigenstates which doesn't have this symmetry?

I don't mean a specific Hamiltonian. I mean each left-right symmetric Hamiltonian for this system.

Thanks very much. In the second quantized form, the interchange of particles are included in the commutation of fermionic operators. However, I would like to know whether the left-right symmetry of this system demands the eigenstates of the Hamiltonian be in the form ##...

In the picture below we have two identical orbitals A and B and the system has left-right symmetry. I use the notation ##|n_{A \uparrow}, n_{A \downarrow},n_{B \uparrow},n_{B \downarrow}>## which for example ##n_{A \uparrow}## indicates the number of spin-up electrons in the orbital A. I would...

Thank you very much for pointing the parity inversion concept which seems that resolved my confusion about the symmetric system. As I underestood by your comment, the mirror symmetry means that ##H(x_1,x_2)=H(-x_1,-x_2)## which doesnt demand the condition ##x_1=-x_2##.

My reasoning is based on the mirror symmetry of the system. In the figure, I sent in my previous post, could you please explain the possibility of having instantly the two electrons in the non symmetric configuration while the system is symmetric?

Orbitals are nothing except for wave functions which are functions of ##x##. Therefore, I think my reasoning runs again in this regard.

I attached a non-symmetric configuration of electrons in the orbital in the mirror-symmetric system. Do you mean that we can have such an electronic position?

I pointed that we are talking about the specific case which two electrons with opposite spins are on two orbitals. In this case, because of symmetry we have $x_1=-x_2$ and $\alpha=\beta$ and we can infer that the spatial part is zero.