Recent content by hokhani

Thanks, but as I read in some references, internal energy is equal to the sum of the average kinetic and potential energies of all molecules. So, at higher places the potential energy of particles increases and hence, the internal energy must increase. Otherwise I think this definition of the...

Suppose that we just put a cylinder of an ideal gas in a higher place. Does it's internal energy increase?

The notes you sent formalizes the time reversal in a nice approach in accord with Sakurai. The many body book of Bruus is a good reference which covers condense matter topics. However, in spite of several editions it contains a few such problems. Do you know any other alternative text at this...

I think if we accept the complex conjugation only acts on the numbers there is no ambiguity in this term, because at least the explicit form of the operator is clear. In contrary, the form of ##H^*## is representation-dependent and while the representation is not specified, there is no sense of...

Eq. (7.18) in the text, introduces the time reversal dependent Hamiltonian as: ##H\psi(r)=[-\frac{1}{2m}(\nabla_r+ieA)^2+V(r)]\psi(r)##. So, here ##H## is in the positon space as ##H=H_r=\langle r|\hat{H}| r \rangle.##

Right, since in that text the Hamiltonian were represented in position space where ##H_r## is diagonal, I implicitlly assumed diagonal q-representation for ##\hat{H}## as well as properties of ##\Theta##, ##\Theta(c|q\rangle)=c^*|q\rangle##.

Thanks, it was the key point here. If we work in the basis ##|q\rangle## where ##\Theta |q\rangle =|q\rangle##, then we have: ##\Theta H |\rangle=\sum_q H_q^* \langle q|\rangle^* |q\rangle## and ##H\Theta|\rangle=\sum_q H_q \langle q|\rangle^* |q\rangle## so, ##[H,\Theta]=0## corresponds to...

From the Eq. (7.18) in that text, it seems that by ##H## the authors mean the Hamiltonian in the position space. Also, as I remember, it was discussed previously in one of my last threads that ##H## is an operator and writing in the functional form ##H(variable)## doesn't make sense.

Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016...

Thanks: right, your method was very nice, but I would like to try this method which still I am stuck in that.

Everything is ok if we have ##(LK)^{\dagger}=L^{\dagger} K##. Is this equation correct?

The calculations give ##|L_z=0\rangle =\sqrt{\frac{3}{8}} (|L_x=2\rangle +|L_x=-2\rangle ) - \frac{1}{2} |L_x=0\rangle## and we expect the same expansion for y-component. Also, as you said, the probability of having one of the configurations, say ##|L_x=2, L_y=2, L_z=0\rangle##, is zero...

Thanks, but doing calculation considering ##A^{\dagger} = KL^{\dagger}## we have: ## |\alpha \rangle =\hat{A}^{\dagger} |\phi\rangle =K L^{\dagger}=K \begin{bmatrix} 0 & i \\ -i & 0 \end{bmatrix} \begin{bmatrix} \phi_1 \\ \phi_2 \end{bmatrix}=K\begin{bmatrix} i\phi_2 \\ -i\phi_1...

Let's calculate as follows: ##|\alpha \rangle =\hat{A}^{\dagger} |\phi\rangle =K^{\dagger} L^{\dagger}=K^{\dagger} \begin{bmatrix} 0 & i \\ -i & 0 \end{bmatrix} \begin{bmatrix} \phi_1 \\ \phi_2 \end{bmatrix}=K^{\dagger} \begin{bmatrix} i\phi_2 \\ -i\phi_1 \end{bmatrix} ## so, ##\langle...

Thanks, could you please explain more how did you get this equation? I considered ##|\alpha \rangle =\hat{A}^{\dagger} |\phi \rangle =K^{\dagger} L^{\dagger}=-i\phi_2^*|+\rangle +i \phi_1^* |-\rangle## (supposing naively ##K^{\dagger}=K##), then we have ##\langle \alpha | =\langle \phi|...