Recent content by hokhani

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|...

Right, Right, but how to realize which configuration is not preferred?

All the configurations of ##(L_x,L_y,0) ## are as follows; 4 configurations with ##(L_x,L_y)=(\pm2, \pm2)## and ##L^2=8 \hbar^2## 4 configurations with ##(\pm1, \pm1)## and ##L^2=2 \hbar^2## 4 configurations with ##(\pm1, \pm2)## and ##L^2=5 \hbar^2## 4 configurations with ##(\pm2, \pm1)## and...

An antilinear operator ##\hat{A}## can be considered as, ##\hat{A}=\hat{L}\hat{K}##, where ##\hat{L}## is a linear operator and ##\hat{K} c=c^*## (##c## is a complex number). In the Eq. (26) of the text https://bohr.physics.berkeley.edu/classes/221/notes/timerev.pdf the equality ##(\langle \phi...

Even if the system be in an equal superposition of all of these 25 configurations, we have ##\langle L^2 \rangle =4 \hbar^2## which again doesn't result in the expected value.

For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values...

I look for the error in the term ##\langle x |P| x \rangle=-i\hbar \frac{\partial}{\partial x} \delta(0)## which seems undefined. Thanks for regarding the preview check, but I don't know why my preview sometimes doesn't work!

Right, it is undefined. However, still another problem remains: The eigenfunctions of the particle in box, ##\psi_m(x)##, should form a complete basis and any function, ##\psi(x)##, seems to be expanded as ##\psi(x)=\sum_m a_m \psi_m(x)## while ##\psi_m(x)## is zero outside the box!

For infinite walls I did that in the post #8.

Very nice separation of the Hamiltonian. If we take ##\psi(x)=\sum_m a_m \psi_m## where ##\psi_m## are the eigenfunction of the Hamiltonian, then it seems that we have ##[V(x),P]\psi(x)=0## since inside the box (##|x|<L##) the potential is zero ##V(x)=0## and for ##|x| \geq L## we have ##\psi_m=0##!

I would like to know how to calculate the ##[\hat{H}, \hat{P}]## for a particle in a 1D box? At the first glance it seems that they commute but they don't get diagonalized in identical basis. How to calculate this commutation?