Recent content by mathsisu97

$$ W_{n,n} = \int_0^{2 \pi} \frac{1}{\sqrt{2 \pi}} e^{-inx} V_0 \cos(x) \frac{1}{\sqrt{2 \pi}} e^{inx} dx $$ $$ = 0 $$ $$ W_{n, -n} = \int_0^{2 \pi} \frac{1}{\sqrt{2 \pi}} e^{-inx} V_0 \cos(x) \frac{1}{\sqrt{2 \pi}} e^{-inx} dx $$ $$ = \frac{a n ( \sin(4 \pi n) + i \cos( 4 \pi n) - i...

So we get $$ f(\vec{J}_z) | jm \rangle = \sum_{k=0}^{\infty} \frac{1}{k!} \vec{J}_z^k |jm \rangle $$ $$ = \sum_{k=0}^{\infty} \frac{1}{k!} m^k |jm \rangle $$ So the propagator is $$ = \sum_{k=0}^{\infty} \frac{1}{k!} m^k |jm \rangle = \exp(-i \omega t \vec{J}_z) $$ Doesn't quite seem right or...

A function of the ##J_z## is a rotation? Sorry for not getting the point! I am currently looking back on my notes on Total Angular momentum and not see what youre are getting at with the arbitrary function application. Thanks for all the help so far

Ahh yes sorry! I already had that ##\vec{J}^2 |jm \rangle = \hbar^2j(j+1) |jm \rangle ## and ##\vec{J}_z |jm \rangle = \hbar m|jm \rangle ## . I know that for the orbital angular momentum quantum numbers ##L## and ##L_z## we get the Spherical Harmonics from which we can derive the Legendre...

For part (a) I see what you are hinting at. I still do not understand part (c) or what you have said. Applying ## J_z ## to ##|jm>## gives the eigenstates. I am trying to draw parallel with the free particle propagator but not getting anywhere.

To show that when ##[J^2, H]=0 ## the propagator vanishes unless ##j_1 = j_2## , I did (##\hbar =1##) $$ K(j_1, m_1, j_2 m_2; t) = [jm, e^{-iHt}]= e^{iHt} (e^{iHt} jm e^{-iHt}) - e^{-iHt} jm $$ $$ = e^{iHt}[jm_H - jm] $$ So we have $$ \langle j_1 m_1 | [jm, e^{-iHt} ] | j_2 m_2 \rangle $$ $$ =...