Recent content by lambdadandbda

sorry, I left a ##t_0## in the equation.

ah thanks! it was simple, factoring out as you said: $$\psi(x,t_0) =\frac{1}{\sqrt{2}}e^{\frac{-iE_1}{\hbar}t_0}(\psi_1(x) +e^{\frac{-i(E_2-E_1)}{\hbar}t_0}\psi_2(x))$$ then ##-1 = e^{-i\pi} ## gives ##t_0 = \frac{\hbar\pi}{E_2-E_1}##

thanks but I don't understand what is wrong, is ##e^{-i\pi}=-1## correct?

I updated the thread with a solution attempt, sorry, I'm still new to the forum.

I can write $$\psi(x,t_0) =\frac{1}{\sqrt{2}}(e^{\frac{-iE_1}{\hbar}t_0}\psi_1(x) +e^{\frac{-iE_2}{\hbar}t_0}\psi_2(x))$$ for the second coefficient to be -1 i need ## -1=e^{-i\pi}=e^{\frac{-iE_2}{\hbar}t_0} ## so ##t_0=\frac{\pi\hbar}{E_2}## and the above equation becomes $$\psi(x,t_0)...

Beautiful! Having it written like this it's easy to see that ##\alpha## is a periodic function of period ##\pi## with maximum at 0 and ##\pi## and symmetric about 0, ##\pi/2## and ##\pi/2## and hence with minimum at ##\pi/2##. Thank you very much, have a nice day!

thanks! I got confused using the tangent because of the asyntotes at ##\pi/2## so I would prefer a proof without it. Anyhow I cant find a clear way to state the result using symmetry, for sure we can see that ##\alpha## has maxima at ##\theta^*= 0,\pi## but I don't know how to formulate the...

I'm doing special relativity in undergrad and I have the following problem: Let a particle of mass M travelling at speed ##\beta = 1/2## (##\gamma = 2/\sqrt 3 \ \ c=1##) decay in to two photons: ##A \rightarrow 1+2## 1) Calculate energy and moment of the photons in the reference frame of the...