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dan815
 one year ago
Best ResponseYou've already chosen the best response.0ah its jut an intro question

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1Cool, looks like we just plug and chug into \[\alpha^2 + \beta^2 = 1\] for that first one.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0i think using alpha=x+iy = r*e^itan^1(y/x)

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1Well here we're given that \[\alpha = e^{i \gamma} \cos \theta\] and beta as well

dan815
 one year ago
Best ResponseYou've already chosen the best response.0right okay so i just have to show it satisfies that eqn

dan815
 one year ago
Best ResponseYou've already chosen the best response.0it has to because that complex e^itheta is 1 rotated around so modulus of that =1

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1\[\alpha^2 = \alpha^* \alpha = e^{i \gamma  i \gamma} \cos^2 \theta = \cos^2 \theta\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.0i didnt think of just using conjugate like that

dan815
 one year ago
Best ResponseYou've already chosen the best response.0is this a property though? A B = A B ?

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1Well we could find out I guess

dan815
 one year ago
Best ResponseYou've already chosen the best response.0AB^2 = (AB)* BA = A* B* B A = A B

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1\[\sqrt{a^*ab^*b} = \sqrt{a^*a}\sqrt{b^*b}\] checks out

dan815
 one year ago
Best ResponseYou've already chosen the best response.0what does it mean by same "physical" state

dan815
 one year ago
Best ResponseYou've already chosen the best response.0i can see that multiplying by anything of modulus 1 will not change that state vector property* of the w> = A0> + B 1> A^2 + B^2=1

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1it's the same physical state because a complex phase doesn't change it, \[e^{i \theta} a^2 = a^2\]

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1So yeah I think that's what you're saying

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1I know this to be true, in fact I know I can always choose my eigenstates to be real and I can prove it.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0so like grapinh PSI has no meaning?

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1Graphing anything has no meaning, however complex phase on a wave function has meaning and it's important to QM

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1It's just when you're looking at pure states that you don't have interference so it doesn't do anything haha

dan815
 one year ago
Best ResponseYou've already chosen the best response.0ok i see i guess multiplying by modulus 1 is just like multiplying some equation by 2 on both sides

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1Not quite, the phase is more of a relative thing that gives you the interference patterns from being in two different states simultaneously

dan815
 one year ago
Best ResponseYou've already chosen the best response.0so since both states are rotated same equally

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1The thing is, the particle has a probability of being in multiple states at once and those PROBABILITIES interfere with each other with the complex phase.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0its just the same sphere with a diff coordinate system

dan815
 one year ago
Best ResponseYou've already chosen the best response.0ah i see since probability didnt change it didnt change anything with the complex phase?

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1No no I think this is something I can't explain you have to hear Alan Adams explain it, I'm just a mortal.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0okay i see this complex phase wrt to prob isnt really makign sense right now, ill come back to that then

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1The 'idea' of where a particle could have traveled is what interferes with the particle's path.

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1and because of this, the complex phase plays a role during this interference, but right now we're not doing anything complicated so let's just ignore this until we come to it and let's move forward.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0oh right i do remember this its kinda weird

dan815
 one year ago
Best ResponseYou've already chosen the best response.0we even calculate the wave collision pattern

dan815
 one year ago
Best ResponseYou've already chosen the best response.0since we assume the particles is all places with some prob

dan815
 one year ago
Best ResponseYou've already chosen the best response.0i am kind of getting an idea, so i should be thinking of the wave patterns that can arise with some changes in this angle

dan815
 one year ago
Best ResponseYou've already chosen the best response.0or maybe a phase just means the same wave pattern just happened somewhere else on the wall now