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horor question :v

ah its jut an intro question

Cool, looks like we just plug and chug into \[|\alpha|^2 + |\beta|^2 = 1\] for that first one.

i think using alpha=x+iy = r*e^itan^-1(y/x)

|dw:1444631085836:dw|

Well here we're given that \[\alpha = e^{i \gamma} \cos \theta\] and beta as well

right okay so i just have to show it satisfies that eqn

it has to because that complex e^itheta is 1 rotated around so modulus of that =1

\[|\alpha|^2 = \alpha^* \alpha = e^{i \gamma - i \gamma} \cos^2 \theta = \cos^2 \theta\]

|dw:1444631358098:dw|

oo ok

i didnt think of just using conjugate like that

is this a property though?
|A B| = |A| |B| ?

Well we could find out I guess

|AB|^2 = (AB)* BA = A* B* B A = |A| |B|

\[\sqrt{a^*ab^*b} = \sqrt{a^*a}\sqrt{b^*b}\] checks out

k cool

next qn

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what does it mean by same "physical" state

So yeah I think that's what you're saying

hmmm

so like grapinh PSI has no meaning?

graphing*

ok i see i guess multiplying by modulus 1 is just like multiplying some equation by 2 on both sides

y=x
2y=2x same thing

what does that mean

so since both states are rotated same equally

its just the same sphere with a diff coordinate system

ah i see since probability didnt change it didnt change anything with the complex phase?

The 'idea' of where a particle could have traveled is what interferes with the particle's path.

oh right i do remember this its kinda weird

we even calculate the wave collision pattern

since we assume the particles is all places with some prob

|dw:1444632321197:dw|

or maybe a phase just means the same wave pattern just happened somewhere else on the wall now

|dw:1444632415530:dw|