State space stuff @kainui
http://prntscr.com/8qcmml

- dan815

State space stuff @kainui
http://prntscr.com/8qcmml

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

horor question :v

- dan815

ah its jut an intro question

- Kainui

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- dan815

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

- dan815

|dw:1444631085836:dw|

- Kainui

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

- dan815

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

- dan815

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

- Kainui

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

- dan815

|dw:1444631358098:dw|

- dan815

oo ok

- dan815

i didnt think of just using conjugate like that

- dan815

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

- Kainui

Well we could find out I guess

- dan815

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

- Kainui

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

- dan815

k cool

- dan815

next qn

- dan815

http://prntscr.com/8qcptc

- dan815

what does it mean by same "physical" state

- dan815

i can see that multiplying by anything of modulus 1 will not change that state vector property* of the
|w> = A|0> + B |1>
|A|^2 + |B|^2=1

- Kainui

it's the same physical state because a complex phase doesn't change it, \[|e^{i \theta} a|^2 = |a|^2\]

- Kainui

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

- dan815

hmmm

- Kainui

I know this to be true, in fact I know I can always choose my eigenstates to be real and I can prove it.

- dan815

so like grapinh PSI has no meaning?

- dan815

graphing*

- Kainui

Graphing anything has no meaning, however complex phase on a wave function has meaning and it's important to QM

- Kainui

It's just when you're looking at pure states that you don't have interference so it doesn't do anything haha

- dan815

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

- dan815

y=x
2y=2x same thing

- Kainui

Not quite, the phase is more of a relative thing that gives you the interference patterns from being in two different states simultaneously

- dan815

what does that mean

- dan815

so since both states are rotated same equally

- Kainui

The 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

its just the same sphere with a diff coordinate system

- dan815

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

- Kainui

No no I think this is something I can't explain you have to hear Alan Adams explain it, I'm just a mortal.

- dan815

okay i see this complex phase wrt to prob isnt really makign sense right now, ill come back to that then

- Kainui

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

- Kainui

and 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

oh right i do remember this its kinda weird

- dan815

we even calculate the wave collision pattern

- dan815

since we assume the particles is all places with some prob

- dan815

i 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

|dw:1444632321197:dw|

- dan815

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

- dan815

|dw:1444632415530:dw|