Quantum State Vectors - Dirac Notation

Hey all, I'm slightly lost on the meaning and methods of the dirac notation for quantum state vectors. Specifically, pic related #1.2a.

It's telling me to find a normalized vector |φ> that is orthogonal to each |ψ>.

I'm not quite sure what I'm supposed to be solving for here. I know that: = = 1 AND = = 0.

So... so I pick some |φ> = a|+> + b|-> and say: = 0 = (because they are orthogonal?) and solve?


I know that this board isn't for homework problems to be solved, and that's not what I'm looking for explicitly. I'm wondering what resources you guys would recommend to get a handle on this dirac notation and the concepts behind these things.

>mfw physicist can't into elementary linear algebra

>mfw physicists still teach QM with bra-cuck notation.

It's so stupid. It's throwing me off. Matrix notation is so much better.

You have to set two equations to optain your a and b constants. First equation is from orthogonality and second comes from normalisation to 1; so you have
|xi>=a|+>+|b>
=0
=1
From here it is straight forward to optain a and b.

You can learn about this from j.j.sakurai qm, it starts with dirac notation and it explains it good.

Yeah that's kind of what I figured.

But How would, given equation 1 in the OP pic, = 1 help us considering we know the coefficients in front of |+> and |-> in |ψ> already?

You need to do normalisation for new orthogonal vector, vectors that are given in op are already normalised.
Maybe i didnt understand your question, if that is the case could you rephrase it?

Perhaps I read your response wrong. It makes sense now.

We are taking the new vector we found, φ, and doing:

=1
=0

Where ψ is given in OP pic in the problem.


Right?

what?
for example, the answer for the first one is
root(2/3)I+> + i*root(1/3)I->

you'll find that it is normalized and orthogonal to the first vector

Thats correct

I think I'm still too dumb to get it...

yeah you have a relationship between alpha and beta, and you realize that you have to normalize them, so do it.. lol

dummy way to do it is set alpha to 1, figure out what beta must be to hold that boxed relationship you have, now you have your vector, now surely you can normalize it, yeah?

I'm just laughing because I went through the quantum gauntlet last year and now I'm DONE with that shit, have fun getting assraped by griffiths, sorry m8

>yeah you have a relationship between alpha and beta, and you realize that you have to normalize them, so do it.. lol
I'm not sure how to do this. How do you normalize some X* + Y* = 0?

>now you have your vector, now surely you can normalize it, yeah
I could do this if I was able to get here.


*sigh*

>I'm not sure how to do this. How do you normalize some X* + Y* = 0?
sigh

you have two boxed relationships, from which you can find your wanted vector

one easy way to do this is to pick one (an easy one) of the infinite solutions for alpha and beta, then to normalize it

for example, if alpha is one, then... beta equals... i over root(2)....
now you have a vector of the form alphaI+> + betaI-> and the only problem is that it isn't normalized...

if you can't do this you'll either be doing 14 hour physics days for the year and succeeding, or failing, up to you bro but be prepared

>if you can't do this you'll either be doing 14 hour physics days for the year and succeeding, or failing, up to you bro but be prepared
Typically it's the 14 hour physics days because I'm a brainlet, as you can see.

I will try what you suggested. I guess the term that is giving me issues is "normalize".

lmao is linear algebra not a required course for physicists or something? this shit is trivial

My professor for linear algebra wasn't good at all, and more importantly I'm a huge brainlet. Sucks, but whatever.

>normalize

it just means the length of the vector is 1 bro
or in this context, as a verb, it means "to make the length of the vector 1" generally by finding the magnitude of the vector and dividing each component by that

brainlets gotta stick together, i won't even make fun of you for this

No I get what normalize means in terms of vectors, that's not the problem.

>one easy way to do this is to pick one (an easy one) of the infinite solutions for alpha and beta, then to normalize it
This is what's confusing. But I haven't sat down and tried doing this yet so I'm sure when I do it I will understand it. To me I this seems like I'm normalizing some equation, not a vector. Which doesn't make sense. I'm sure that isn't the case.

So let me go attempt it. Being a brainlet sucks m8.

What book is that?

>To me I this seems like I'm normalizing some equation, not a vector.

you use the equation to find values for alpha and beta then you make those your vector coefficients and normalize the vector, you don't normalize an equation

you can do it m8
and really they should rename QM "applied linear algebra" so um may want to touch up on that

I honk I did it! Also I get it now. Honestly what tripped me up was solving a and b and not realizing I should just "pick a value for one and solve", like middle school algebra system of equations shit. Doh!

Think* not honk. Good lord. Also apologies for the rotated pic, isn't like that on my phone.

you messed up a negative sign early on, you get 5/10 points

Where?

also, if you had checked your answer this would have become apparent and easily fixed, so I'm going to give you 2/10

wow i'm a brainlet yeah okay wow WHEW LAD you did it 10/10

To obtain a vector orthogonal to |phi>, note that |xi>=a|+> + b|->. Then, =a* (1/3)^2+b*i (2/3)^2=0 which implies that a*^2(1/3)=b*^2(2/3).

Observe that a*^2+b*^2=1 by completion. Use this to solve for a and b.

Don't know what user is talking about, there's no sign error.

Griffiths is so fucking easy, what is there to assrape you?

Giffiths is easy as shit. I would like to see what you make of Sakurai when - if - you get there.

where's his sign error?

it's a compact notation and it's easy to deal with when you get used to it. I don't understand why people bitch about it so much.