does |z> = |+z> + |-z> ?
No, |+z> + |-z> = |z> - |z> = 0.
What is [itex] | z \rangle [/itex]?
In quantum mechanics, |z> is a state vector read at "ket z".
It describes the state a particle is in.
Yeah, but if [itex] | z_+ \rangle, | z_- \rangle [/itex] are basekets of the Hilbert space you consider, your equation would be a definition of [itex] | z \rangle[/itex]
is it a valid/correct definition?
I'm in an intro quantum class and I need to calculate:
so far, we have just been calculating things like: <+or-phi|+or-psi>
Now we are asked to calculate things like:
Which i read as that is the amplitude of something in either the +x or -x state being in the -z state.
No since |z>=(1,0) in the z basis and |-z>= (0,1) in the z basis you could a. never get a scalar under addition and you could not get an answer of the zero vector since these vectors are linearly independent and form a complete basis. For your previous post you need to calculate what |-z> is in the x basis or what |x> is in the z basis to compute the inner product. Griffiths QM or Liboff are good sources for this. As is Nielsen and Chuang
Hope that helps
|z> is not equal to |+z>
I thought |+z>=(1,0)
For a QM-Interpretation you have to normalize the statevector, so that
[itex] \langle z | z \rangle = 1 [/itex]. If [itex] \langle z_+ | z_+ \rangle = 1 [/itex] and [itex] \langle z_- | z_- \rangle = 1[/itex]-, this not the case here.
I wouldn't say "amplitude" but "probability": [itex] |\langle -z | x \rangle|^2 [/itex] is the probability to measure "z-spin-down" on a particle of which you know it is in state "x-spin-up".
<-z|-z> = 1
<-z|-z> means what is the "probability" amplitude that a particle in state |-z> will be in state |-z>
Yeah, I just wanted to say that your probability of finding z in state z from the equation above would be [itex] 2^2=4 [/itex], so you have to normalize.
you dont have to write the plus explicity. |z>=|+z>=(1,0) which is not equal to -|-z>=(0,-1). The negatives are part of the nomenclature and do not have algebraic signifigance.
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