Does |z> equal |+z> plus |-z> in quantum mechanics?

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does |z> = |+z> + |-z> ?
 
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No, |+z> + |-z> = |z> - |z> = 0.
 
Logarythmic said:
No, |+z> + |-z> = |z> - |z> = 0.

umm..., no.
 
What is [itex]| z \rangle[/itex]?
 
Last edited:
Hargoth said:
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]
 
Hargoth said:
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:

<-z|x>

Which i read as that is the amplitude of something in either the +x or -x state being in the -z state.
 
Agnostic said:
does |z> = |+z> + |-z> ?

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
 
jonestr said:
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)
 
Agnostic said:
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:

<-z|x>

Which i read as that is the amplitude of something in either the +x or -x state being in the -z state.
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".
 
Hargoth said:
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 don't 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 significance.