View Full Version : does |z> = |+z> + |-z> ?
Agnostic
Sep30-06, 12:11 PM
does |z> = |+z> + |-z> ?
Logarythmic
Sep30-06, 12:18 PM
No, |+z> + |-z> = |z> - |z> = 0.
Agnostic
Sep30-06, 01:33 PM
No, |+z> + |-z> = |z> - |z> = 0.
umm..., no.
Hargoth
Sep30-06, 02:37 PM
What is | z \rangle ?
Agnostic
Sep30-06, 03:03 PM
What is | z \rangle ?
In quantum mechanics, |z> is a state vector read at "ket z".
It describes the state a particle is in.
Hargoth
Sep30-06, 03:25 PM
Yeah, but if | z_+ \rangle, | z_- \rangle are basekets of the Hilbert space you consider, your equation would be a definition of | z \rangle
Agnostic
Sep30-06, 04:46 PM
Yeah, but if | z_+ \rangle, | z_- \rangle are basekets of the Hilbert space you consider, your equation would be a definition of | z \rangle
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.
jonestr
Sep30-06, 04:46 PM
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
Agnostic
Sep30-06, 05:00 PM
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)
Hargoth
Sep30-06, 08:01 PM
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
\langle z | z \rangle = 1 . If \langle z_+ | z_+ \rangle = 1 and \langle z_- | z_- \rangle = 1-, this not the case here.
I wouldn't say "amplitude" but "probability": |\langle -z | x \rangle|^2 is the probability to measure "z-spin-down" on a particle of which you know it is in state "x-spin-up".
Agnostic
Sep30-06, 08:12 PM
For a QM-Interpretation you have to normalize the statevector, so that
\langle z | z \rangle = 1 . If \langle z_+ | z_+ \rangle = 1 and \langle z_- | z_- \rangle = 1-, this not the case here.
I wouldn't say "amplitude" but "probability": |\langle -z | x \rangle|^2 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>
Hargoth
Sep30-06, 08:15 PM
Yeah, I just wanted to say that your probability of finding z in state z from the equation above would be 2^2=4 , so you have to normalize.
jonestr
Sep30-06, 08:52 PM
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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