Understanding the Difference Between |z>, |+z>, and |-z> in Quantum Mechanics

  • Thread starter Thread starter Agnostic
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around the quantum mechanics concepts of state vectors, specifically the relationships and definitions of the states |z>, |+z>, and |-z>. Participants are exploring the implications of these definitions and their mathematical representations.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants question whether |z> can be expressed as the sum of |+z> and |-z>, with some asserting that this is incorrect due to linear independence. Others discuss the definitions of state vectors and their implications in calculations, particularly in relation to inner products and normalization.

Discussion Status

The discussion is active, with participants providing insights into the definitions and properties of quantum states. Some have offered guidance on the need for normalization and the interpretation of probabilities versus amplitudes in quantum mechanics. Multiple interpretations of the relationships between the states are being explored.

Contextual Notes

Participants are working within the context of an introductory quantum mechanics course, which may impose certain constraints on their understanding and the depth of discussion. There are references to specific textbooks for further reading, indicating a reliance on established sources for clarification.

Agnostic
Messages
110
Reaction score
0
does |z> = |+z> + |-z> ?
 
Physics news on Phys.org
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)
 
  • #10
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".
 
  • #11
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>
 
  • #12
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.
 
  • #13
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.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
Replies
15
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K