Matrix representation in x and y basis for spin operators

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Homework Help Overview

The discussion revolves around finding the matrix representation of the spin operators ##\mathbb{S}_+## and ##\mathbb{S}_-## in the ##|\pm y\rangle## and ##|\pm x\rangle## bases, contrasting this with the more common ##z## basis representation.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the method of finding projections along different axes, particularly questioning how to transition from the ##z## basis to the ##x## and ##y## bases.
  • Some participants suggest using the known relationships between the bases and express curiosity about the implications of these transformations.
  • Questions arise regarding the definitions and meanings of the bases, particularly the concept of a "z basis."

Discussion Status

Participants are actively exploring different interpretations and methods for representing the spin operators in alternative bases. Some guidance has been offered regarding the mathematical expressions needed for these representations, but no consensus has been reached on the best approach.

Contextual Notes

There is a noted lack of clarity around the terminology used for the bases, particularly the concept of the "z basis," which some participants challenge. The discussion also reflects the constraints of typical homework assignments that often focus on the ##z## basis.

Robben
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Homework Statement



How can I find the matrix representation of ##\mathbb{S}_+## and ##\mathbb{S}_-## in the ##|\pm y\rangle## or ##|\pm x\rangle## basis?

Homework Equations



##
\mathbb{\hat{S}}_+|s,m\rangle = \sqrt{s(s+1)-m(m+1)}\hbar|s,m+1\rangle
##

The Attempt at a Solution



The book almost always ask to find something in the ##z-##basis but rarely asks to find something in ##x## or ##y## basis. I know how this matrix representation will look with the ##z-##basis, so I am wondering how the matrix representation will look in the ##x## or ##y## basis instead of ##z##?
 
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You find the projections along the unit vectors like normal.
If you already have the projection along the z axis, then what will the projections along the x and y axes look like?
 
Simon Bridge said:
You find the projections along the unit vectors like normal.
If you already have the projection along the z axis, then what will the projections along the x and y axes look like?

Can you elaborate please? Will it just then be a reflection about the axis?
 
Well the way I'm reading it, he's asking for a component.
"the z basis" does not really mean anything.

You can find, for eg, the z component of angular momentum right?
You know how the x and y components are given the z component.
If you don't, look it up.
 
Simon Bridge said:
Well the way I'm reading it, he's asking for a component.
"the z basis" does not really mean anything.

You can find, for eg, the z component of angular momentum right?
You know how the x and y components are given the z component.
If you don't, look it up.

So I will just use ##|+x\rangle = \frac{1}{\sqrt{2}}(|+z\rangle +|-z\rangle)## as my basis and solve it the way I solved for the ##z## basis?
 
Well, again, there is no such thing as a z basis that I know of.
I think you need to revisit the definitions - what is it that S+ and S- do?
 
Simon Bridge said:
Well, again, there is no such thing as a z basis that I know of.
It's a common expression when talking about spin. The z basis for a spin-1/2 particle is the set ##\left\{ |+z\rangle, |-z\rangle\right\}##, with "spin up" corresponding to a spin along +z, and "spin down" to a spin along -z.

Robben said:
How can I find the matrix representation of ##\mathbb{S}_+## and ##\mathbb{S}_-## in the ##|\pm y\rangle## or ##|\pm x\rangle## basis?
You can express an operator ##\hat{O}## in, e.g., the x basis, as a matrix using
$$
\mathbf{O}_x = \begin{pmatrix} \langle +x | \hat{O} | +x \rangle & \langle +x | \hat{O} | -x \rangle \\
\langle -x | \hat{O} | +x \rangle & \langle -x | \hat{O} | -x \rangle \end{pmatrix}
$$
For that, you need of course to find how to apply the operator in the x basis, i.e., ##\hat{O} | +x \rangle##. This you can find by expressing ## | +x \rangle## in terms of ##| +z \rangle## and ##| -z \rangle##.

Alternatively, you can calculate the rotation matrix ##\mathbf{R}## that brings you from the z basis to the x basis, and find the operator matrix using
$$
\mathbf{O}_x = \mathbf{R} \mathbf{O}_z \mathbf{R}^T
$$
 
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DrClaude said:
It's a common expression when talking about spin. The z basis for a spin-1/2 particle is the set ##\left\{ |+z\rangle, |-z\rangle\right\}##, with "spin up" corresponding to a spin along +z, and "spin down" to a spin along -z.You can express an operator ##\hat{O}## in, e.g., the x basis, as a matrix using
$$
\mathbf{O}_x = \begin{pmatrix} \langle +x | \hat{O} | +x \rangle & \langle +x | \hat{O} | -x \rangle \\
\langle -x | \hat{O} | +x \rangle & \langle -x | \hat{O} | -x \rangle \end{pmatrix}
$$
For that, you need of course to find how to apply the operator in the x basis, i.e., ##\hat{O} | +x \rangle##. This you can find by expressing ## | +x \rangle## in terms of ##| +z \rangle## and ##| -z \rangle##.

Alternatively, you can calculate the rotation matrix ##\mathbf{R}## that brings you from the z basis to the x basis, and find the operator matrix using
$$
\mathbf{O}_x = \mathbf{R} \mathbf{O}_z \mathbf{R}^T
$$

Very helpful! Thank you very much!
 

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