Quantum Mechanics: Basis

  • Thread starter Robben
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  • #1
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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##?
 

Answers and Replies

  • #2
Simon Bridge
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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?
 
  • #3
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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?

Can you elaborate please? Will it just then be a reflection about the axis?
 
  • #4
Simon Bridge
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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.
 
  • #5
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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?
 
  • #6
Simon Bridge
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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?
 
  • #7
DrClaude
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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.

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
$$
\mathbf{O}_x = \mathbf{R} \mathbf{O}_z \mathbf{R}^T
$$
 
  • #8
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2
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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