Sping Matrices and Commutation Relations

  • Thread starter Rahmuss
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Homework Statement


Check that the spin matrices (Equations 4.145 and 4.147) obey the fundamental commutation relations for angular momentum, Equation 4.134.


Homework Equations


Eq. 4.147a --> [tex]S_{x} = \frac{\hbar}{2}\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}[/tex]

Eq. 4.147b --> [tex]S_{y} = \frac{\hbar}{2}\begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix}[/tex]

Eq. 4.145 --> [tex]S_{z} = \frac{\hbar}{2}\begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}[/tex]

[tex]\left[ S_{x}, S_{y}\right] = i\hbar S_{z}[/tex]

[tex]\left[ S_{y}, S_{z}\right] = i\hbar S_{x}[/tex]

[tex]\left[ S_{z}, S_{x}\right] = i\hbar S_{y}[/tex]


The Attempt at a Solution



Well, from Eq. 4.147 I have [tex]S_{x} = \frac{\hbar}{2}\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}[/tex] and also [tex]S_{y} = \frac{\hbar}{2}\begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix}[/tex]

So [tex]\left[ S_{x}, S_{y} \right] = i\hbar S_{z} \Rightarrow[/tex]

[tex]\left[ \frac{\hbar}{2}\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}, \frac{\hbar}{2}\begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix} \right] \Rightarrow[/tex]

[tex]\frac{\hbar^{2}}{4}\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix} - \frac{\hbar^{2}}{4}\begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} \Rightarrow[/tex]

[tex]\frac{\hbar^{2}}{4}\begin{pmatrix}i & 0 \\ 0 & -i \end{pmatrix} - \frac{\hbar^{2}}{4}\begin{pmatrix}-i & 0 \\ 0 & i \end{pmatrix} = \frac{\hbar^{2}}{4}\begin{pmatrix}2i & 0 \\ 0 & -2i \end{pmatrix}[/tex]

It doesn't quite seem right to me because the answer shows:

[tex]\left[ S_{x}, S_{y}\right] = i\hbar S_{z}[/tex]

But I can't see how I'd get that from where I'm going.
 
Last edited:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Your answer IS i*hbar*S_z.
 
  • #3
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Err... hey, you're right. :D

This idiot thanks you.
 
Last edited:

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