I'm really confused about a problem involving a single electron which eventually wants me to calculate Clebsch-Gordan coefficients. I think this is probably because, I've only ever seen examples done where there are two particles and are required to add their angular momentum.

I'm told I have an electron in a state of orbital angular momentum $l=2$.

I therefore know that its possible total angular momenta are $j=5/2$ or $j=3/2$.

I also know that there are six possible states for $m$ which are $5/2, 3/2, 1/2, -1/2, -3/2, -5/2$.

But I've been asked to construct the state vectors $\psi_{j,m}$ with total angular momentum $j=5/2$ and corresponding 3-components $m=5/2$ and $m=3/2$ as linear combinations of state vectors where I know the values of $S_{z}$ and $L_{z}$.

Firstly, "3-components" have never been mentioned on my course, so I'm a bit confused about what that means. And we just don't seem to have been given any information about how to do this for a single electron. Can anyone help get me started?


1 Answer 1


It is first important to note that it doesn't matter if you write a state in the $| l \; m_l\; s \; m_s \rangle$ basis or the $| l \;s\; j \; m \rangle$, you'll always have that $m = m_l + m_s$.

This means that

$$| l = 2, m_l = 2,s=1/2, m_s=1/2 \rangle = | l=2, s=1/2, j=5/2, m=5/2 \rangle$$ because these are the only kets in each of the bases that have $m = m_l + m_s = 5/2$.

From that you can use the lowering operators to construct the other states.

For example, if you want to write $| l=2, s=2, j=5/2, m=3/2 \rangle$ in the $| l \; m_l\; s \; m_s \rangle$ basis, you only need to apply the $j_-=L_-+S_-$ operator to both sides of the equation.

In the left side you apply $$ (L_-+S_- )| l \, m_l\, s \, m_s \rangle = L_-| l \, m_l\, s \, m_s \rangle+S_-| l \, m_l\, s \, m_s \rangle, $$ with $$ L_-| l \; m_l\; s \; m_s \rangle = \hbar\sqrt{l(l+1)-m_l(m_l-1)}| l \; m_l-1\; s \; m_s \rangle $$ and $$ S_-| l \; m_l\; s \; m_s \rangle = \hbar\sqrt{s(s+1)-m_s(m_s-1)}| l \; m_l\; s \; m_s-1 \rangle. $$.

On the right side $$ j_- | l \;s\; j \; m \rangle = \hbar\sqrt{j(j+1)-m(m-1)} | l \;s\; j \; m-1 \rangle $$

Put everything together and you'll find what you're after.

  • $\begingroup$ I think I'm really confused about what's going on here, any suggestions of where I could look to have another go at starting to build from the ground up? $\endgroup$
    – user37154
    Nov 3, 2014 at 21:27
  • 1
    $\begingroup$ I found section 7.1 in this notes quite useful for a full-disclosure treatment of the subject. I will expand my answer to try and clarify. $\endgroup$
    – Asaf
    Nov 3, 2014 at 21:46
  • $\begingroup$ @user37154 I've updated my answer. Let me know if something is unclear. $\endgroup$
    – Asaf
    Nov 3, 2014 at 22:32
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    $\begingroup$ How come you have $s=2$ in $|l=2,s=2,j=5/2,m=3/2⟩$, I thought $s=1/2$ always for a fermion? (Thanks for expanding, think I'm starting to get it) $\endgroup$
    – user37154
    Nov 3, 2014 at 22:33
  • $\begingroup$ It was a typo :P $\endgroup$
    – Asaf
    Nov 3, 2014 at 22:35

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