Consider the Lie algebra $\mathfrak{sl}(2,\mathbb{C})$ with its "standard basis" $h,e,f$ (see here). Let further $V_\lambda := \{v\in V\mid hv=\lambda v\}$, where $V$ is just an arbitrary $\mathfrak{sl}(2,\mathbb{C})$-module ($\lambda$ is usually called a weight and $V_\lambda$ a weight space).


Let $V$ be an irreducible $\mathfrak{sl}(2,\mathbb{C})$-module. Choose a maximal vector, say $v_0\in V_\lambda$; set $v_{-1}:=0,\, v_i := \frac{1}{i!}f^iv_0$ for $i\ge 0$, then: $$ \begin{array}{l} {h v_{m}=(\lambda-2 m) v_{m},} \\ {e v_{m}=m(\lambda+1-m) v_{m-1},} \\ {f v_{m}=v_{m+1}.} \end{array} $$ Proof. See Humphreys' book, Lemma 7.2.

This looks to me suspiciously similar to something that one encounters when studying the representation theory of $\mathfrak{su}(2)$ . One usually will at some point establish the result (in physics notation...): $$ \begin{array}{l} {M_{3}|\lambda, m\rangle= m|\lambda, m\rangle ,} \\ { M_{\pm}|\lambda, m\rangle=\sqrt{\lambda(\lambda+1)-m(m \pm 1)}|\lambda, m \pm 1\rangle ,} \end{array} $$ where $M_\pm := M_1\pm iM_2$ and the $M_i$ are from a irreducible representation of $\mathfrak{su}(2)$. When I saw this, my immediate thought was that since $\mathfrak{sl}(2,\mathbb{C}) \cong \mathfrak{su}(2)\otimes \mathbb{C}$ we probably have $v_m \propto |\lambda,m\rangle$, and the operators $h,e,f$ in some way related to $M_3$ and $M_\pm$.

I think the only thing that is acutally necessary to bring the Lemma from above into the second form below is a rescaling of the $v_m$ by some factor that depends on $\lambda$ and $m$. In some old lecture notes I found $$u_m := \kappa_m v_m,\quad \kappa_m/\kappa_{m-1} :=\frac{1}{\sqrt{m(\lambda-m)}} \Longrightarrow \left\{\begin{aligned} h u_{m} &=(\lambda-2 m) u_{m}, \\ e u_{m} &=\sqrt{m(\lambda+1-m)} u_{m-1}, \\ f u_{m-1} &=\sqrt{m(\lambda+1-m)} u_{m}. \end{aligned}\right.$$ This is closer to what I'm looking for, but still not quite there... Does anybody have an idea how exactly the rescaling has to look like or if this is even possible?


1 Answer 1


Yeah, it's possible and standard. This is the language mismatch that, innocent as it might be, gets physics students across our divide shake their head at math books and walk away mumbling.

The mismatch of notations is that what mathematicians call going up is what physicists call going down―they focus on the eigenvalues of h. I am interpolating from the standard physics notation to yours, setting $\hbar=1$, etc... At the su(2) Lie algebra level, $$ [M_3,M_\pm]=\pm M_\pm , \qquad [M_+,M_-]=2M_3, $$ so that $$ e=M_+, \qquad f=M_-, \qquad h=2 M_3, \qquad \mathbf {M}^2 =j(j+1) \implies 2j=\lambda . $$ You then have, as per the linked WP discussion, for $|j,m\rangle$ normalized to unit length, $$ \begin{array}{l} {M_{3}|j, m\rangle= m| j, m\rangle ,} \\ { M_{\pm}| j, m\rangle=\sqrt{ j( j+1)-m(m \pm 1)}| j, m \pm 1\rangle ,} \end{array} $$ so that your maximal vector annihilated by $e=M_+$ is $$ v_0 = | j, j\rangle ~~~ \implies \\ v_1 = \sqrt{2j} | j, j-1\rangle, \qquad v_2 = \sqrt{\frac{2j(2j-1)}{2}}| j, j-2\rangle, \\ v_k=\frac{f^k}{k!}| j, j\rangle = c_k |j,j-k\rangle ,\qquad v_{ 2j }= c_{2j} | j, - j\rangle, $$ the last vector being annihilated by the action of one more f.

You are then invited to evaluate the constants $c_k=\sqrt{(2j+1)/k-1} ~ c_{k-1}$. For starters, find that $c_{2j}=1$. Then, $\sqrt{\frac{2j(2j-1)...(2j+1-k)}{1\cdot 2\cdot 3 ... k}}$, so that $c_k=\sqrt{{{2j} \choose k}}$ .

Test these formulas for $j=3/2$, so that $v_0=|3/2,3/2\rangle, ..., v_3=|3/2,-3/2\rangle$, to appreciate the formal structure. The physics kets have unit normalization, but most of the $v_k$s don't.

  • $\begingroup$ Very nice, thank you very much! I will go through the calculations tomorrow (especially finding the relation between the $c_k$) and if everything adds up, I'll accept the answer. $\endgroup$
    – Sito
    Feb 14, 2020 at 20:47
  • $\begingroup$ I was able to reproduce everything up to $c_k=\sqrt{2j \choose k}$. My idea for this one was to observe that $$v_k = \frac{1}{k!}\sqrt{j(j+1)-j(j-1)} \sqrt{j(j+1)-j(j-2)}\dots \sqrt{j(j+1)-j(j-k)}|j,j-k\rangle,$$ which leads to $$c_k = \frac{1}{k!}\prod_{\ell=1}^k\sqrt{j(j+1)-j(j-\ell)} = \frac{\sqrt{j^k \Gamma(k+2)}}{k!}.$$ But this doesn't reduce to your result... Is there something wrong in my approach? $\endgroup$
    – Sito
    Feb 15, 2020 at 15:10
  • $\begingroup$ $c_1=\sqrt{2j}=c_{2j-1}$. Does it check? You may fiddle with j=3/2 explicitly. $\endgroup$ Feb 15, 2020 at 15:36
  • $\begingroup$ Since there is a $j^k$ term in my solution one can directly see that it doesn't work fro $k=2j$, so the limit is certainly wrong.. It checks out for $c_k$, but not for $c_{2j-k}$.. $\endgroup$
    – Sito
    Feb 15, 2020 at 15:39
  • $\begingroup$ Sorry, forgot to mark it. Took me a while, but I figured it out in the end. Thanks for the help! $\endgroup$
    – Sito
    Feb 18, 2020 at 16:45

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