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I'm just going over a few past exams for tomorrow, and I've come across a question that I'm having quite a bit of difficulty with.

Let $\left|0\right\rangle$ denote the Fock vacuum state so that $b_j \left|0\right\rangle = 0$, for all $j$. For any positive integer $N$, show that the state $(b_1^{\dagger})^N \left|0\right\rangle$ is a maximal weight state of $gl(3)$ formed by $a_{jk} = b_j^{\dagger} b_k$

Conceptually, I'm just kinda unsure what I'm meant to be doing. Any help would be awesome. :)

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  • $\begingroup$ Are you certain that the problem specifies $gl(3)$? It seems odd to me that the number $3$ is singled-out here. $\endgroup$
    – joshphysics
    Commented Nov 12, 2013 at 6:54
  • $\begingroup$ It does indeed specify $gl(3)$ - it's pretty much all we've worked with in my course. $\endgroup$
    – Jack
    Commented Nov 12, 2013 at 7:00
  • $\begingroup$ Hmmm. Is the context similar to your question here: physics.stackexchange.com/questions/80291/…? In particular, are there precisely 3 distinct bosonic creation operators $b_1^\dagger,b_2^\dagger,b_3^\dagger$ in the operator algebra? $\endgroup$
    – joshphysics
    Commented Nov 12, 2013 at 7:13
  • $\begingroup$ I am fairly certain, yes. That was an assignment question of mine, but I was able to work that out. What we've done in lectures is that, like you've said, we have three distinct boson creation operators, and three distinct boson annihilation operators. Stop me if I'm going overboard here, but, in class, the vectors we've been talking about (in Dirac notation) are of the form $\left|n l m\right\rangle$, with $n = 0, 1, 2, ...$, $l = n, n-2, ...$ and $m = -l, -l + 1, ..., l - 1, l$. We define $n$ and $l$ to be greater than or equal to zero. $\endgroup$
    – Jack
    Commented Nov 12, 2013 at 7:33

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A vector in a (polynimial) $gl(3)$ representation is highest weight if it is annihilated by the raising root operators $a_jk = b_j^{\dagger}b_k$, $k>j$.

In our case, the relevant operators are $a_{12}$, $a_{23}$, and $a_{13}$. We do not need to check the third case, because $a_{13} = [ a_{12}, a_{23}]$ is given by the commutator of the two other operators. Now, the check is fairly easy:

$a_{12} b_1^N | 0 \rangle = (b_1^{\dagger})^{N+1} b_2| 0 \rangle = 0$

$a_{23 }b_1^N | 0 \rangle = (b_2^{\dagger})(b_1^{\dagger})^{N} b_3| 0 \rangle = 0$

Of course the vectors $b_1^N | 0 \rangle$ for different $N$ will belong to distinct representations . Also, please notice that these vectors do not generate all the $gl(3)$ irreducible representations, because the weights of these highest weight vectors will be $(n_1, n_2, n_3) = (N, 0, 0)$ where, $n_i$ is the eigenvalue of the $i$-th number operator $N_i = b_i^{\dagger}b_i$

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  • $\begingroup$ Sorry to be pedantic, I'm just kind of not seeing something. So, we have that $a_{12} = b_1^{\dagger} b_2$, right?? So, then, you've got $b_1^{\dagger} b_2 (b_1^{\dagger})^N | 0 \rangle$, and then you just use a commutator to shift the $b_2$ and $(b_1^{\dagger})^N$ ?? Or am I over-complicating matters?? $\endgroup$
    – Jack
    Commented Nov 12, 2013 at 10:05
  • $\begingroup$ Yes, we use commutators to shift $b_2$. In fact all the commutators are trivial and the operators commute because the raising operators do not contain $b_1$ which is the only operator not commuting with $b_1^{\dagger}$ $\endgroup$
    – David Bar Moshe
    Commented Nov 12, 2013 at 10:22

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