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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. :)

joshphysics
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Jack
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1 Answers1

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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 $N_i = b_i^{\dagger}b_i$

David Bar Moshe
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