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On the last page, Esquire asked whether a non-square matrix could represent a hermitian operator. A hermitian matrix must be square and, more-over, must be equal to its “transpose conjugate” (which is the same as a hermitian conjugate). The transpose conjugate is attained by transposing the matrix (flipping it along the diagonal) and taking the complex conjugate of each element. See pages 443-444. To get some insight into why this is so, try finding <math><\alpha |\hat{Q} \beta></math> and <math><\hat{Q} \alpha | \beta ></math> where <math>\displaystyle \hat{Q}= \left(\begin{array}{ccc} 1 & 0 & 1
1 & 0 & 1
1 & 0 & 1 \end{array} \right)</math>, <math>\displaystyle \vec{\alpha} = \left(\begin{array}{c} \alpha_1
\alpha_2
\alpha_3 \end{array} \right)</math>, and <math>\displaystyle \vec{\beta} = \left(\begin{array}{c} \beta_1
\beta_2
\beta_3 \end{array} \right)</math>.

I get <math>(\alpha_1 + \alpha_2 + \alpha_3)^*(\beta_1 + \beta_3)</math> and <math>(\alpha_1 + \alpha_3)^* (\beta_1 + \beta_2 + \beta_3)</math>. These are clearly not equal. If you don't find that experiment convincing, try using more general matrix elements (a la <math>q_{1,2}</math>); you should come to the conclusion that <math>q_{m,n}=(q_{n,m})^*</math> for all m and n for any hermitian matrix. If the matrix isn't square, there are m and n such that one of those terms exists and the other doesn't. In fact, as Spherical Chicken said on the last page, you'll find that you can't even calculate both of those quantities with a non-square matrix. Hopefully some of this was helpful (as opposed to redundant).

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