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Unitary operators

Suppose we have a basis set $ \{ \vert a_{1} > \, \vert a_{2}> ... \vert a_{n}> \}$ that is complete. These states might be eigenstates of some hermitian operator, say A. If we have another hermitian operator B, then the eigenstates of B could also serve as a complete basis set, lets call it $ \{ \vert b_{1} > \, \vert b_{2}> ... \vert b_{n}> \} $. We define the operator $ U = \sum_{k} \vert a_{k}><b_{k}\vert $, and its adjoint $ U^{\dag } = \sum_{k} \vert b_{k}><a_{k}\vert $. Note that

 \begin{displaymath}U U^{\dag } = \sum_{kk'} \vert a_{k}><a_{k'}\vert \delta_{kk'} =
\sum_{k} \vert a_{k}><a_{k}\vert= I,
\end{displaymath} (1.46)

and we can show, in the same manner, that $ U^{\dag } U = I$. In other words, the adjoint of operator U is its inverse U-1. Operators that have this property are called unitary operators.

Consider any Hilbert space operator $ X=\sum_{nm} \vert a_{n}><a_{m}\vert
{\underline X}_{nm} $ where $ {\underline X} $ is the matrix representation of X with respect to basis $ \{ \vert a_{1} > \, \vert a_{2}> ... \vert a_{n}> \}$. Alternatively, we can express X in terms of the $ \{ \vert b_{1} > \, \vert b_{2}> ... \vert b_{n}> \} $basis. $ X=\sum_{nm} \vert b_{n}><b_{m}\vert {\underline X'}_{nm} $ where $ {\underline X'} $ is the matrix representation of X in the |bn> basis, or

 \begin{displaymath}\sum_{nm} \vert a_{n}><a_{m}\vert {\underline X}_{nm}
=\sum_{nm} \vert b_{n}><b_{m}\vert {\underline X'}_{nm}.
\end{displaymath} (1.47)

We multiply this relation by the bra <al| on the left, and by ket |ap> on the right to get

 \begin{displaymath}\sum_{nm} \delta_{ln} \delta_{mp} {\underline X}_{nm} =
\sum_{nm} <a_{l}\vert b_{n}><b_{m}\vert a_{p}> {\underline X'}_{nm}
\end{displaymath} (1.48)

but, $ <a_{l}\vert b_{n}> = {\underline U}_{an} $ and $ <b_{m}\vert a_{p}>= <a_{p}\vert b_{m}>^{*}
= {\underline U}^{*}_{pm} = {\underline U}^{\dag }_{mp} $or

 \begin{displaymath}{\underline X}_{lp}=\sum_{nm} {\underline U}_{an} {\underline X'}_{nm}
{\underline U}^{\dag }_{mp}.
\end{displaymath} (1.49)

In matrix notation this equation is simply written

 \begin{displaymath}{\underline X} = {\underline U}\, {\underline X'} \,{\underline U}^{\dag }
\end{displaymath} (1.50)

that is, operator X, whose matrix representation is given by $ {\underline X} $ in the |an> basis, is related to the matrix representation of X in the |bn> basis, $ {\underline X'} $, by the above relation. (Note: ${\underline U}$ is the matrix representation of operator X in the |an> representation).

Matrix equations that have this structure are called similarity transformations, and have important properties. On of such properties is that the eigenvalues of a matrix are invariant under a unitary similarity transformation. That means even though $ {\underline X} $, and $ {\underline X'} $ are different matrices, their eigenvalues are the same since they are related by a unitary similarity transformation. This implies that we can always find ${\underline U}$ for any operator representation $ {\underline X} $so that the transformed matrix is diagonal. It's eigenvalues are then distributed on the diagonal.


next up previous
Next: About this document ... Up: Bra-ket notation Previous: Matrix representation of operators
Bernard Zygelman
1999-09-21