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

$$U U^{\dag } = \sum_{kk'} \vert a_{k}><a_{k'}\vert \delta_{kk'} = \sum_{k} \vert a_{k}><a_{k}\vert= I,$$ (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 |b_n> basis, or

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

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

$$\sum_{nm} \delta_{ln} \delta_{mp} {\underline X}_{nm} = \sum_{nm} <a_{l}\vert b_{n}><b_{m}\vert a_{p}> {\underline X'}_{nm}$$ (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

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

In matrix notation this equation is simply written

$${\underline X} = {\underline U}\, {\underline X'} \,{\underline U}^{\dag }$$ (1.50)

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