Unitary transformation

A unitary transformation of a set of vectors to another set of vectors preserves the lengths of the vectors and the angles between the vectors.

In other words, a unitary transformation is a rotation of axes in the Hilbert space. This implies that if the transformation involves matrices of eigenvectors, the eigenvalues of the eigenvectors are preserved. Consider 2 complete sets of orthonormal bases:

$basis\;1:\;\;\;\left\langle\phi_i\vert\phi_j\right\rangle=\delta_{ij}\;\;\;\sum_{i=1}^n\vert\phi_i\rangle\langle\phi_i\vert=I$

$basis\;2:\;\;\;\left\langle\phi_q^{'}\vert\phi_s^{'}\right\rangle=\delta_{qs}\;\;\;\sum_{q=1}^n\vert\phi_q^{'}\rangle\langle\phi_q^{'}\vert=I$

where I is the identity matrix.

The relation between the two basis sets are

$\vert\phi_q^{'}\rangle=I\vert\phi_q^{'}\rangle=\sum_{i=1}^n\vert\phi_i\rangle\langle\phi_i\vert\phi_q^{'}\rangle=\sum_{i=1}^n\vert\phi_i\rangle u_{iq}=\sum_{i=1}^n\vert\phi_i\rangle (U)_{iq}\;\;\;\;\;\;\;\;(68)$

where $u_{iq}=\langle\phi_i\vert\phi_q^{'}\rangle$ are elements of the transformation matrix U.

We say that $\vert\phi_i\rangle$ is transformed to $\vert\phi_q^{'}\rangle$ by $u_{iq}$ . Similarly, we have $\vert\phi_s^{'}\rangle=\sum_{j=1}^n\vert\phi_j\rangle u_{js}$ .

The reverse transformation of eq68 is:

$\vert\phi_i\rangle=I\vert\phi_i\rangle=\sum_{q=1}^n\vert\phi_q^{'}\rangle\langle\phi_q^{'}\vert\phi_i\rangle=\sum_{q=1}^n\vert\phi_q^{'}\rangle u_{iq}^*=\sum_{q=1}^n\vert\phi_q^{'}\rangle (U^\dagger)_{qi}\;\;\;\;\;\;\;\;(69)$

Question

Show that $U^{-1}=U^{\dagger}$ .

Answer

$I=\delta_{ij}=\langle\phi_i\vert\phi_j\rangle=\sum_{q=1}^n\langle\phi_i\vert\phi_q^{'}\rangle\langle\phi_q^{'}\vert\phi_j\rangle$ $=\sum_{q=1}^nu_{iq}u_{jq}^*=\sum_{q=1}^n(U)_{iq}(U^\dagger)_{qj}=(UU^\dagger)_{ij}$

$I=UU^\dagger$

$U^{-1}I=U^{-1}UU^\dagger$

$U^{-1}=U^\dagger\;\;\;\;\;\;\;\;(70)$

Similarly,

$I=\delta_{qs}=\langle\phi_q^{'}\vert\phi_s^{'}\rangle=\sum_{i=1}^n\langle\phi_q^{'}\vert\phi_i\rangle\langle\phi_i\vert\phi_s^{'}\rangle$ $=\sum_{i=1}^nu_{iq}^*u_{is}=\sum_{i=1}^n(U^\dagger)_{qi}(U)_{is}=(U^\dagger U)_{qs}\;\;\;\;\;\;\;\;(71)$