I'm trying to prove that unitary transformations preserve orthonormality for quantum chemistry course. This is how I tried but it seems that I can't manage.
Usual equations (definitions) regarding unitary transformations and their matrix elements:
$U^{\dagger}=U^{-1}$
$U^{\dagger}U=UU^{\dagger}=UU^{-1}=U^{-1}U=I$
$(U^{\dagger})_{ij}=U_{ji}^{*}$
$∑_{k}U_{ik}^{\dagger}U_{kl}=δ_{il}$
$∑_{k}U_{ki}^{*}U_{kl}=δ_{il}$
To prove that unitary transformation preserve orthonormality, I expressed two orthonormal vectors $|n⟩$ and $|m⟩$ ( of course $⟨m|n⟩=δ_{mn}$ ) in different basis where basis is changed by unitary transformation:
$|n⟩=∑_{k}|k⟩U_{nk}$
$|m⟩=∑_{l}|l⟩U_{ml}$
$⟨m|=∑_{l}⟨l|U_{lm}^*$
Taking inner product of these orthonormal vectors and then substituting their transformed form should yield Kronecker delta.
$⟨m|n⟩=∑_{l}∑_{k}U_{lm}^*⟨l|k⟩U_{nk}=∑_{l}∑_{k}U_{lm}^*δ_{lk}U_{nk}=∑_{k}U_{km}^*U_{nk}=∑_{k}U_{mk}^{\dagger}U_{nk}$
But, indices doesn't seem right comparing to definitons regarding matrix elements of unitary transformations.
Where did this go wrong? Should something here be done in a different manner?
Best Answer
A unitary matrix $U$ has orthonormal columns and orthonormal rows. The property of having orthonormal columns can be written as $U^\dagger U=I$ or, equivalently, as $$ (1) \quad \sum_{k} U^*_{ki} U_{kj} = \sum_{k} U_{ki} U^*_{kj} = \delta_{ij} $$ The property of having orthonormal rows is $UU^{\dagger} = I$ or, equivalently, $$ (2) \quad \sum_{k} U^*_{ik} U_{jk} = \sum_{k} U_{ik} U^{*}_{jk} = \delta_{ij} $$ In bra-ket notation, a (column) vector $v$ is represented w.r.t. canonical base $\{ \mathbf{e}_i\}$ as $$ | v \rangle = \sum_k v_k | \mathbf{e}_k \rangle = \sum_k v_k | k \rangle $$ where we identified $| \mathbf{e}_k \rangle = | k \rangle$. We also have that $| v \rangle^\dagger = \langle v |$.
Let $x,y$ be two distinct orthogonal vectors, so $\langle x | y \rangle = \langle y | x \rangle = 0$. More precisely, $$ |x\rangle = \sum_k x_k |k\rangle, \quad |y\rangle = \sum_h y_h |h\rangle, \quad \langle x| y \rangle = \sum_k x_k^* y_k $$ We now apply the basis change: $\mathcal{C} \rightarrow \mathcal{B}$ via unitary transform $U$. That is, each $\{ \mathbf{e}_i \}$ becomes $\mathbf{b}_i = U \mathbf{e}_i$, which means that $\mathbf{e}_i = U^{-1} \mathbf{b}_i = U^{\dagger} \mathbf{b}_i$. Writing the vectors in the new basis gives $$ |x\rangle = \sum_k x_k | \mathbf{e}_k \rangle = \sum_{k} x_k U^\dagger | \mathbf{b}_k \rangle, \quad |y\rangle = \sum_h y_h U^{\dagger} |\mathbf{b}_h\rangle $$ and $\langle x| = \sum_k x_k^* \langle \mathbf{b}_k | U$. We can now compute $$ \langle x | y \rangle = \sum_k \sum_h x^*_k y_h \langle \mathbf{b}_k| \underbrace{U U^\dagger}_{=I} | \mathbf{b}_h\rangle = \sum_k \sum_h x^*_k y_h \underbrace{\langle \mathbf{b}_k| \mathbf{b}_h\rangle}_{=\delta_{kh}} = \sum_k x_k^* y_k = 0 $$ which proves that orthogonality is preserved. Normality is preserved since the length of a vector is preserved ($U$ is an isometry). In fact, if $w = Ux$, then $$ \| w \| = \| U x \| = \sqrt{ \langle x | U^{\dagger} U | x \rangle } = \sqrt{ \langle x | x \rangle } = \| x \|. $$ and this is true also in the case $\| U^{\dagger} x \|$, since $UU^{\dagger} = I$.