Ex5 1

$\def\abs#1{|#1}\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\mathbf{Exercise\ 5.1}$

Show that any linear transformation $U$ that takes unit vectors to unit vectors preserves orthogonality: if subspaces $S_1$ and $S_2$ are orthogonal, then so are $U S_1$ and $U S_2$.

Does this make sense?

Laker4 22 Jan 2014 01:50

If $U$ takes unit vectors to unit vectors it preserves the inner product because $<x|x> = 1=<x|U^\dagger U|x>$
So then if $|x>$ and $|y>$ are orthogonal then $<x|y>$ = 0 and $<x|U^\dagger U|y> = 0$ as well.

Not sure if this is sufficient but best I came up with so far.

Orthogonality preserved by unit vector preserving transformations

Anders B Madsen 18 Dec 2021 20:17

If the linear transformation $U$ preserves unit vectors, we only know that it preserves the inner product if the inner product has value 1.

But if unit vectors are preserved, then by linearity the norm is also preserved:

(1)

\begin{equation} ||Ux|| = ||x|| \end{equation}

Furthermore, orthogonality will be preserved by the properties of the Hermitian inner product.
To prove this, let $\braket xy = 0$. We want to show that $\braket{Ux}{Uy} = 0$.

We compute

(2)

\begin{align} ||U(x + y)||^2 &= \braket{U(x + y)}{U(x + y)} \\ &= ||Ux||² + \braket{Ux}{Uy} + \braket{Uy}{Ux} + ||Uy||² \\ &= ||x||² + 2\operatorname{Re}(\braket{Ux}{Uy}) + ||y||²\,. \end{align}

But by the norm preserving property we also have

(3)

\begin{align} ||U(x + y)||^2 &= ||x+y||² \\ &= ||x||² + ||y||²\,, \end{align}

since $x,y$ are orthogonal (Pythagoras). This means that $\operatorname{Re}(\braket{Ux}{Uy}) = 0$.

Likewise,

(4)

\begin{align} ||U(x + iy)||^2 &= ||Ux||² + i\braket{Ux}{Uy} -i \braket{Uy}{Ux} + ||Uy||² \\ &= ||x||² + 2i\operatorname{Im}(\braket{Ux}{Uy}) + ||y||²\,. \end{align}

But since

(5)

\begin{align} ||U(x + iy)||^2 = ||x+iy||² = ||x||² + ||y||²\,, \end{align}

we must also have $\operatorname{Im}(\braket{Ux}{Uy}) = 0$.

This shows $\braket{Ux}{Uy} = 0$.