Unitary Spaces

Theorem 1.

Schwarz inequality in a unitary space \(\,V(C):\)

\[|\,\langle x,y\rangle\,|^2\ \ \leq\ \ \langle x,x\rangle\,\langle y,y\rangle\,,\qquad x,y\in V\,,\]

becomes equality if and only if \(\,\) the vectors \(\,x,y\ \) are linearly dependent:

\[|\,\langle x,y\rangle\,|^2\ \,=\ \,\langle x,x\rangle\,\langle y,y\rangle \qquad\Leftrightarrow\qquad x,y\ \ \text{linearly dependent}\,.\]

Proof.

\(\ \Rightarrow\,:\ \) Assume that vectors \(\,x,y\ \) are linearly independent.

Then \(\ y\neq\theta,\ \) and moreover (since every non-trivial linear combination of linearly independent vectors is a non-zero vector) for every \(\,\alpha\in C:\)

\[ \begin{align}\begin{aligned}x-\alpha\,y\ \,=\ \,1\cdot x\,-\,\alpha\cdot y\ \neq\ \theta\,,\\\langle\,x-\alpha\,y,\,x-\alpha\,y\,\rangle\ >\ 0\,.\end{aligned}\end{align} \]

If we take \(\ \,\alpha\ =\ \displaystyle\frac{\langle x,y\rangle^*}{\langle y,y\rangle}\ \) as in the proof of Schwarz inequality, we obtain

\[\langle x,x\rangle\,\langle y,y\rangle\ \ >\ \ |\,\langle x,y\rangle\,|^2\,.\]

That is, we get the implication

\[x,y\ \ \text{linearly independent} \qquad\Rightarrow\qquad |\,\langle x,y\rangle\,|^2\ \ \neq\ \ \langle x,x\rangle\,\langle y,y\rangle\,,\]

which by contraposition is equivalent to

\[|\,\langle x,y\rangle\,|^2\ \ =\ \ \langle x,x\rangle\,\langle y,y\rangle \qquad\Rightarrow\qquad x,y\ \ \text{linearly dependent}\,.\]

\(\ \Leftarrow\,:\ \) Assume that vectors \(\,x,y\ \) are linearly dependent.

Then \(\ \,y=\alpha\,x\ \) or \(\ \,x=\beta\,y\ \,\) for some \(\ \alpha,\beta\in C.\)

In the first case:

\[\begin{split}\begin{array}{l} |\,\langle x,y\rangle\,|^2\ \,=\ \,|\,\langle x,\,\alpha\,x\rangle\,|^2\ \,=\ \, |\,\alpha\,\langle x,x\rangle\,|^2\ \,=\ \,|\alpha|^2\ \langle x,x\rangle^2\,, \\ \langle x,x\rangle\,\langle y,y\rangle\ \,=\ \, \langle x,x\rangle\,\langle\alpha\,x,\,\alpha\,x\rangle\ \,=\ \, \langle x,x\rangle\ \alpha^*\alpha\,\langle x,x\rangle\ \,=\ \, |\alpha|^2\ \langle x,x\rangle^2\,, \end{array}\end{split}\]

an in the second one:

\[\begin{split}\begin{array}{l} |\,\langle x,y\rangle\,|^2\ \,=\ \, |\,\langle\beta\,y,\,y\rangle\,|^2\ \,=\ \, |\,\beta^*\,\langle y,y\rangle\,|^2\ \,=\ \,|\beta|^2\ \langle y,y\rangle^2\,, \\ \langle x,x\rangle\,\langle y,y\rangle\ \,=\ \, \langle\beta\,y,\,\beta\,y\rangle\,\langle y,y\rangle\ \,=\ \, \beta^*\beta\ \langle y,y\rangle\ \langle y,y\rangle\,=\ \, |\beta|^2\ \langle y,y\rangle^2\,. \end{array}\end{split}\]

Hence, in both cases \(\ \ |\,\langle x,y\rangle\,|^2\ \,=\ \,\langle x,x\rangle\,\langle y,y\rangle\,.\)

In this way we proved the remaining implication

\[ \begin{align}\begin{aligned}x,y\ \ \text{linearly dependent} \qquad\Rightarrow\qquad |\,\langle x,y\rangle\,|^2\ \ =\ \ \langle x,x\rangle\,\langle y,y\rangle\,.\\\;\end{aligned}\end{align} \]

Lemma 1. presents necessary and sufficient conditions for a linear operator to be the zero operator \(\,\mathcal{O},\) \(\\\) which is defined as \(\ \mathcal{O}\,x=\theta.\)

Lemat 1. \(\\\)

Let \(\ F\in\text{End}(V)\,\) be a linear operator defined on the space \(\ V=V(K)\,,\ K\in\{R,C\}\,.\ \) \(\\\)

  1. If \(\ F\ \) is defined on a unitary space \(\,V(C)\ \) \(\\\) or a Euclidean space \(\,V(R)\,,\ \) then

    \[F\ =\ \mathcal{O}\qquad\Leftrightarrow\qquad \langle\,x,Fy\,\rangle\,=\,0\quad\text{for all}\ \ x,y\in V\,.\]

  2. If \(\ F\ \) is a Hermitian operator: \(\ F^+=\,F\,,\ \) \(\\\) defined on a Euclidean space \(\,V(R),\ \) then

    \[F\ =\ \mathcal{O}\qquad\Leftrightarrow\qquad \langle\,x,Fx\,\rangle\,=\,0\quad\text{for all}\ \ x\in V\,.\]

  3. If \(\ F\ \) is defined on a unitary space \(\,V(C)\,,\ \) then

    \[F\ =\ \mathcal{O}\qquad\Leftrightarrow\qquad \langle\,x,Fx\,\rangle\,=\,0\quad\text{for all}\ \ x\in V\,.\]

Proof.

  1. \(\Rightarrow\ :\ \) If \(\ F=\mathcal{O}\,,\ \,\) then \(\ \,\langle\,x,Fy\,\rangle\,=\, \langle\,x,\mathcal{O}\,y\,\rangle\,=\, \langle\,x,\theta\,\rangle\,=\,0\,.\)

    \(\Leftarrow\ :\ \) Assume that \(\,\langle\,x,Fy\,\rangle\,=\,0\ \) for all \(\ x,y\in V\,.\)

    Then for \(\ x=Fy\ \) we have \(\ \langle\,Fy,Fy\,\rangle=0\,,\ \) and thus \(\ Fy=\theta\ \) for all \(\ y\in V.\ \\\) This means that \(\ F=\mathcal{O}\,.\)

  2. \(\Rightarrow\ :\ \) The same proof as in point 0.

    \(\Leftarrow\ :\ \) Assume that \(\ \langle x,Fx\rangle\,=\,0\ \) for all \(\ x\in V(R).\)

    In particular, if we put \(\ x+y\ \) in place of \(\ x\ ,\ \) we obtain \(\ \,\langle\,x+y,F(x+y)\,\rangle\,=\,0\,,\ \ x,y\in V\,,\ \ \) that is

    (1)\[ \begin{align}\begin{aligned}\langle\,x+y,F(x+y)\,\rangle\,=\,\langle\,x+y,Fx+Fy\,\rangle\,=\\=\ \langle\,x,Fx\,\rangle\,+\,\langle\,x,Fy\,\rangle\,+\, \langle\,y,Fx\,\rangle\,+\,\langle\,y,Fy\,\rangle\,=\\=\ \langle\,x,Fy\,\rangle\,+\,\langle\,y,Fx\,\rangle\,=\,0\,,\quad x,y\in V\,.\end{aligned}\end{align} \]

    Since \(\,F\ \) is a Hermitian operator on a real space,

    (2)\[\langle\,y,Fx\,\rangle\ =\ \langle\,Fy,x\,\rangle\ =\ \langle\,x,Fy\,\rangle\,.\]

    Substitution of the equality (2) into (1) gives \(\ \langle\,x,Fy\,\rangle=0\,,\ \ x,y\in V\,,\ \\\) which \(\,\) by the point 0. \(\,\) is equivalent to \(\ F=\mathcal{O}.\\\)

  3. \(\Rightarrow\ :\ \) The same proof as in point 0.

    \(\Leftarrow\ :\ \) Assume that \(\ \langle x,Fx\rangle\,=\,0\ \) for all \(\ x\in V(C).\)

    We make two substitutions, similar to the one in point 1.: \(\ x\rightarrow x+y\ \,\) and \(\ \,x\rightarrow x+i\,y\,.\ \) Then

    \[\begin{split}\begin{array}{lcr} & \left\{\ \begin{array}{r} \langle\,x,Fy\,\rangle\,+\,\langle\,y,Fx\,\rangle\,=\,0 \\ \langle\,x,F(iy)\,\rangle\,+\,\langle\,iy,Fx\,\rangle\,=\,0 \end{array}\right. & \quad x,y\in V\,, \\ \\ \text{and thus} & \left\{\ \begin{array}{r} \langle\,x,Fy\,\rangle\,+\,\langle\,y,Fx\,\rangle\,=\,0 \\ \langle\,x,Fy\,\rangle\,-\,\langle\,y,Fx\,\rangle\,=\,0 \end{array}\right. & \quad x,y\in V\,. \end{array}\end{split}\]

    Adding the two last equalities we obtain \(\ \langle\,x,Fy\,\rangle=0\,,\ \ x,y\in V\,,\ \) which means that \(\,F=\mathcal{O}.\,\) Note that in a complex space \(V\,\) the assumption on the operator \(\ F\ \) to be Hermitian is not necessary. \(\\\)

Corollary. \(\\\)

If one of the two following conditions holds: \(\\\)

  1. \(\ F\ \) and \(\ G\ \) are Hermitian linear operators: \(\ F^+=\,F\,,\ \ G^+=\,G\,,\) \(\\\) defined on a Euclidean space \(\,V(R)\,,\) \(\\\)

  2. \(\ F\ \) and \(\ G\ \) are linear operators defined on a unitary space \(\,V(C)\,,\) \(\\\)

then \(\qquad\quad F\ =\ G\quad\Leftrightarrow\quad \langle\,x,Fx\,\rangle\,=\,\langle\,x,G\,x\,\rangle \quad\text{for all}\ \ x\in V\,.\)

Indeed, if the equality \(\ \ \langle\,x,Fx\,\rangle=\langle\,x,G\,x\,\rangle \ \) holds for all \(\ x\in V\ \ \) then \(\\ \\\) \(\ \ \langle\,x,(F-G)\,x\,\rangle\,=\,0\,,\ \ x\in V\,,\ \) where in case 1.: \(\ \ (F-G)^+=F^+-G^+=F-G\,.\ \\ \\\) Hence \(\ \ F-G=\mathcal{O}\,,\ \ \) and thus \(\ \ F=G.\)

Now we state and prove an important criterion for a linear operator to be Hermitian:

Theorem 2.

If \(\,F\ \) is a linear operator defined on a unitary space \(\,V(C)\,,\ \,\) then

\[F=F^+\qquad\Leftrightarrow\qquad \langle\,x,Fx\,\rangle\in R\quad\text{for all}\ \ x\in V\,.\]

Proof. \(\,\) Because \(\ \ \langle\,x,F^+x\,\rangle\ =\ \langle\,Fx,x\,\rangle\ =\ \langle\,x,Fx\,\rangle^*\,,\ \ x\in V\,,\ \\\) Lemma 1. implies equivalence of the following conditions:

\[ \begin{align}\begin{aligned}F\ =\ F^+\\\langle\,x,Fx\,\rangle\ =\ \langle\,x,F^+x\,\rangle\,,\ \ x\in V\,,\\\langle\,x,Fx\,\rangle\ =\ \langle\,x,Fx\,\rangle^*\,,\\\langle\,x,Fx\,\rangle\in R\,,\ \ x\in V\,.\end{aligned}\end{align} \]

In quantum mechanics, the states of quantum system are represented by vectors from certain unitary space \(\,V(C)\ \) of states, and measurable physical quantities of the system correspond to linear operators defined on this space. It is assumed that if \(\,\|x\|=1\,,\ \) then the expression \(\,\langle\,x,Fx\,\rangle\ \) represents the mean value of the quantity \(\,F\ \) in the state \(\,x.\ \) This postulate makes sense if and only if the expression takes real values for all \(\,x\in V.\ \) Such a condition is fulfilled only by Hermitian operators. Hence, only these operators can represent physical quantities. \(\\\)

Theorem 3.

Let \(\,U\,\) be a linear operator defined on a Euclidean or unitary space \(\,V(K),\ K\in\{R,C\}.\ \,\) Then the following conditions are equivalent:

  1. \(\ U^+U=I\,,\ \) where \(\,I\ \) is the identity operator: \(\ \,Ix=x,\ x\in V\,;\)

  2. \(\ \langle\,Ux,Uy\,\rangle\,=\,\langle x,y\rangle\quad \text{for all}\ \,x,y\in V\,;\)

  3. \(\ \|\,Ux\,\|\,=\,\|x\|\quad\text{for every}\ \,x\in V\,.\)

Proof.

Note that

\[\begin{array}{l} U^+U=I\quad\Rightarrow\quad\langle\,Ux,Uy\,\rangle\,=\, \langle\,U^+U\,x,y\,\rangle\,=\,\langle\,Ix,y\,\rangle\,=\, \langle x,y\rangle\,,\quad x,y\in V \end{array}\]

and

\[\begin{split}\begin{array}{lcl} \langle\,Ux,Uy\,\rangle\,=\,\langle x,y\rangle & \quad\Rightarrow & \quad \|\,Ux\,\|^{\,2}\,=\,\langle\,Ux,Ux\,\rangle\,=\,\langle x,x\rangle\,=\,\|x\|^2 \\ \\ & \quad\Rightarrow & \quad\|\,Ux\,\|\,=\,\|x\|\,,\quad x\in V\,. \end{array}\end{split}\]

This proves the implications \(\ \,\text{1.}\,\Rightarrow\,\text{2.}\ \,\) and \(\ \,\text{2.}\,\Rightarrow\,\text{3.}\ \,\) Now it remains to show that \(\ \,\text{3.}\,\Rightarrow\,\text{1.}\)

\begin{eqnarray*} \|\,Ux\,\| & = & \|x\| \\ \|\,Ux\,\|^{\,2} & = & \|x\|^2 \\ \langle\,Ux,Ux\,\rangle & = & \langle x,x\rangle \\ \langle\,x,\,U^+U\,x\,\rangle & = & \langle x,Ix\rangle \end{eqnarray*}

The operators \(\ U^+U\ \ \) and \(\ I\ \) are Hermitian: \(\ (U^+U)^+=U^+U,\ \ I^+=I.\ \) Hence, corollary to Lemma 1. implies that both in a Euclidean (real) and unitary (complex) space \(\,V\ \) the equality \(\,U^+U=I\ \) holds.


© Copyright 2018, Jan Aksamit, Joanna Marzec, and Marcin Kostur. Last updated on Jan 29, 2022. Created using Sphinx 4.0.1.