Loewner Order

linear-algebra operator-theory quantum-mechanics

Definition

Loewner Order

Let $\mathcal{H}$ be a complex Hilbert space, and let

$$\mathsf{Herm}(\mathcal{H})=\{A:\mathcal{H}\to \mathcal{H}\mid A=A^{†}\}$$

be the set of Hermitian operators on $\mathcal{H}$. The Loewner order is the partial order $\le$ on $\mathsf{Herm}(\mathcal{H})$ defined by

$$A\le B⟺B-A\ge 0,$$

where $B-A\ge 0$ means that $B-A$ is a positive semidefinite operator:

$$⟨x|(B-A)|x⟩\ge 0\forall |x⟩\in \mathcal{H}.$$

Quadratic Form Characterisation

The definition can be read directly through expectation values:

$$A\le B⟺⟨x|A|x⟩\le ⟨x|B|x⟩\forall |x⟩\in \mathcal{H}.$$

Thus $B$ is larger than $A$ exactly when every state vector assigns at least as large an expectation value to $B$ as to $A$.

Order Properties

On $\mathsf{Herm}(\mathcal{H})$, the Loewner order is a partial order.

Partial order

For Hermitian operators $A,B,C\in \mathsf{Herm}(\mathcal{H})$:

1. Reflexivity: $A\le A$.

2. Antisymmetry: If $A\le B$ and $B\le A$, then $A=B$.

3. Transitivity: If $A\le B$ and $B\le C$, then $A\le C$.

$A-A=0\ge 0$.

Reflexivity follows from

For antisymmetry, $A\le B$ and $B\le A$ imply

$$⟨x|(B-A)|x⟩\ge 0\text{and}⟨x|(A-B)|x⟩\ge 0$$

for all $|x⟩$. Hence $⟨x|(B-A)|x⟩=0$ for all $|x⟩$, so $B-A=0$.

For transitivity, $A\le B$ and $B\le C$ imply

$$C-A=(C-B)+(B-A).$$

The sum of positive semidefinite operators is positive semidefinite, so $C-A\ge 0$.

Positive Cone

The positive semidefinite operators form the positive cone of the order:

$$A\ge 0⟺0\le A.$$

The interval between $0$ and the identity is

$$[0,I]=\{E\in \mathsf{Herm}(\mathcal{H})\mid 0\le E\le I\}.$$

This interval is exactly the set of measurement effects.

Relation to Eigenvalues

In finite dimensions,

$$A\le B⟺{\lambda }_i(B-A)\ge 0\forall i,$$

where ${\lambda }_i(B-A)$ are the eigenvalues of the Hermitian operator $B-A$.

Equivalently, diagonalise the difference $B-A$ and check whether all eigenvalues are non-negative.

Not an Entrywise Order

For matrices, $A\le B$ in the Loewner order does not mean

$$A_{ij}\le B_{ij}\forall i,j.$$

It means instead that the difference $B-A$ has non-negative quadratic form.

Matrix comparison

Let

$$A=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),B=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right).$$

Then

$$B-A=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right).$$

The eigenvalues of $B-A$ are $0$ and $2$, so $B-A\ge 0$. Hence

$$A\le B.$$

Quantum Measurement Use

For a measurement effect $E$,

$$0\le E\le I.$$

For every normalised quantum state $|\psi ⟩$, the Loewner bounds imply

$$0\le ⟨\psi |E|\psi ⟩\le ⟨\psi |I|\psi ⟩=1.$$

So the Loewner order is the formal reason why effects produce valid Born-rule probabilities.