Positive Semidefinite Operator

linear-algebra quantum-mechanics

Definition

Positive Semidefinite Operator

Let $\mathcal{H}$ be a complex Hilbert space. A linear operator $A:\mathcal{H}\to \mathcal{H}$ is positive semidefinite if

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

In this case one writes

$$A\ge 0.$$

In quantum theory, positive semidefinite operators are the operators that can assign non-negative probabilities through expressions of the form $⟨\psi |A|\psi ⟩$.

Basic Consequences

If $A\ge 0$, then $A$ is Hermitian:

$$A=A^{†}.$$

Hence all eigenvalues of $A$ are real. In finite dimensions, positivity is equivalent to non-negative eigenvalues:

$$A\ge 0⟺\lambda \ge 0\forall \lambda \in \mathrm{spec}(A).$$

Spectral Form

In finite dimensions, a Hermitian operator $A$ has a spectral decomposition

$$A=\sum _i{\lambda }_i{P}_i,$$

where each $P_i$ is an orthogonal projection. Then

$$A\ge 0⟺{\lambda }_i\ge 0\forall i.$$

This gives a direct test: diagonalise the operator and check whether all spectral weights are non-negative.

Factorisation

An operator is positive semidefinite iff it can be written as

$$A=B^{†}B$$

for some linear operator $B$.

Indeed,

$$⟨x|B^{†}B|x⟩=⟨Bx|Bx⟩=\Vert B|x⟩{\Vert }^2\ge 0.$$

In particular, every positive semidefinite operator has a positive square root $A^{1/2}$ such that

$$A=A^{1/2}{A}^{1/2}.$$

Operator Order

For Hermitian operators $A$ and $B$, define

$$A\le B⟺B-A\ge 0.$$

This is the Loewner order. A measurement effect is an operator satisfying

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

The lower bound ensures non-negative probabilities. The upper bound ensures probabilities at most $1$.

Matrix Example

Let

$$A=\left(\begin{array}{cc}2& 0\\ 0& 3\end{array}\right).$$

For $x=(x_1,x_2{)}^T\in {\mathbb{C}}^2$,

$$⟨x|A|x⟩=2|x_1{|}^2+3|x_2{|}^2\ge 0.$$

Therefore $A\ge 0$.

By contrast,

$$B=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$$

is Hermitian but not positive semidefinite, since for $x=(0,1{)}^T$,

$$⟨x|B|x⟩=-1<0.$$