Math 202B: Lecture 9

Let V be a finite-dimensional Hilbert space and let \mathcal{A}=\mathrm{End}(V) be the algebra of linear operators on V. We have shown that the unique faithful tracial state on \mathcal{A} is

\tau(A)=\frac{1}{\dim V} \sum\limits_{x \in X} \langle x,Ax \rangle,

the average of the diagonal matrix elements of A \in \mathcal{A} with respect to any orthonormal basis X \subset V. Equivalently, the unique Frobenius scalar product on \mathcal{A}=\mathrm{End}(V) is

\langle A,B \rangle_F = \frac{1}{\dim V}\sum\limits_{x \in X} \langle Ax,Bs \rangle.

We also showed that \tau is not an algebra homomorphism, so that algebra homomorphisms \mathrm{End}(V) \to \mathbb{C} simply do not exist.

We can also go the other way and consider states which are not necessarily faithful or tracial. That is, we consider linear functionals \sigma \colon \mathcal{A} \to \mathbb{C} which are required to satisfy

\sigma(I) = 1 \quad\text{and}\quad \sigma(A^*A) \geq 0,

but nothing more. The corresponding sesquilinear form

\langle A,B \rangle_\sigma=\sigma(A^*B)

is Hermitian, satisfies \langle I,I \rangle_\sigma=\sigma(I)=1, and the left Frobenius identity holds,

\langle AB,C \rangle_\sigma=\sigma((AB)^*C)=\sigma(B^*A^*C)=\langle B,A^*C\rangle_\sigma.

In the case where \mathcal{A}=\mathcal{F}(X) is the function algebra of a finite set X, states on \mathcal{A} are in bijection with probability measures on X. Faithful states correspond to probability measures whose support is all of X, i.e. \mathbb{P}(x) >0 for all x \in X. We will now obtain an analogous classification of states on \mathcal{A}=\mathrm{End}(V). In this noncommutative setting, the role of the probability measure \mathbb{P} on X is played by a special kind of operator on P on V called a density operator.

Let \sigma be an arbitrary linear functional on \mathcal{A}=\mathrm{End}(V). Equip \mathcal{A} with the Hilbert-Schmidt scalar product. By the Riesz representation theorem, there is R \in \mathcal{A} such that

\sigma(A) = \langle R,A \rangle_{HS} = \mathrm{Tr}(R^*A), \quad A \in \mathcal{A}.

Equivalently, setting P=R^* we have

\sigma(A)=\mathrm{Tr}(PA), \quad A \in \mathcal{A}.

Proposition 9.1. We have \sigma(I)=1 if and only if \mathrm{Tr}(P)=1.

Proof: \sigma(I)=\mathrm{Tr}(P). \square

Proposition 9.2. We have \sigma(A^*A) \geq 0 for all A \in \mathcal{A} if and only if P=Q^*Q for some Q \in \mathcal{A}.

Proof: If P=Q^*Q, then

\sigma(A^*A)=\mathrm{Tr}(PA^*A)=\mathrm{Tr}(Q^*QA^*A)=\mathrm{Tr}(QA^*AQ^*)=\langle AQ,AQ\rangle_{HS} \geq 0.

Conversely, suppose \sigma(A^*A) \geq 0 for all A \in \mathcal{A}. Then, choosing an orthonormal basis X\subset V we have that

\mathrm{Tr}(PA^*A) =\sum\limits_{x \in X} \langle x,PA^*Ax \rangle \geq 0, \quad A \in \mathcal{A}.

Taking A=E_{yx} to be an elementary operator, we have

\mathrm{Tr}(PA^*A)=\mathrm{Tr}(PE_{xy}E_{yx}) = \sum\limits_{z \in X} \langle z,PE_{xx}z\rangle = \langle x,Px \rangle,

so that

\langle x,Px \rangle \geq 0 \quad\text{ for all } x \in X.

Thus, nonnegativity of \sigma means that all diagonal matrix elements of P, in any orthonormal basis X \subset V, are nonnegative. Writing P=H+iK with H,K \in \mathcal{A} selfadjoint, we have

\langle x,Px \rangle = \langle x,Hx \rangle + i\langle x,Kx \rangle.

Thus, the fact that x,Px \rangle is real for all x \in X is enough to force P to be selfadjoint.

What remains now is to show that a selfadjoint operator P which satisfies \langle v,Pv \rangle \geq 0 for all v \in V can be factored as P=Q^*Q for some Q \in \mathcal{A}. To do this we use the spectral theorem: since P is selfadjoint we have

P=\sum\limits_{\lambda \in \Lambda} \alpha_\lambda P_\lambda,

where |\Lambda| is a finite nonempty set of cardinality at most |X|=\dim V, and $P_\lambda,$ \lambda \in \Lambda are selfadjoint orthogonal idempotents: for all \lambda,\mu \in \Lambda we have

P_\lambda^*=P_\lambda, \quad P_\lambda P_\mu =\delta_{\lambda\mu}.

Indeed, this is the algebraic form of the spectral theorem for selfadjoint operators in \mathrm{End}(V). The geometric form of the theorem is that there exists an orthogonal decomposition

V=\bigoplus_{\lambda \in \Lambda} V_\lambda

of V into nonzero subspaces such that P restricted to V_\lambda is a scalar multiple \alpha_\lambda of the identity operator. The connection between the algebraic statement and the geometric one is that P_\lambda is the orthogonal projection of V onto V_\lambda.

Now if e_\lambda is a unit vector in the image V_\lambda of P_\lambda then for any \lambda \in \Lambda we have

e_\lambda,Pe_\lambda \rangle = \langle e_\lambda,\alpha_\lambda e_\lambda\rangle \geq 0,

so \alpha_\lambda is a nonzero nonnegative number, aka a positive number. Thus we can define

Q=\sum\limits_{\lambda \in \Lambda}\sqrt{\alpha_\lambda}P_\lambda

and we have P=Q^*Q. \square

What we have proved is that every state on the algebra \mathcal{A}=\mathrm{End}(V) has the form

\tau(A)=\mathrm{Tr}(PA), \quad A \in \mathcal{A}

where \mathrm{Tr}(P)=1 and P=Q^*Q for some Q \in \mathcal{A}. Any P \in \mathcal{A}=\mathrm{End}(V) with these two properties is called a density operator.

Thus, we have shown that states on the endomorphism algebra \mathrm{End}(V) of a finite-dimensional Hilbert space V are in bijection with density operators on V. This is reminiscent of our earlier theorem that states on the function algebra \mathcal{F}(X) of a finite set X are in bijection with probability measures on X. In fact, density operators are the noncommutative analogue of probability measures in a very precise way. In the proof above, we saw that the condition that there exists a factorization P=Q^*Q is equivalent to the condition that

P=\sum\limits_{\lambda \in \Lambda} \alpha_\lambda P_\lambda

where \Lambda is a nonempty set of cardinality bounded by \dim V and P_\lambda are pairwise orthogonal selfadjoint idempotents weighted by positive numbers \alpha_\lambda. This gives

\mathrm{Tr}(P) = \sum\limits_{\lambda \in \Lambda} \alpha_\lambda (\dim V_\lambda),

where V_\lambda is the eigenspace of P which P_\lambda projects onto. Together with the condition \mathrm{Tr}(P)=1, this means the density operator P induces a probability measure \mathbb{P} on \Lambda defined by

\mathbb{P}(\lambda) = \alpha_\lambda (\dim V_\lambda).

Problem 9.1. Let P \in \mathrm{End}(V) be a density operator. Show that the corresponding state \tau_P on \mathrm{End}(V) is a faithful state if and only if \mathrm{rank}(P)=\mathrm{dim}(V).

Published by Jonathan Novak

Mathematician at UC San Diego.

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