Trace Rule for Expectation Values

For a density operator $\rho$ and a bounded observable $A$, the expectation value is

$$\langle A\rangle_\rho = \operatorname{Tr}(\rho A).$$

For a projective outcome $a$ with projector $P_a$,

$$p(a) = \operatorname{Tr}(\rho P_a).$$

For a general measurement effect $E_i$,

$$p(i) = \operatorname{Tr}(\rho E_i).$$

These are not three unrelated rules. The probability formulas are trace rules applied to effects, and the expectation formula is the probability-weighted average of the outcomes of an observable. Together they extend the pure-state Born rule to pure states, randomized ensembles, open subsystems, and general measurements.

The derivations below use finite matrices. The final section explains what changes for unbounded observables such as position and momentum.

Helpful background. Density operators explains positive, trace-one states; expectation values explains the average of measurement outcomes.

Trace essentials

In a finite-dimensional Hilbert space, the trace of an operator $X$ is

$$\operatorname{Tr}X = \sum_n \langle n\rvert X\lvert n\rangle,$$

where $\{\lvert n\rangle\}$ is any orthonormal basis.

Basis independence

Let another orthonormal basis be related by

$$\lvert m'\rangle = \sum_n U_{nm}\lvert n\rangle$$

for a unitary matrix $U$. Then

$$\begin{aligned} \sum_m \langle m'\rvert X\lvert m'\rangle &= \sum_{m,n,k} U_{nm}^*U_{km} \langle n\rvert X\lvert k\rangle \\ &= \sum_{n,k} \delta_{nk} \langle n\rvert X\lvert k\rangle \\ &= \sum_n \langle n\rvert X\lvert n\rangle. \end{aligned}$$

The trace may be calculated in any convenient basis without changing the answer.

Cyclicity

For finite matrices,

$$\operatorname{Tr}(XY) = \operatorname{Tr}(YX),$$

and

$$\operatorname{Tr}(XYZ) = \operatorname{Tr}(ZXY) = \operatorname{Tr}(YZX).$$

Cyclicity permits cyclic rotations, not arbitrary reordering. In general,

$$\operatorname{Tr}(XYZ) \ne \operatorname{Tr}(XZY).$$

This distinction matters when observables and states do not commute.

Rank-one trace identity

For any vectors $\lvert u\rangle,\lvert v\rangle$ and operator $A$,

$$\operatorname{Tr} \left( \lvert u\rangle\langle v\rvert A \right) = \langle v\rvert A\lvert u\rangle.$$

To verify it, insert an orthonormal basis:

$$\begin{aligned} \operatorname{Tr} \left( \lvert u\rangle\langle v\rvert A \right) &= \sum_n \langle n\vert u\rangle \langle v\rvert A\lvert n\rangle \\ &= \langle v\rvert A \left( \sum_n \lvert n\rangle\langle n\rvert \right) \lvert u\rangle \\ &= \langle v\rvert A\lvert u\rangle. \end{aligned}$$

This identity is the cleanest bridge between trace notation and bra–ket notation.

Recovering the pure-state formula

A normalized pure state has

$$\rho_\psi = \lvert\psi\rangle\langle\psi\rvert.$$

The rank-one identity gives

$$\begin{aligned} \operatorname{Tr}(\rho_\psi A) &= \operatorname{Tr} \left( \lvert\psi\rangle\langle\psi\rvert A \right) \\ &= \langle\psi\rvert A\lvert\psi\rangle. \end{aligned}$$

The density-operator rule therefore reproduces the familiar pure-state expectation value exactly.

For a rank-one outcome

$$P_\phi = \lvert\phi\rangle\langle\phi\rvert,$$

the same identity gives

$$\begin{aligned} \operatorname{Tr}(\rho_\psi P_\phi) &= \lvert \langle\phi\vert\psi\rangle \rvert^2. \end{aligned}$$

This is the ordinary Born probability.

Component formula

In a chosen orthonormal basis,

$$\begin{aligned} \operatorname{Tr}(\rho A) &= \sum_i (\rho A)_{ii} \\ &= \sum_{i,j} \rho_{ij}A_{ji}. \end{aligned}$$

This formula is useful for direct matrix calculations. It also shows that off-diagonal entries of $\rho$ contribute whenever $A$ has matching off-diagonal matrix elements.

If the basis diagonalizes $A$,

$$A = \sum_a aP_a,$$

then

$$\langle A\rangle_\rho = \sum_a a\operatorname{Tr}(\rho P_a).$$

For a nondegenerate eigenbasis, this becomes

$$\langle A\rangle_\rho = \sum_a a\,\rho_{aa}.$$

In a nondegenerate measurement eigenbasis, each diagonal entry is one outcome probability. For a degenerate eigenvalue, add the entries belonging to its entire eigenspace.

Ensemble average

If

$$\rho = \sum_i p_i \lvert\psi_i\rangle\langle\psi_i\rvert,$$

then linearity gives

$$\begin{aligned} \operatorname{Tr}(\rho A) &= \sum_i p_i \operatorname{Tr} \left( \lvert\psi_i\rangle\langle\psi_i\rvert A \right) \\ &= \sum_i p_i \langle\psi_i\rvert A\lvert\psi_i\rangle. \end{aligned}$$

The trace rule combines quantum expectation within each preparation branch with the classical average over branches. Different ensembles with the same $\rho$ necessarily give the same expectation for every $A$.

Outcomes, moments, and variance

Let the spectral decomposition of a Hermitian observable be

$$A=\sum_a aP_a.$$

The outcome probabilities are

$$p(a) = \operatorname{Tr}(\rho P_a),$$

and therefore

$$\begin{aligned} \sum_a a\,p(a) &= \sum_a a\operatorname{Tr}(\rho P_a) \\ &= \operatorname{Tr}(\rho A). \end{aligned}$$

The expectation is an ensemble average over possible outcomes. It need not itself be an eigenvalue or a possible result of one trial.

For any function $f$ defined on the spectrum,

$$\langle f(A)\rangle_\rho = \operatorname{Tr} \left[ \rho f(A) \right] = \sum_a f(a)p(a).$$

In particular,

$$\langle A^n\rangle_\rho = \operatorname{Tr}(\rho A^n),$$

and the variance is

$$\begin{aligned} \operatorname{Var}_\rho(A) &= \left\langle \left( A-\langle A\rangle_\rho I \right)^2 \right\rangle_\rho \\ &= \operatorname{Tr}(\rho A^2) - \left[ \operatorname{Tr}(\rho A) \right]^2. \end{aligned}$$

Reality and bounds

If $A=A^\dagger$, then $\operatorname{Tr}(\rho A)$ is real:

$$\begin{aligned} \operatorname{Tr}(\rho A)^* &= \operatorname{Tr} \left[ (\rho A)^\dagger \right] \\ &= \operatorname{Tr}(A\rho) \\ &= \operatorname{Tr}(\rho A). \end{aligned}$$

If the spectrum lies between $a_{\min}$ and $a_{\max}$, then

$$a_{\min} \le \langle A\rangle_\rho \le a_{\max}.$$

This follows because the expectation is a convex average of the eigenvalues. Likewise,

$$\operatorname{Var}_\rho(A)\ge0$$

because it is the expectation of a positive operator.

These inequalities are powerful error checks. An answer outside the spectral range or a negative variance signals a calculation or normalization mistake.

Projective measurement probabilities

For projectors satisfying

$$P_aP_b=\delta_{ab}P_a, \qquad \sum_a P_a=I,$$

the trace rule gives

$$p(a)=\operatorname{Tr}(\rho P_a).$$

The probabilities are nonnegative because both $\rho$ and $P_a$ are positive. They are normalized:

$$\begin{aligned} \sum_a p(a) &= \operatorname{Tr} \left( \rho\sum_aP_a \right) \\ &= \operatorname{Tr}\rho =1. \end{aligned}$$

For a degenerate outcome, $P_a$ projects onto the full eigenspace. The probability is the sum of the density-matrix populations in any orthonormal basis of that eigenspace.

POVM probabilities

A general POVM has effects

$$E_i\ge0, \qquad \sum_iE_i=I.$$

Its outcome probabilities are

$$p(i)=\operatorname{Tr}(\rho E_i).$$

To see nonnegativity directly, write

$$\begin{aligned} \operatorname{Tr}(\rho E_i) &= \operatorname{Tr} \left( \rho^{1/2}E_i\rho^{1/2} \right) \\ &\ge0, \end{aligned}$$

because $\rho^{1/2}E_i\rho^{1/2}$ is positive. Completeness again gives

$$\sum_i p(i)=1.$$

The effects determine outcome probabilities. They do not alone determine the conditional output states of the measured system; that requires an instrument. See POVMs: First Encounter and Generalized Measurements Overview.

Reduced states and local observables

For a joint state $\rho_{AB}$, an observable local to $A$ is

$$M_A\otimes I_B.$$

If

$$\rho_A = \operatorname{Tr}_B\rho_{AB},$$

then

$$\operatorname{Tr}_{AB} \left[ \rho_{AB}(M_A\otimes I_B) \right] = \operatorname{Tr}_A(\rho_A M_A).$$

This equality is the defining operational property of the partial trace: the reduced state reproduces every trace-rule prediction local to the subsystem. Correlation observables such as $A\otimes B$ still require the joint state.

See Reduced Density Matrices for the construction and Subsystems and Local Observables for the operator embedding.

Qubit matrix example

Write a general qubit density matrix as

$$\rho = \begin{pmatrix} p&c\\ c^*&1-p \end{pmatrix}.$$

Using the Pauli matrices,

$$\begin{aligned} \langle X\rangle_\rho &= 2\operatorname{Re}c, \\ \langle Y\rangle_\rho &= -2\operatorname{Im}c, \\ \langle Z\rangle_\rho &= 2p-1. \end{aligned}$$

The $Z$ expectation depends on the populations, while $X$ and $Y$ expose the real and imaginary parts of the coherence.

Equivalently, if

$$\rho = \frac12 \left( I+\mathbf r\cdot\boldsymbol\sigma \right)$$

and

$$A = a_0I+\mathbf a\cdot\boldsymbol\sigma,$$

then

$$\langle A\rangle_\rho = a_0+\mathbf r\cdot\mathbf a.$$

This compact formula follows from $\operatorname{Tr}(\sigma_i\sigma_j)=2\delta_{ij}$.

For example, take

$$\rho=\begin{pmatrix}0.6&0.2+0.1i\\0.2-0.1i&0.4\end{pmatrix}, \qquad A=2I+3Z.$$

The Bloch vector is $(0.4,-0.2,0.2)$, whose squared length is $0.24<1$, so this is a valid mixed state. The two possible readings of $A$ are $5$ and $-1$, with probabilities $0.6$ and $0.4$. Thus

$$\langle A\rangle=0.6(5)+0.4(-1)=2.6, \qquad \operatorname{Var}(A)=9(1-0.2^2)=8.64.$$

The mean lies between the possible readings, but is not itself a reading.

Unsharp qubit measurement

For $0\le\eta\le1$, define two effects

$$E_\pm = \frac12 \left( I\pm\eta Z \right).$$

They are positive and satisfy $E_++E_-=I$. If the state’s Bloch component is

$$r_z=\operatorname{Tr}(\rho Z),$$

then

$$p(\pm) = \frac12 \left( 1\pm\eta r_z \right).$$

At $\eta=1$, this is the projective $Z$ measurement. At $\eta=0$, both outcomes have probability $1/2$ regardless of the state. Intermediate values model a noisy or unsharp binary measurement at the level of effects. If the recorded numbers are $+1$ and $-1$, their mean is $\eta r_z$ and their second moment is $1$. The first-moment operator is $\eta Z$, but its square is $\eta^2 I$, not the measurement’s second-moment operator $I$. For a POVM use $\sum_i x_i^n E_i$ for the $n$th moment of the recorded values $x_i$; powers of one observable describe its spectral measurement.

Infinite-dimensional qualification

In an infinite-dimensional Hilbert space, a density operator is trace class. If $A$ is bounded, then $\rho A$ is trace class and $\operatorname{Tr}(\rho A)$ is well-defined.

For an unbounded self-adjoint $A$, start with its spectral projections $P_A(\Delta)$. They define the probability measure

$$\mu_\rho(\Delta)=\operatorname{Tr}[\rho P_A(\Delta)].$$

A finite expectation exists when $\int |a|\,d\mu_\rho(a)<\infty$, and then $\langle A\rangle_\rho=\int a\,d\mu_\rho(a)$. The notation $\operatorname{Tr}(\rho A)$ is often used for this spectral average, but interpreting it as the trace of an ordinary operator product requires additional conditions.

For a pure state, finite absolute first moment means $\psi\in D(|A|^{1/2})$; it need not imply $\psi\in D(A)$. Finite second moment means $\psi\in D(A)$ and gives $\langle A^2\rangle=\|A\psi\|^2$, without requiring $\psi\in D(A^2)$. Thus “the mean exists” and “the vector $A\psi$ exists” are different assertions. See Expectation values for the spectral formulation.

Cyclicity is safe for a bounded operator times a trace-class operator, and for two Hilbert–Schmidt factors. Unbounded factors require separate domain and convergence checks; finite-dimensional rearrangements alone do not justify them.

The representation-independent formulation is developed in States as Positive Linear Functionals.

Common pitfalls

Cyclic does not mean commutative. A trace permits cyclic rotation of finite matrix factors, not an arbitrary exchange of two factors.

A mean is not a single outcome. Check the eigenvalues to identify possible readings, and use the trace to average their probabilities.

Coherences can contribute. Diagonal populations suffice when the observable is diagonal in the same basis. Otherwise retain the off-diagonal terms in $\sum_{ij}\rho_{ij}A_{ji}$.

Local and joint predictions differ. A reduced state determines a local observable; a correlation between two subsystems generally needs their joint state.

Exercises

1. Prove the rank-one trace identity

$$\operatorname{Tr} \left( \lvert u\rangle\langle v\rvert A \right) = \langle v\rvert A\lvert u\rangle.$$

Solution

Insert an orthonormal basis:

$$\begin{aligned} \operatorname{Tr} \left( \lvert u\rangle\langle v\rvert A \right) &= \sum_n \langle n\vert u\rangle \langle v\rvert A\lvert n\rangle \\ &= \langle v\rvert A \left( \sum_n \lvert n\rangle\langle n\rvert \right) \lvert u\rangle \\ &= \langle v\rvert A\lvert u\rangle. \end{aligned}$$

2. Starting from matrices, show that

$$\operatorname{Tr}(\rho A) = \sum_{i,j}\rho_{ij}A_{ji}.$$

Solution

By matrix multiplication,

$$(\rho A)_{ii} = \sum_j\rho_{ij}A_{ji}.$$

Summing the diagonal entries gives

$$\operatorname{Tr}(\rho A) = \sum_i(\rho A)_{ii} = \sum_{i,j}\rho_{ij}A_{ji}.$$

3. Let $A$ have smallest and largest eigenvalues $a_{\min},a_{\max}$. Prove that every density operator obeys

$$a_{\min} \le \operatorname{Tr}(\rho A) \le a_{\max}.$$

Solution

With $A=\sum_a aP_a$, define

$$p(a)=\operatorname{Tr}(\rho P_a).$$

These probabilities are nonnegative and sum to one. Therefore

$$\operatorname{Tr}(\rho A) = \sum_a a\,p(a)$$

is a convex average of the eigenvalues and lies between their minimum and maximum.

4. Show that

$$\operatorname{Var}_\rho(A) = \operatorname{Tr}(\rho A^2) - \operatorname{Tr}(\rho A)^2$$

is nonnegative.

Solution

Let

$$\delta A = A-\langle A\rangle_\rho I.$$

Because $(\delta A)^2$ is positive,

$$\operatorname{Tr} \left[ \rho(\delta A)^2 \right] \ge0.$$

Expanding and using $\operatorname{Tr}\rho=1$ gives

$$\operatorname{Tr} \left[ \rho(\delta A)^2 \right] = \operatorname{Tr}(\rho A^2) - \operatorname{Tr}(\rho A)^2.$$

Thus the variance is nonnegative.

5. Prove that a POVM $\{E_i\}$ gives normalized, nonnegative probabilities.

Solution

For each effect,

$$p(i) = \operatorname{Tr} \left( \rho^{1/2}E_i\rho^{1/2} \right) \ge0$$

because the operator inside the trace is positive. Completeness gives

$$\begin{aligned} \sum_i p(i) &= \operatorname{Tr} \left( \rho\sum_iE_i \right) \\ &= \operatorname{Tr}\rho =1. \end{aligned}$$

6. For

$$\rho = \begin{pmatrix} p&c\\ c^*&1-p \end{pmatrix},$$

derive the three Pauli expectations.

Solution

Direct multiplication gives

$$\begin{aligned} \operatorname{Tr}(\rho X) &= c+c^* = 2\operatorname{Re}c, \\ \operatorname{Tr}(\rho Y) &= i(c-c^*) = -2\operatorname{Im}c, \\ \operatorname{Tr}(\rho Z) &= p-(1-p) = 2p-1. \end{aligned}$$

The signs agree with the convention

$$Y = \begin{pmatrix} 0&-i\\ i&0 \end{pmatrix}.$$

7. An unsharp detector has effects $E_\pm=(I\pm\eta Z)/2$ and records the numbers $\pm1$. For $\eta=3/5$ and $\langle Z\rangle=1/5$, calculate its outcome probabilities, mean, and variance. Compare the variance with that of a spectral measurement of the operator $M=\eta Z$ on the same state. Why do the means agree but the variances differ?

Solution

The probabilities are $p(+)=14/25$ and $p(-)=11/25$. Thus the recorded mean is $3/25$, the second moment is one, and

$$\operatorname{Var}(x)=1-(3/25)^2=\frac{616}{625}.$$

For a spectral measurement of $M$, the possible values are $\pm3/5$, with probabilities $3/5$ and $2/5$. Its mean is again $3/25$, but

$$\operatorname{Var}(M)=\eta^2-\eta^2\langle Z\rangle^2 =\frac{216}{625}.$$

The first-moment operator $\sum_\pm(\pm1)E_\pm=M$ determines the mean, but the detector’s second-moment operator is $\sum_\pm(\pm1)^2E_\pm=I$, rather than $M^2$. The extra readout variance is $1-\eta^2=16/25$. At $\eta=1$ the two measurements coincide; at $\eta=0$ the detector still outputs a fair random sign while $M=0$ always reads zero.

8. Verify the reduced-state identity in a product basis:

$$\operatorname{Tr}_{AB} \left[ \rho_{AB}(M_A\otimes I_B) \right] = \operatorname{Tr}_A(\rho_A M_A).$$

Solution

Write the joint matrix elements as

$$\rho_{i\mu,j\nu} = \langle i,\mu\rvert \rho_{AB} \lvert j,\nu\rangle.$$

Then

$$\operatorname{Tr}_{AB} \left[ \rho_{AB}(M_A\otimes I_B) \right] = \sum_{i,j,\mu} \rho_{i\mu,j\mu} (M_A)_{ji}.$$

The reduced-state entries are

$$(\rho_A)_{ij} = \sum_\mu \rho_{i\mu,j\mu}.$$

Substitution gives

$$\sum_{i,j} (\rho_A)_{ij}(M_A)_{ji} = \operatorname{Tr}_A(\rho_A M_A).$$

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific (2014), Ch. 3.
• A. S. Holevo, Probabilistic and Statistical Aspects of Quantum Theory, 2nd ed., Edizioni della Normale (2011), Ch. 2.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010), Secs. 2.2 and 2.4.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018), Ch. 2.