Trace Rule for Expectation Values

For a density operator $\rho$ and an 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.

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

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

Only in the measurement eigenbasis may one read the outcome probabilities directly from the diagonal entries.

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 post-measurement states; 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

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

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 $\operatorname{Tr}(\sigma_i\sigma_j)=2\delta_{ij}$.

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.

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.

Unbounded observables require more care. The expectation exists only when the state has the required finite moment and the operator-domain conditions are satisfied. Formally writing $\operatorname{Tr}(\rho A)$ does not guarantee convergence. Cyclic trace manipulations also require the relevant products to be trace class.

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

Practical workflow

For a finite-dimensional expectation or probability:

1. Confirm that $\rho$ is positive and has trace one.
2. Put $\rho$ and the observable or effect in the same basis.
3. Check that the operators act on the same Hilbert space.
4. Multiply in the written order and take the trace.
5. Use a diagonal basis when it simplifies the calculation.
6. For local quantities, either include the identity factor or use the reduced state.
7. Check that probabilities lie in $[0,1]$ and sum to one.
8. Check that Hermitian expectations are real and lie in the spectral range.
9. Distinguish an expectation value from a possible single-shot outcome.
10. For unbounded operators, verify existence rather than relying on formal cyclicity.

Common mistakes

• Treating $\operatorname{Tr}(\rho A)$ as an operator instead of a number.
• Assuming the expectation must be an eigenvalue.
• Reading diagonal entries as probabilities before choosing the measurement basis.
• Reordering factors inside a trace rather than rotating them cyclically.
• Replacing $\operatorname{Tr}(\rho A^2)$ with $\operatorname{Tr}(\rho A)^2$.
• Using an arbitrary positive operator as a POVM effect without checking the complete set sums to identity.
• Confusing POVM probabilities with a post-measurement update rule.
• Dropping the identity in a local observable on a composite space.
• Using a reduced state to compute a joint correlation.
• Applying finite-dimensional cyclicity to unbounded products without checking trace-class conditions.

Where deeper treatment lives

Born Rule develops the probability postulate. Expectation Values focuses on physical interpretation in state-vector language. Variance and Standard Deviation develops moments and uncertainty.

Density-Matrix Formulation collects the postulates in density-operator language. A compact lookup is available at Reference Formula: Expectation Value.

References

• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press (1955), Ch. IV.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer (1994), Chs. 1 and 4.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific (2014), Ch. 3.
• 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.
• A. S. Holevo, Probabilistic and Statistical Aspects of Quantum Theory, 2nd ed., Edizioni della Normale (2011), Ch. 2.

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. For the unsharp effects

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

derive $p(\pm)$ and check the limits $\eta=0$ and $\eta=1$.

Solution

Using $\operatorname{Tr}\rho=1$ and $r_z=\operatorname{Tr}(\rho Z)$,

$$\begin{aligned} p(\pm) &= \operatorname{Tr}(\rho E_\pm) \\ &= \frac12 \left( 1\pm\eta r_z \right). \end{aligned}$$

At $\eta=0$, both probabilities are $1/2$. At $\eta=1$,

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

which are the projective $Z$-measurement probabilities.

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

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

Solution

Write the joint matrix elements as

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

Then

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

The reduced-state entries are

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

Substitution gives

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