The positive operator-valued measure (POVM) is a mathematical formalism/tool representing a measurement operation [NC10, Sec. 2.2.6; WN17, Sec. 1.5.1] that satisfies Postulate 3.

Suppose a measurement described by measurement operators $\mathbf{M}_m$, where $m$ takes value from a finite set, is performed on a quantum system in the state $|\psi\rangle$, then the probability of outcome $m$ is given by

$\Pr\{m|\psi\} = \langle\psi|\mathbf{M}_m^\dagger\mathbf{M}_m|\psi\rangle = \text{tr}(\mathbf{M}_m^\dagger\mathbf{M}_m|\psi\rangle\langle\psi|).$

Suppose we define

$\mathbf{E}_m \triangleq \mathbf{M}_m^\dagger\mathbf{M}_m,$

then $\mathbf{E}_m$ satisfies 1️⃣ the completeness relation $\sum_m\mathbf{E}_m=\mathbf{I}$ and 2️⃣ $\Pr\{m|\psi\} = \langle\psi|\mathbf{E}_m|\psi\rangle = \text{tr}(\mathbf{E}_m|\psi\rangle\langle\psi|) \geq 0 \implies \mathbf{E}_m$ is positive-semidefinite.

• The general equality $\Pr\{m\} = \text{tr}(\mathbf{E}_m\rho)$ for some density matrix $\rho$ is known as the POVM version of the Born rule.
• The operators $\mathbf{E}_m$ are called the POVM elements associated with the measurement.
• The complete set $\{\mathbf{E}_m\}$ is known as a POVM.

When the POVM elements are projectors, such as in the preceding example, the POVM is called a projection-valued measure (PVM) [Hay17, p. 7].

Watch edX Lecture 1.6 “Generalized measurements” of Quantum Cryptography by CaltechDelftX QuCryptox.

For Hermitian matrix $\mathbf{A}$, the following statements are equivalent, and any one can serve as the definition of the positive definiteness of $\mathbf{A}$ [Mey00, Sec. 7.6; Woe16, Sec. 7.4]:

• $\vec{x}^\ast\mathbf{A}\vec{x}>0$, for all $\vec{x}\neq\vec{0}$. Note $\vec{x}^\ast\mathbf{A}\vec{x}$ is called a quadratic form.

  In Dirac notation, $\langle x|\mathbf{A}|x\rangle\gt0$, for all $|x\rangle\neq\vec{0}$. Note $|0\rangle\color{red}{\neq}\vec{0}$.

• The eigenvalues of $\mathbf{A}$ are all positive.
• $\mathbf{A}$ can be put in the form $\mathbf{A}=\mathbf{B}^\ast\mathbf{B}$ or $\mathbf{A}=\mathbf{B}\mathbf{B}^\ast$, where $\mathbf{B}$ is a nonsingular matrix.

Similarly, the following statements are equivalent, and any one can serve as the definition of the positive semidefiniteness of $\mathbf{A}$:

• $\vec{x}^\ast\mathbf{A}\vec{x}\geq0$, for all $\vec{x}\neq\vec{0}$.

  In Dirac notation, $\langle x|\mathbf{A}|x\rangle\geq0$, for all $|x\rangle\neq\vec{0}$.

• The eigenvalues of $\mathbf{A}$ are all nonnegative.

The discussion here follows from the discussion of the positive operator-valued measure (POVM).

The POVM cannot be used to determine the post-measurement state because the post-measurement state may not be pure [WN17, p. 13]. Instead, a Kraus operator representation of the POVM, known as projective measurement, is necessary to specify the post-measurement state.

Projective measurements are repeatable, and the outcome observed as a result of projective measurement is deterministic [MM12, Sec. 2.7], as explained below.

Projective measurements are often discussed in terms of an observable, which is defined as a Hermitian operator acting on the state space of a system.

Every observable $\mathbf{M}$ has a spectral decomposition [Mey00, p. 517]:

where $\lambda_i$ are the eigenvalues of $\mathbf{M}$, and $\mathbf{M}_i$ are so-called spectral/orthogonal projectors onto the null space of $(\mathbf{M}-\lambda_i\mathbf{I})$. More precisely, 1️⃣ $\mathbf{M}_i$ are Hermitian and idempotent, 2️⃣ $\sum_i\mathbf{M}_i=\mathbf{I}$, and 3️⃣ $\mathbf{M}_i$ are mutually orthogonal.

Clearly, $\mathbf{M}_i$ are valid POVM elements.

A (non-unique) Kraus operator representation or Kraus decomposition of $\mathbf{M}$ is defined as [WN17, Definition 1.5.2]:

where $\mathbf{A}_i^\dagger\mathbf{A}_i = \mathbf{M}_i$.

• If we define $\mathbf{A}_i\triangleq\mathbf{M}_i$, then $\mathbf{A}_i^\dagger\mathbf{A}_i = \mathbf{M}_i^\dagger\mathbf{M}_i = \mathbf{M}_i^2 = \mathbf{M}_i$.
• If we define $\mathbf{A}_i\triangleq\mathbf{U}\sqrt{\mathbf{M}_i}$, where $\mathbf{U}$ is any unitary matrix, then $\mathbf{A}_i^\dagger\mathbf{A}_i = \sqrt{\mathbf{M}_i}^\dagger\mathbf{U}^\dagger\mathbf{U}\sqrt{\mathbf{M}_i} = \mathbf{M}_i$.
• Thus, there is no unique Kraus representation/decomposition.

With Kraus operators, we can now define projective measurement:

Since the default Kraus operators are the same as the orthogonal projectors, most authors skip discussion of Kraus operators altogether 🤷‍♂️.

Measuring the observable $\mathbf{M}$ is equivalent to performing a projective measurement with respect to the decomposition $\mathbf{I} = \sum_i\mathbf{M}_i$, where the measurement outcome $i$ corresponds to eigenvalue $\lambda_i$.

In Example 2, the observation that the projective measurement preserves the quantum state is not a coincidence. In fact, building on Definition 1, applying measurement $\mathbf{A}_i$ to $|\psi'\rangle$, i.e., applying $\mathbf{A}_i$ to $|\psi\rangle$ twice gives us [MM12, p. 153]:

$\begin{split} |\psi''\rangle = \dfrac{\mathbf{A}_i|\psi'\rangle}{\sqrt{\langle\psi'|\mathbf{A}_i|\psi'\rangle}} = \dfrac{\mathbf{A}_i|\psi\rangle}{\sqrt{\langle\psi'|\mathbf{A}_i|\psi'\rangle\langle\psi|\mathbf{A}_i|\psi\rangle}} = \dfrac{\mathbf{A}_i|\psi\rangle}{\sqrt{\frac{\langle\psi|\mathbf{A}^\dagger_i}{\sqrt{p_i}}\mathbf{A}_i\frac{\mathbf{A}_i|\psi\rangle}{\sqrt{p_i}}p_i}} = |\psi'\rangle, \end{split}$

where $p_i\triangleq\langle\psi|\mathbf{A}_i|\psi\rangle$.

As defined in Practical 1 of COMP 5074, a qubit is a quantum state with two possible outcomes. Mathematically, a qubit is a two-dimensional Hilbert space.

A pure state is the quantum state of a qubit that we can precisely define at any point in time [Qis21, Sec. 1].

• For example, a qubit, say $|q\rangle$, that started out as $|0\rangle$ becomes $|+\rangle$ when it passes through a Hadamard gate.
• When we measure $|q\rangle$ in the computational basis (Z-basis), we get $|0\rangle$ at a probability of 0.5, or $|1\rangle$ at the same probability.
• Regardless of our measurement, in the ideal condition, we can say with absolute certainty that $|q\rangle$ has quantum state $|+\rangle$, and $|q\rangle$ is an example of a pure state.

In non-ideal conditions, a qubit can assume a certain quantum state at a certain probability, and other quantum states at other probabilities, regardless of measurements.

• For example, consider the two-qubit entangled state in Fig. 1:

  $|\psi_{AB}\rangle = \frac{1}{\sqrt{2}}(|0_A0_B\rangle + |1_A1_B\rangle),$

  where the subscripts $A$ and $B$ label the qubits associated with registers $q_1$ and $q_0$ respectively.

• Since the qubits of $q_1$ and $q_0$ are entangled, measuring a $|0_A\rangle$ in $q_1$ causes $|0_B\rangle$ to be measured in $q_0$. Similarly, measuring a $|1_A\rangle$ in $q_1$ causes $|1_B\rangle$ to be measured in $q_0$.
• Equivalently, the qubit of $q_0$, namely $\psi_B$, assumes values $|0_B\rangle$ and $|1_B\rangle$ at equal probability.
• However, this is NOT to say $\psi_B$ is a superposition of $|0_B\rangle$ and $|1_B\rangle$, i.e., we CANNOT express $\psi_B$ as $\frac{1}{\sqrt{2}}(|0_B\rangle+|1_B\rangle)$.
• $\psi_B$ is a mixture or ensemble of the states $|0_B\rangle$ and $|1_B\rangle$ [KLM07, Sec. 3.5].
• $\psi_B$ is an example of a mixed state, i.e., a statistical ensemble of quantum states [Qis21, Sec. 2].

A direct but rather clunky representation of a mixed state is the set of pairs:

$\{(|\psi_1\rangle,p_1), \ldots, (|\psi_n\rangle,p_n)\},$

where each pair $(|\psi_i\rangle,p_i)$ means the pure state $|\psi_i\rangle$ appears at probability $p_i$, for $i=1,\ldots,n$. Note $\sum_{i=1}^n p_i=1$.

A more compact and useful representation of a mixed state can be obtained from the type of Hilbert-space operators called density operators [KLM07, Sec. 3.5.1].

• The matrix representation of a density operator is a density matrix.

• The density matrix representing the pure state $|\psi\rangle$ is defined as the outer product

  $\rho \triangleq |\psi\rangle\langle\psi|.$

• The density matrix representing the mixed state $\{(|\psi_1\rangle,p_1), \ldots, (|\psi_n\rangle,p_n)\}$ is defined as the weighted sum of outer products:

  $\rho \triangleq \sum_{i=1}^n p_i|\psi_i\rangle\langle\psi_i|.$

  Note $\psi_i$ do not need to be basis states.

  Example 1

  Revisiting the example in Fig. 1, the density matrix of $|\psi_B\rangle$ is thus

  $\begin{aligned} \rho_B & = \frac{1}{2}|0_B\rangle\langle 0_B| + \frac{1}{2}|1_B\rangle\langle1_B| = \frac{1}{2}\begin{bmatrix} 1 \\ 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \end{bmatrix} + \frac{1}{2}\begin{bmatrix} 0 \\ 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \end{bmatrix} \\ & = \frac{1}{2}\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + \frac{1}{2}\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} = \frac{1}{2}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \frac{1}{2}\mathbf{I}. \end{aligned}$

• It is important to know the effect of measuring a mixed state.

  Suppose the pure state $|\psi_i\rangle\langle\psi_i|$, where $i=1,\ldots,n$, is measured in the computational (Z) basis, then the probability of getting $|0\rangle$ as the measurement outcome is [WN17, Sec. 1.2.1]:

  $\Pr\{0|i\} = |\langle\psi_i|0\rangle|^2 = \langle\psi_i|0\rangle\langle\psi_i|0\rangle = \langle0|\psi_i\rangle\langle\psi_i|0\rangle.$

  Similarly, the probability of getting $|1\rangle$ as the measurement outcome is $\Pr\{1|i\} = \langle1|\psi_i\rangle\langle\psi_i|1\rangle$.

  Now suppose we are measuring the mixed state $\rho = \sum_i p_i|\psi_i\rangle\langle\psi_i|$. By the law of total probability, the probability of getting $|0\rangle$ is now 

  $\Pr\{0\} = \sum_i p_i\Pr\{0|i\} = \sum_i p_i\langle0|\psi_i\rangle\langle\psi_i|0\rangle = \langle0| \cdot \sum_i p_i|\psi_i\rangle\langle\psi_i| \cdot |0\rangle = \langle0|\rho|0\rangle.$

  Similarly, the probability of getting $|1\rangle$ is $\Pr\{1\} = \langle1|\rho|1\rangle$.

  Thus in general, the probability of measuring mixed state $\rho$ and getting basis vector $|i\rangle$ is

  $\Pr\{i\} = \langle i|\rho|i\rangle.$

  Example 2

  Continuing from Example 1, let us find out the probability of getting $|+\rangle$ when $\rho = \frac{1}{2} |0\rangle\langle0| + \frac{1}{2}|1\rangle\langle1|$ is measured in the Hadamard (X) basis.

  The probability of getting $|+\rangle$ is

  $\langle+|\rho|+\rangle = \begin{bmatrix}1/\sqrt{2} & 1/\sqrt{2}\end{bmatrix}\begin{bmatrix}1/2 & 0 \\ 0 & 1/2\end{bmatrix}\begin{bmatrix}1/\sqrt{2} \\ 1/\sqrt{2}\end{bmatrix} = \dfrac{1}{2}.$

  Similarly, the probability of getting $|-\rangle$ is $\langle-|\rho|-\rangle=\dfrac{1}{2}$.

  In case there is any confusion that mixed state $\rho$ is equivalent to $|+\rangle$, consider what happens when $|+\rangle$ is measured in the X basis: the measurement outcome is simply $|+\rangle$ at a probability of 1.

  Example 3

  [Leo10, Sec. 1.2]

• A quick way to test whether a density matrix represents a pure state or a mixed state is to use the trace operator [Gra21, p. 11; NC10, Theorem 2.5]:

  $\rho$ represents a pure state if $\text{tr}(\rho^2)=1$.

  $\rho$ represents a mixed state if $\text{tr}(\rho^2)\lt1$.

  ⚠ $\rho$ is positive-semidefinite (i.e., $\rho$ is a positive operator) and $\text{tr}(\rho)\equiv1$ for any density matrix $\rho$.

  A useful property of $\text{tr}$ is available through the observation that since $\langle\psi|\mathbf{A}|\psi\rangle$ is a scalar,

  $\langle\psi|\mathbf{A}|\psi\rangle = \text{tr}(\langle\psi|\mathbf{A}|\psi\rangle),$

  for matrix $\mathbf{A}$ and ket $|\psi\rangle$. Applying the cyclic property of $\text{tr}$, we get $\text{tr}(\langle\psi|\mathbf{A}|\psi\rangle) = \text{tr}(\mathbf{A}|\psi\rangle\langle\psi|)$, and consequently [WN17, Exercise 1.2.3]:

  $\text{tr}(\mathbf{A}|\psi\rangle\langle\psi|) = \langle\psi|\mathbf{A}|\psi\rangle.$

  (1)

If the explanation above is hard to grasp, see if edX Lecture 1.2 “The density matrix” of Quantum Cryptography by CaltechDelftX QuCryptox helps.

When applied with unitary operator $\mathbf{U}$, state $|\psi\rangle$ evolves into $\mathbf{U}|\psi\rangle$. Accordingly, the density matrix of the mixed state $\rho = \sum_{i=1}^n p_i|\psi_i\rangle\langle\psi_i|$ evolves into

Our adventure with the trace operator continues because an important application of the trace operator is determining the probability of getting result $m$ when applying measurement operator $\mathbf{M}_m$ to mixed state $\rho=\sum_i p_i|\psi_i\rangle\langle\psi_i|$.

By definition, the probability of getting result $m$ given initial state $|\psi_i\rangle$ is [NC10, p. 99]:

$\Pr\{m|i\} = |\mathbf{M}_m|\psi_i\rangle|^2 = \langle\psi_i|\mathbf{M}_m^\dagger\mathbf{M}_m|\psi_i\rangle = \text{tr}(\mathbf{M}_m^\dagger\mathbf{M}_m|\psi_i\rangle\langle\psi_i|).$

The last equality is due to Eq. (1). By the law of total probability,

Omitting the derivation in [NC10, pp. 99-100], the density matrix of the system after making measurement $\mathbf{M}_m$ and obtaining result $m$ is:

By the law of total probability,

$\rho = \sum_m\Pr\{m\}\rho_m = \sum_m\text{tr}(\mathbf{M}_m^\dagger\mathbf{M}_m\rho)\dfrac{\mathbf{M}_m\rho\mathbf{M}_m^\dagger}{\text{tr}(\mathbf{M}_m^\dagger\mathbf{M}_m\rho)} = \sum_m\mathbf{M}_m\rho\mathbf{M}_m^\dagger,$

implying

$\sum_m\mathbf{M}_m\mathbf{M}_m^\dagger = \mathbf{I},$

which is called the completeness equation. ⚠ Note $\mathbf{M}_m$ is unitary if and only if $m$ has one value.

Postulates of quantum mechanics

Using the language of the density operator, we can phrase the fundamental postulates of quantum mechanics as [NC10, pp. 98-102; Gra21, p. 11-12]:

Postulate 3 defines general/generalised measurement [NC10, Box 2.5], which consists of 1️⃣ a rule describing the probabilities of different measurement outcomes, and 2️⃣ a rule describing the post-measurement state.

• For applications where measurement is made once at the conclusion of some experiment, the main items of interest are the probabilities rather than the post-measurement state. For these applications, the mathematical tool of positive operator-valued measure (POVM) is applicable.
• To determine the post-measurement state but in a way that the measurement is repeatable and the outcome is deterministic, we use projective measurement.