$\require{cancel} \newcommand{\Ket}[1]{\left|{#1}\right\rangle} \newcommand{\Bra}[1]{\left\langle{#1}\right|} \newcommand{\Braket}[1]{\left\langle{#1}\right\rangle} \newcommand{\Rsr}[1]{\frac{1}{\sqrt{#1}}} \newcommand{\RSR}[1]{1/\sqrt{#1}} \newcommand{\Verti}{\rvert} \newcommand{\HAT}[1]{\hat{\,#1~}} \DeclareMathOperator{\Tr}{Tr}$

Density Matrix

^{First created in September 2018}

Ref:

• https://physics.stackexchange.com/questions/322712/measurement-on-density-operator

• https://pages.uoregon.edu/svanenk/solutions/Mixed_states.pdf

Operators

An operator is a mathematical device mapping a vector space to another vector space. Various types of operators differ in their physical processes and mathemtical properties.

Here are some types of operators:

• Transformation - Reversibly transforms a state vector to another state vector in the same state space. Examples include quantum logic gates, which transform a qubit system state into another qubit system state. They are unitary transformations ($U^\dagger=U^{-1}$), preserving the length of the vector.

• Projector - Irreversibly measures (or "collapses") a state vector onto an eigenspace of the Projector. i.e. For a unit vector $\Ket\alpha,~P=\Ket\alpha\Bra\alpha$ projects the state vector $\Ket{\psi}$ onto $\Ket{\alpha}$, $P\Ket\psi=\Braket{\alpha\Verti\psi}\cdot\Ket\alpha$. The probability amplitude is $\lvert P\Ket\psi\rvert=\Braket{\alpha\Verti\psi}\le 1.$ The posterior state (of unit length) is $\frac{P\Ket{\psi}}{\lvert P\Ket\psi\rvert}.$ A projector is idempotent, so $P^n=P.$ A projector is Hermitian, so $P^\dagger=P.$

• Density Operator - A description of a mixed state in terms of a probability weighted combination of pure stats: $\rho=\sum_{i=1}^n p_i\Ket{\psi_i}\Bra{\psi_i}$. A density operator is also called a state operator. It is Hermitian, so $\rho^\dagger=\rho.$ A pure state can be expressed as a density matrix with $n=1$. (Although of similar form, a projector $\Ket j\Bra j$ is a measurement along the $\Ket j$ basis vector, while $\Ket{\psi_i}\Bra{\psi_i}$ is the state vector to measure.)

A Density Operator, be it finite-dimensional or infinite-dimensional, must satisfy three conditions:

1. Hermitian: $\rho=\rho^\dagger.$
2. Positive semi-definite: $\Braket{\phi\Verti\rho\Verti\phi}\ge 0~$ for any $\Ket\phi.$
3. Trace one (rank-one): $\Tr(\rho)=1.$

Any operators applied on a density operator must preserve the above properties in order to be valid.

(Note: $\rho$ is not idempotent, so being rank-one does not mean single-dimension.)

Projection of Density Matrix

A projector is in the form $\displaystyle P_i=\Ket{i}\Bra{i}.~~$ This form implies idempotence, so $P_i^n=P_i$, and Hermiticity, so $P_i^\dagger=P_i.$

A measurement$~~A=\sum_i\lambda_i P_i=\sum_i\lambda_i\Ket{i}\Bra{i},~$where the $\Ket{i}$s are mutually orthonormal.

For one of these orthonormal vector $\Ket j,~A\Ket{j} =\left(\sum_i\lambda_i\Ket{i}\Bra{i}\right)\Ket{j} =\lambda_j\Ket{j} ,$ so $\Ket{j}$ is an eigenvector with eigenvalue $\lambda_j$.

In quantum mechanics, $\lambda_j$ is the probability amplitude, but in qubit measurement, $\lambda_j$ is often a label rather than having a physical interpretation.

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For a mixed state containing $\Ket{\psi_1},\Ket{\psi_2},\ldots,\Ket{\psi_n}$, its density operator is defined as $\rho=\sum_k p_k\Ket{\psi_k}\Bra{\psi_k}$, where $p_k\in\mathbb{R},~0<p_k\le 1,~\sum_{k=1}^n p_k=1$ and $\Ket{\psi_k}$ are unique pure state. $\rho$ is Hermitian, so $\rho^\dagger=\rho.$

For projector $P_i=\Ket{i}\Bra{i},$ $\boxed{P_i\rho P_i=\beta_i~P_i~,~~\text{where}~ \beta_i=\sum_k p_k~\big|\Braket{i\Verti\psi_k}\big|^2}.$

$\beta_i$ is the success rate (probability with a pure state or proportion in a mixed-state ensemble) of the measurement $P_i$ on the mixed state.

Proof: $~ P_i\rho P_i =\Ket{i}\Bra{i} \left(\sum_k p_k\Ket{\psi_k}\Bra{\psi_k}\right) \Ket{i}\Bra{i} =\Ket{i}\left(\sum_{k}p_k\big(\Braket{i\Verti\psi_k}\Braket{\psi_k\Verti i}\big)\right)\Bra{i} =\beta_i~\Ket{i}\Bra{i} =\beta_i~P_i~,~~$where $\beta_i=\sum_k p_k~\big|\Braket{i\Verti\psi_k}\big|^2~,$

Note: $~ P_i\rho P_i=\beta_i P_i~~$ does not mean that $P_i\rho$ gives you $\beta_i$. The former is a matrix and the latter is a scaler. One may take $\beta_i$ as an eigenvalue of $P_i\rho$ with "eigenmatrix" (if there is such a thing) $P_i$. To be more serious and proper, the columns of $P_i$ are eigenvectors.

Expectation Value as the Average of Pure States

While the expectation value of measurement $A$ on a pure state $\Ket\psi$ is $\Braket{\psi\lvert A\rvert\psi}$, the same on a mixed state is $\Tr(\rho A),~$where $\rho$ is the density operator.

Proof: For each of the pure state $\Ket{\psi_k}$ in the mixed state $\rho=\sum_k p_k\Ket{\psi_k}\Bra{\psi_k}$, the expectation value of measurement $A$ is $\Braket{A}_k\equiv\Braket{\psi_k\lvert A\rvert\psi_k}.$

So the expectation value of measuring the mixed state is $\Braket{A}_\rho =\sum_k p_k\Braket{A}_k=\sum_k p_k\Bra{\psi_k}\left(\sum_i\lambda_i\Ket{i}\Bra{i}\right)\Ket{\psi_k} =\sum_k p_k\Braket{\psi_k\lvert A\rvert\psi_k} =\sum_i\lambda_i\left(\sum_k p_k\left(\Braket{\psi_k\Verti i}\Braket{i\Verti\psi_k}\right)\right) =\sum_i\lambda_i\beta_i ~,~$ which is the expected value.

$\therefore\boxed{\Braket{A}_\rho =\sum\nolimits_i\lambda_i\beta_i =\Tr(\rho A) .}~~~~$ Consider $A=I,~~\boxed{\Tr(\rho)=1.}$.

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As pure state is just mixed state with a single term, $\Tr(\rho A)$ being the expectation value of measurement applies to pure states as well. So $\Tr(\rho A)$ is considered more general than $\Braket{\psi\Verti A\Verti \psi}$, but the latter is more fundamental and therefore easier to conceptualise.

Note: "Expectation Value" is used in the context of Quantum Mechenics. "Expectation value" is in the form of $\Braket{A}_\psi=\Braket{\psi\lvert A\rvert\psi}$ representing the "expected value" of the measurement result.

Example: Trace of Projector

Let us use projector $P_x={1\over2}\begin{bmatrix}1&1\\1&1\end{bmatrix}$ on the $X$-axis (measure along $\Ket+$), with the mixed state $\rho =p_0\Ket0\Bra0+p_1\Ket1\Bra1 =\begin{bmatrix} p_0 & 0\\ 0 & p_1 \end{bmatrix} .$

The expected value is $\Tr(\rho P_x) =\Tr \left( \begin{bmatrix} p_0 & 0\\ 0 & p_1 \end{bmatrix} {1\over2} \begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix} \right) =\Tr \left( {1\over2} \begin{bmatrix} p_0 & p_0\\ p_1 & p_1 \end{bmatrix} \right) ={p_0+p_1\over2} .$

Pure State as a Mixed State

While a pure state is simpler, the mixed state is taken as a more general form. So a pure state $\Ket\psi$ is often casted as a density matrix $\rho=\Ket\psi\Bra\psi$, as with mixed states.

When a unitary transformation is applied to the pure state $U\Ket\psi$, its counterpart expression in mixed state form is $(U\Ket\psi)~(\Bra\psi U^\dagger)=U\rho U^\dagger.$