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$\require{cancel} \newcommand{\Ket}[1]{\left|{#1}\right\rangle} \newcommand{\Bra}[1]{\left\langle{#1}\right|} \newcommand{\Braket}[1]{\left\langle{#1}\right\rangle} \newcommand{\Rsr}[1]{\frac{1}{\sqrt{#1}}} \newcommand{\RSR}[1]{1/\sqrt{#1}} \newcommand{\Verti}{\rvert} \newcommand{\HAT}[1]{\hat{\,#1~}} \DeclareMathOperator{\Tr}{Tr}$
First created in July 2018
A system in a mixed state is a classical statistical ensemble of different pure states. The word "classical" means that the probability is real-valued between 0 and 1, not a probability amplitude in the superposition sense.
The uncertainty may be due to lack of knowledge of the state of a qubit or the shear number of them in the ensemble. If the system is opened, for example, entangled with another, it is considered to be in a mixed state, even the supersystem containing all such entangled subsystems as a whole is in a pure state.
A mixed state is represented by a density matrix $\rho=\sum_{i=1}^n p_i\Ket{\psi_i}\Bra{\psi_i}$.
However, the matrix form can be decomposed into different Ket-Bra forms. This means multiple mixed states with different decompositions of state vectors $\Ket{\psi_i}$ and probabilities $p_i$ can behave the same in terms of measurement.
In other words, the density matrix contains all information about the mixed state in terms of measurement (but still the decomposition cannot be determined).
All these mixed states corresponding to the same density matrix and therefore behave the same in measurement are treated as the same mixed state.
In order to recover the decomposition ($p_i$ and $\Ket{\psi_i}$) from the density matrix, additional information is required. For example, if we know that $\Ket{\psi_i}$s are orthonormal, $p_i$ and $\Ket{\psi_i}$ can be recovered as the eigenvalues and their corresponding eigenvectors of the density matrix.
Two ensembles $\left\{p_i\Ket{\psi_i}\right\}$ and $\left\{p'_j\Ket{\psi'_j}\right\}$ compose into the same density matrix if and only if there exists a unitary $U$ such that $\sqrt{p_i}\Ket{\psi_i}=\sum_j u_{ji}\sqrt{p'_j}\Ket{\psi'_j}.$
Proof: Let $~~ \rho=\sum_i p_i\Ket{\psi_i}\Bra{\psi_i},~~ \rho'=\sum_j p'_j\Ket{\psi'_j}\Bra{\psi'_j},$
$M=\Big[\sqrt{p_1}\Ket{\psi_1}~~\sqrt{p_2}\Ket{\psi_2}~~\ldots~~\sqrt{p_n}\Ket{\psi_n}\Big],~~ M'=\Big[\sqrt{p'_1}\Ket{\psi'_1}~~\sqrt{p'_2}\Ket{\psi'_2}~~\ldots~~\sqrt{p'_n}\Ket{\psi'_n}\Big],~$ and $~U=[u_{ij}]$.
From the above, $MM^\dagger=\rho$ and $M'M'^\dagger=\rho'$.
The condition $\Ket{\psi_i}\sqrt{p_i}=\sum_j u_{ji}\Ket{\psi'_j}\sqrt{p'_j}$ is equivalent to $M=M'U.$
If follows that $MU^\dagger=M',~~ \rho'=M'M'^\dagger=(MU^\dagger)(MU^\dagger)^\dagger=MU^\dagger UM^\dagger=MM^\dagger=\rho .$
Conversely, when $\rho=\rho',~~ MM^\dagger=M'M'^\dagger,~~ M=M'\left(M'^\dagger(M^\dagger)^{-1}\right),~~ M=M'W,~~$where $W=M'^\dagger(M^\dagger)^{-1}.$
$WW^\dagger =(M'^\dagger(M^\dagger)^{-1})(M'^\dagger(M^\dagger)^{-1})^\dagger =(M'^\dagger(M^\dagger)^{-1})(M'^\dagger(M^{-1})^\dagger)^\dagger =M'^\dagger(M^\dagger)^{-1}M^{-1}M'\\ =M'^\dagger(MM^\dagger)^{-1}M' =M'^\dagger(M'M'^\dagger)^{-1}M' =M'^\dagger(M'^\dagger)^{-1}M'^{-1}M' =I .$
So $W=M'^\dagger(M^\dagger)^{-1}$ is the unitary $U$ satisfying the condition $M=M'U.$
Some implications of the one-density-matrix-many-decompositions situation:
$\rho$ does not contain all information of the quantum system. (It has a unitary degree of freedom.)
$p_k$ may not always be interpreted as probability. (It just corresponds to the probability of a possible decomposition.)
If $\Ket{\psi_i}$s are orthonormal, then $p_k$ can be determined as the eigenvalues of $\rho$, and information about the quantum system is all known.
On the last point, even when $\Ket{\psi_i}$s are not orthonormal, we can unitarily transform them to an orthonormal basis.
A Hermitian matrix is a normal matrix, defined as $A^\dagger A=AA^\dagger.$ ($\because~H^\dagger=H$ so $H^\dagger H=HH=HH^\dagger.$) A density matrix is Hermitian and therefire is normal.
Being "normal", a density matrix can be decomposed into orthogonal eigenbasis:
$\rho=U\Lambda U^\dagger,$
where $U$ is unitary and $\Lambda$ is a diagonal matrix of $\rho$'s eigenvalues.
(See Eigendecomposition.)
Also, the columns of a unitary matrix form an orthonormal basis. Let us call it $\{\Ket{u_i}\}$.
(See Unitary matrix.)
Therefore, the density matrix can be decomposed into $\rho=\sum_i\lambda_i\Ket{u_i}\Bra{u_i}.~$
Because $\rho\Ket{u_i}=\left(\sum_i\lambda_i\Ket{u_i}\Bra{u_i}\right)\Ket{u_i}=\lambda_i\Ket{u_i},~$ $\{\Ket{u_i}\}$ is the eigenbasis of $\rho$.
Therefore, we can conclude that any density matrices can be decomposed into a linear combination of its orthonormal eigenbasis.
Notes:
Whether $\Ket\varphi$ is the "true" decomposition or not, we can study it as if it is, because it would behave the same in measurement to other possible decompositions of the same mixed state.
If the eigenvalues are nondegenerate, the decomposition is unique. In case of degeneracy, one can always choose an arbitrary orthonormal basis in each degenerated subspace.
If $\{\Ket{\psi_i}\}$ is already orthonormal, the unitary is $I$.
While a pure state is a "single-term mixed state", there is a measurable distinction between a (multi-term) mixed state and a pure state in superposition.
For example, an unpolarised photon ensemble in a mixed state with equal distribution over all pure states would pass through a polariser 50% of the time regardless of the orientation of the polariser (as there bound to be some in the unpolarised ensemble that will pass). On the other hand, a polarised photon in a pure state would pass through a polariser with a probability depending on its orientation, and there is polariser that will allow 100% of such pure state photon to pass through, which can never happen in a multi-term mixed state.
To illustrate, $\Rsr2(\Ket0+\Ket1)$ is a pure state. The horizontal polariser will let 50% of its kind to pass, so will a vertical one. But the diagnoal one will pass through 100% of them.
An unpolarised stream of photons is in a mixed state, a classical statistical ensemble presumably evenly distributed in $[0,2\pi)$. Their combined probability is $\displaystyle\lim_{n\rightarrow\infty}{1\over n}\sum_{k=0}^{n-1}\big(\cos(2k\pi/n)\Ket0+\sin(2k\pi/n)\Ket1\big)=1.$
In continuous form: $\displaystyle\int_0^{2\pi}d\theta{1\over\sqrt{2\pi}}\left(\cos\theta\Ket0+\sin\theta\Ket1\right)=1.$
The projection on $\Ket0$ is $~~ \int_0^{2\pi}d\theta{1\over\sqrt{2\pi}}\cos\theta\Ket0$ with probability $\int_0^{2\pi}d\theta{1\over{2\pi}}\cos^2\theta ={1\over{4\pi}}\left(\theta+{1\over2}\sin(2\theta)\right)\big|_0^{2\pi} ={1\over2}.$
Project on $\Ket1$ will yield the same result of ${1\over2}$. By linear combination, this will hold true for any arbitrary projections.
Therefore, this stream of mixed state photons will have a projection on $\Rsr2(\Ket0+\Ket1)$ with a 50% probability, not 100%.
The above behaviour of measuring $\Ket0$ and $\Ket1$ with 50-50 chance cannot be expressed as $\Ket\xi=({1\over2}\Ket0+{1\over2}\Ket1)$, not even $\Rsr2(\Ket0+\Ket1)$, as the expected value cannot be obtained by $\Braket{\xi\lvert A\rvert\xi}.$
However, we can express the mixed state by $\rho={1\over2}\Ket0\Bra0+{1\over2}\Ket1\Bra1.$ This is called the Density Matrix, or the Density Operator. (Technically, the density matrix is obtained from the density operator by choice of basis.)
In this case, $\rho= \begin{bmatrix} {1\over2} & 0\\ 0 & {1\over2} \end{bmatrix} .$ To apply projector $P_0=\Ket0\Bra0=\begin{bmatrix}1&0\\0&0\end{bmatrix}$ on $\rho$ means $P_0\rho P_0=\begin{bmatrix}{1\over2}&0\\0&0\end{bmatrix}={1\over2}P_0,$ and the expected value of ${1\over2}$ is obtained.
A simple single-qubit mixed state is $\rho =p_0\Ket0\Bra0+p_1\Ket1\Bra1 =\begin{bmatrix} p_0 & 0\\ 0 & p_1 \end{bmatrix}$ where $p_0+p_1=1$.
The projector on the $X$-axis $\Rsr2(\Ket0+\Ket1)$ is $P_x =\Rsr2(\Ket0+\Ket1)\Rsr2(\Bra0+\Bra1) ={1\over2}(\Ket0\Bra0+\Ket0\Bra1+\Ket1\Bra0+\Ket1\Bra1) ={1\over2} \begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix} .$
Measuring $\rho$ with $P_x$ gives $P_x\rho P_x ={1\over2} \begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix} \begin{bmatrix} p_0 & 0\\ 0 & p_1 \end{bmatrix} {1\over2} \begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix} ={1\over4} \begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix} \begin{bmatrix} p_0 & p_0\\ p_1 & p_1 \end{bmatrix} ={1\over4} \begin{bmatrix} p_0+p_1 & p_0+p_1\\ p_0+p_1 & p_0+p_1 \end{bmatrix} ={p_0+p_1\over2} P_x .$
This means the success rate measuring $\rho$ on $P_x$ is ${p_0+p_1\over2}$, the average of the classical probability of $\Ket0$ and $\Ket1$ in $\rho$.
Generally speaking, a density matrix has $\Tr(\rho)=1$. But if it is a pure state, $\rho^2=\rho$ and therefore $\Tr(\rho^2)=1$. It is otherwise a mixed state, with $\Tr(\rho^2)<1$.
More technically speaking, if $\rho$ is a rank one matrix (columns span a single dimensional space), it is a pure state in the form of $\Ket\psi\Bra\psi$ where $\Ket\psi$ is of unit length, so $\Tr(\rho^2)=\Verti\Ket\psi\Verti^2=1.$
More details can be found in the article Measurement and Operators.
A summary:
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