/ #Mixed States #Density Operator

# Mixed States and General Quantum Operations: A simple view on the framework of Quantum Mechanics

In this text I’ll do some review on the definition of quantum state and quantum measurement, which was introduced here, and reasonably expand the notions into the most general instances of quantum bits.

#### Review

Given a state $$|\psi\rangle = \sum_i \alpha_i |\varphi_i \rangle,$$ it’s possible to perform a Von Neumann measurement on the state’s space $\mathscr{H}$ with respect to the orthonormal basis $B = \{|\varphi_i \rangle\}$. The output would be a label $i$ with corresponding probability $|\alpha_i|^2$ and the system left in state $|\varphi_i\rangle$.

A Von Neumann measurement can be discribed as “complete” or “maximal” if every orthogonal projector $P_i: \sum_i P_i = I$ is of of the form $|\psi_i\rangle \langle\psi_i|$ for a normalized state $|\psi_i\rangle$.

The simplest version of a Von Neumann measurement is a complete measurement in the computational basis, i.e. $P_i = |i\rangle\langle i|$ where $i \in \{0,1\}^n$.

In general, a projective measurement is also known as an observable. An observable is a Hermitian operator $M$ that acts on the system state space. The spectral decomposition of a Hermitian is $$M = \sum_i m_iP_i$$ where $P_i$ is the orthogonal projector onto the eigenspace of M that has real eigenvalue $m_i$. Measuring an observable returns an outcome $i$ corresponding to the eigenvalue $m_i$.

Example: For Pauli observable $Z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$, its spetral decomposition is $Z = |0\rangle\langle 0| - |1\rangle\langle 1|$ It can be seen that two eigenvalues of $Z$ are $1$ and $-1$ corresponding to eigenstates $|0\rangle$ and $|1\rangle$ respectively. The observable contains two measurement projectors $|0\rangle\langle 0|$ and $|1\rangle\langle 1|$ so that a measurement of the Pauli observable $Z$ is a measurement in the computational basis. Reversely speaking, if a system is said to be measured in the computational basis, it’ll under a measurement of $Z$-Pauli measurement.

Of course, any $2\times2$ matrix $A = U^{\dagger}ZU$ for any unitary matrix $U$ also satisfies the criteria of having $1$ and $-1$ as eigenvalues, except that the corresponding eigenstates are different from those of $Z$. This fact helps build a procedure for measuring other Pauli observables, such as $X$ and $Y$. These measurements are given below $$\begin{matrix} \text{Pauli measurement} & U \\ Z & I \\ X & H \\ Y & HS^{\dagger} \end{matrix}$$ You could verify that the measurement of $X$ observable is a complete Pauli measurement with respect to the basis $\{ |+\rangle,|-\rangle \}$

The formalism of observable engenders a convenient way to evaluate the expected value of $m_i$, which relates to physical quantities relevant to our system. In an instance when our system is in state $|\psi\rangle$, this expectation value can be represented as $$E(m_i) = \text{Tr}(M|\psi\rangle\langle \psi|)$$

#### Mixed State and Density Matrix

From the beginning, we have always assumed that a quantum state would be described by a definite column vector. Characterizing the state of a qubit with a vector is, however, only true if we know which state it’s in with perfect certainty. In contrast, for example, if we’re unsure whether a qubit is in either state $|\psi_1\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ or state $|\psi_2\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$, in which form we can represent that qubit. Here the concept of mixed state emerges.

A mixed state, or a mixture of states, is all about your uncertainty of information. While you know exactly what information a pure state offers, a mixed state is just a combination of multiple pieces of information with respective probabilities. In general, we can write a mixed state as a set of multiple state-probability pairs. $$\{(\psi_1,p_1),(\psi_2,p_2),(\psi_3,p_3),…,(\psi_n,p_n)\}$$, where $|\psi_i\rangle$ is a pure state. Nevertheless, it would be really cumbersome if we need do some calculation with such a representation. For that reason, scientists came up with an alternative representation that raises the useful aesthetics of simplicity while keeping mathematical accuracy. It’s called density operator. Like operators describing quantum operations, density operators are in accompany with their matrix representation, or density matrix.

Since a pure state $|\psi\rangle$ is a special instance of a mixed state that there is only one state element, it can be written in density operator as $$\rho = |\psi\rangle \langle\psi|$$ In general, the density operator for such a set of pure states and their corresponding probabilities as in the above is $$\rho = \sum_{i=1}^n p_i |\psi_i\rangle \langle\psi_i|$$ If a mixed state can be written in the form above, it’s separable. Otherwise, it’s entangled. So, what is the difference between a superposed state and a mixed state. For example, in which sense pure state $\psi = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ differs from density operator $\rho = \frac{1}{2}|0\rangle \langle0| + \frac{1}{2}|1\rangle \langle1|$. They are both a combination of $|0\rangle$ and $|1\rangle$ with equal probabilities of $\frac{1}{2}$! The short answer is that these notations describe different types of uncertainty. Because there is more than one way to measure s state, i.e. you can choose any arbitrary basis to perform your measurement in. That choice of basis, in principle, allows you to store and retrieve information. In the context of multi-basis measurement, superposition helps express a state in a basis different from the computational basis. On the other hand, a mixture highlights the presence of probabilistic elements, no matter which basis you choose to observe the state. Please note that density matrices doesn’t match one-to-one with mixed states, so two mixtures might have the same density matrix, just like two pure state that are different by only a global phase.

#### General Quantum Operations

Recalling that $U|\psi\rangle$ is the resulting state as one applies an unitary operator $U$ to $|\psi\rangle$. If described in the form of density operator, it would be $U|\psi\rangle \langle\psi|U^{\dagger}$ - what we need to remember about applying an operator to a mixed state.

So, how about an operator that is measurement operator? For a pure state $|\psi\rangle = \sum_i a_i|\varphi_i\rangle$ where $a_i = \langle \varphi_i| |\psi\rangle = \langle \varphi_i|\psi\rangle$, $$|a_i|^2 = a_i^*a_i = \langle \psi |\varphi_i \rangle \langle \varphi_i | \psi\rangle$$ This equation holds for measurements of density matrix. Considering a measurement of any arbitrary density matrix $\rho = |\psi\rangle \langle \psi|$ in a basis of interest, the probability of the measurement giving a suitable result $\phi$ is given by $$\langle \phi |\psi\rangle \langle \psi| \phi \rangle = \langle \phi |\rho| \phi \rangle$$ The probability is a real number that can be called as the only element and the trace of a $1\times 1$ matrix. This allows us to rewrite the probability as $$\langle \phi |\psi\rangle \langle \psi| \phi \rangle = \text{Tr}(\langle \phi |\psi\rangle \langle \psi| \phi \rangle ) = \text{Tr}(|\phi\rangle\langle \phi| |\psi\rangle\langle \psi|)$$ because of the cyclicity of trace: $\text{Tr}(ABC) = \text{Tr}(BCA) = \text{Tr}(CAB)$.

Now we’re gonna expand the entire things above in this section to the level of a general mixed state. The process of applying operator $U$ to $\rho = \sum_{i=1}^n p_i |\psi_i\rangle \langle\psi_i|$ is, in fact, getting $U$ apply to each state possibility $|\psi_i\rangle$ of $\rho$. Mathematically, such an operation will transform the mixed state into $$\sum_{i=1}^n p_i U|\psi_i\rangle \langle\psi_i|U^{\dagger} = U \left(\sum_{i=1}^n p_i |\psi_i\rangle \langle\psi_i|\right) U^{\dagger} = U\rho U^{\dagger}$$ Similarly, measuring the general mixed state is like adding the all the traces with proper weight. $$\begin{matrix} \displaystyle{\sum_i \text{Tr}(|\phi\rangle\langle\phi| |\psi\rangle\langle\psi|)} \\ \displaystyle{= \text{Tr}\left(\sum_i p_i |\phi\rangle\langle\phi| |\psi\rangle\langle\psi| \right)} \\ \displaystyle{= \text{Tr}\left(|\phi\rangle\langle\phi| \sum_i p_i |\psi\rangle\langle\psi|\right)} \\ \displaystyle{= \text{Tr}(|\phi\rangle\langle\phi|\rho)} \end{matrix}$$ The last two equations regarding applying operators and measuring showcase that the only thing that matters is the density matrix itself, not the precise decomposition to its pure state elements. That being said, two mixtures can have the same density matrix, thus are indistinguishable or completely equivalent even though they were created by two different sets of pure states.

#### Entangled Cat

I work hard so that my cats can live a better live