<aside> 📌 Quite unlike system in classical mechanics, measuring a quantum mechanical system alters its state in an irreversible way.
</aside>
Consider the general qubit:
$\ket{\psi} = \alpha\ket{0} + \beta\ket{1}$
upon measurement, this qubit will be forced into either of the following two situations:
$\begin{matrix}\ket{\psi} \rarr \ket{0} & \ket{\psi} \rarr \ket{1}\end{matrix}$
hence, making the qubit loose its orignal state and making it impossible to determine the orignal values of $\alpha$ & $\beta$.
OPEN SYSTEMS
CLOSED SYSTEMS
Let’s recall a few facts we previously studied:
Time evolution of a closed(physically isolated) Quantum System is governed by the Schrödinger equation:
$i\hbar \large\frac{d}{dt}\small\ket{\psi(t)} = H\ket{\psi(t)}$
We then, derived the equation, giving us the state of the quantum system as a function of time:
$\small{\ket{\psi(t)}} = \large e^{\frac{-iHt}{\hbar}}\small\ket{\psi(0)}$
here, we then defined the Unitary evolution operator:
$U = \large e^{-i(\frac{Ht}{\hbar})}$
Finally, in the chapter “Density Operator for Pure States”, we defined the quantum system in terms of the density operator:
$\rho_t = U\rho_oU^\dagger$
Time-evolution of a quantum system (Dynamics) is trace-preserving, i.e. if the system is initially described by some density operator $\rho_0$ with $Tr(\rho_0)=1$, then after the system has evolved to a final state described by $\rho_t$, then $Tr(\rho_t) = 1$.
Measurement, unlike time evolution is described by trace-decreasing quantum operations. A quantum operation involving measurement is described by a measurement operator (denoted as $M_m$), transforms a density operator according to the equation:
$\rho’ = M_m\rho M_m^\dagger$ where, $Tr(\rho’) \le 1$
$P = P^\dagger$
$P^2 = P$
Product of two commutative operators is also a projection operator, i.e. if,
$P_1P_2 = P_2P_1$, then $P_1*P_2$ is also a projection operator.
Now, let’s establish a few facts:
Two projection operators $P_1$ & $P_2$ are orthogonal if their product is zero. That is, for every state $\ket{\psi}$ $P_1$ & $P_2$ are orthogonal if
$P_1P_2\ket{\psi} = 0$
A set of mutually exclusive measurement results corresponds to a set of orthogonal projection operators that act on the state space of the system. A complete set of orthogonal projection operators is one for which
$\sum_i P_i = I$
Every complete set of orthogonal projectors specifies a measurement that can be realized (at least one of the possible measurement results must be true ). The number of projection operators ($m$) is $\le$ dimension of the hilbert space that defines the system $d$, i.e.:
$m \le d$
for a qubit $\ket{\psi} = \alpha \ket{0} + \beta \ket{1}$, the projection operators corresponding to the mutually exclusive results $\ket{0}$ & $\ket{1}$ are:
$\begin{matrix}P_0 = \ket{0}\bra{0} & P_1 = \ket{1}\bra{1} \end{matrix}$
A given set of projection operators $\{P_1, P_2, P_3, \dots\}$ is said to be mutually orthogonal if:
$P_iP_j = \delta_{ij}P_i$, where $\delta_{ij} = \left\{\begin{matrix}0, & \textrm{i != j} \\ 1, & \textrm{i = j}\end{matrix}\right.$
Sum of two or more projection operators is a projection operator iff they are mutually orthogonal.
Probability of finding the ith outcome on measurement
Let, the dimension of the system be $n$, and consider a set of mutually orthogonal projection operators $\{P_1, P_2, P_3 \dots P_n\}$. Given, that the system is prepared in state $\ket{\psi}$, the probability of finding the ith outcome when a measurement is made, is given by: