Defined by $P$, it is obtained by taking the outer product of a given state $\ket{\psi}$, i.e.
$P = \ket{\psi}\bra{\psi}$
PROPERTIES
A projection operator is Hermitian.
$P = P^\dagger$
If the state $\ket{\psi}$ (via which the operator $P$ is obtained) is normalized, then the projection operator is equal to its own square:
$P = P^2$
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.
Given a vector space that has $n$ dimensions and a basis given by $\ket{1},\ket{2},\dots,\ket{n}$, the operator given by
$P = \sum_{i=1}^{m}\ket{i}\bra{i}$,
projects on to the subspace (of the $n$ dimensional space we just defined) spanned by the set $\ket{1}, \ket{2}, \dots \ket{m}$ .
REWRITING SPECTRAL DECOMPOSITION THEOREM
Recalling the result of the spectral decomposition theorem, any operator $A$ can be represented as follows in terms of its eigenvalues and (some) basis:
$A = \sum_{i=1}^{n}a_i\ket{u_i}\bra{u_i}$
*here, $a_i$ ***turn out to be the various eigenvalues of the operator & $\ket{u_i}$ represents the basis set.
The very same result, can now be rewritten in terms of projection operators:
$A = \sum_{i=1}^{n}a_iP_i$ , here $P_i = \ket{u_i}\bra{u_i}$
Further, since the basis states satisfy the completeness relation (recall from the matrices and operators chapter), we have :
$\sum_{i=1}^{n}\ket{u_i}\bra{u_i} = \sum_{i=1}^{n}P_i = I$
clearly, $I$ being the identity matrix
Example
In most of the scenarios, while studying quantum computing we observe(study) a large number (say $N$) or collection of quantum systems (rather than a single) called an ensemble.
The members(quantum systems) of this ensemble can themselves be found in one of two or more different quantum states.
PROBABILITY DISTRIBUTION
Consider a 2D Hilbert space with a basis vector set $\{\ket{x}, \ket{y}\}$. Further, letโs construct $N$ quantum systems from this basis, where each system may exist in one of the two vector states (normalized):
TWO LEVELS OF PROBABILITY:
โ At the level of single quantum system:
<aside> ๐ Here, the probabilites are calculated via the Born Rule, which states that the probability of obatining a measurement is proportional to the square of its amplitude.
</aside>
Clearly, as studied previously:
โ On the ensemble level:
<aside> ๐ Here, the probabilites are calculated in a classical manner as at the ensemble level we have a statistical mixture.
</aside>
Now, letโs assume that out of the $N$ systems we prepared, $n_a$ quantum systems have the state $\ket{a}$ and the remaining $n_b$ systems have the state $\ket{b}$. ($n_a + n_b = N$). This gives us that the probability of finding a system in state: