Adapted from J.J. Sakurai, UW Physics 248A, and Chem 221A.
Measurement
Having established the postulates of quantum mechanics, what does measurement really mean? Let’s formulate this concept with the idea of kets and operators, while slowly breaking down the words of Paul Dirac:
“A measurement always causes the system to jump into an eigenstate of the dynamical variable that is being measured.”
Let’s say we want to measure something about an observable $\hat A$. Recall that the postulates tell us that we can write the system of interest (before doing any measurements), $\ket \alpha$, as some linear combination of $\hat A$'s eigenvectors. $$\ket \alpha = \sum_{a’}c_{a’}\ket {a’}=\sum_{a’}\ket{a’}\bra{a’}\alpha \rangle$$ And when a measurement is made, the system then goes from a general linear combination of $\hat A$'s eigenvectors to one of the eigenvectors, say $\ket{a’}$. Thus, fundamentally, a measurement changes the state (except if the original system is already an eigenvector of $\hat A$). How do the eigenvectors of $\hat A$ get chosen then? This goes back to the postulate where we claim that the probability of “jumping” into some particular state $\ket{a’}$ is given by, assuming $\ket \alpha$ is normalized: $$\text{Probability for} \ket{a’} = |\bra{a’}\alpha\rangle|^2$$
Projection operators
What does the first equation mean exactly? Does the position of a ket and which way it faces matter? We are used to having the kets and bras arranged in such a way $\bra {}\rangle$ that it always give us a scalar. This is the analogous way of looking at a vector transpose multiply by itself, i.e. $\pmb{v^Tv}$. Let’s say the dimension of $\pmb{v}$ is $m \times 1$, then clearly, $\pmb{v^Tv}$ has dimension $1 \times 1$, a scalar. Let’s say we have $\pmb{vv^T}$, the new dimension is now $m \times m$, a matrix. We call this the outer product, and in dirac notation, it is given by $\ket{}\bra{}$. How does this exactly translate in the language of dirac notation and world of quantum mechanics?
Recall our Stern-Gerlach system and the corresponding states $\ket +$ and $\ket -$ (call them the z axis spin states). Furthermore, we can construct some state $\ket \psi$ as $\ket \psi = a\ket + + b\ket -$. The coefficients can be written as $a = \bra + \psi \rangle$, and $b = \bra - \psi \rangle$ (think about them as vectors and how you can get coefficients of a linear combination of a vector). Looking at the first term, we have some scalar $a$ multiplying by a ket $\ket +$; hence, it doesn’t matter which side we put $a$, we yield the same result.
$$a\ket + = \ket + a = (\ket +)\bra + \psi \rangle$$
However, multiplication of kets preserves distributivity (same as vectors), hence we can move the parenthesis around, which we get
$$ (\ket +)\bra + \psi \rangle = (\ket + \bra +) \ket \psi$$
So far, we have done nothing to the first term other than moving parenthesis around. The term in parenthesis, however, is something new (i.e $(\ket + \bra +)$). Let’s justify it with dimensionality analysis. We know that $a\ket +$ is a ket, and we also know that $(\ket + \bra +)\ket \psi$ is also a ket. However, if we try to say the parenthesis term is also a scalar like $a$, the dimension is wrong. Then, the only object that can give us a ket from a ket is an operator. Performing the same substitution on the second term $b\ket -$, we get
$$\ket \psi = (\ket + \bra +)\ket \psi + (\ket - \bra -)\ket \psi$$
$$= (\ket + \bra + + \ket - \bra -)\ket \psi$$
The term $(\ket + \bra + + \ket - \bra -)$ is a sum of operators, and hence it is also an operator, and particularly, it is the identity operator that preserves the state of a ket.
More concretely,
\[ \begin{aligned}
|+\rangle\langle+|+|-\rangle\langle-| & \doteq\left(\begin{array}{l}
1 \\
0
\end{array}\right)\left(\begin{array}{ll}
1 & 0
\end{array}\right)+\left(\begin{array}{l}
0 \\
1
\end{array}\right)\left(\begin{array}{ll}
0 & 1
\end{array}\right) \\
& \doteq\left(\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right)+\left(\begin{array}{ll}
0 & 0 \\
0 & 1
\end{array}\right) \\
& \doteq\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)
\end{aligned} \]
The individual operators that make up the identity operator is called the projection operator, i.e. $P_+ = \ket + \bra +$ and $P_- = \ket - \bra -$. When a projection operator is acted on a ket, say $\ket \psi$, it produces a ket that is aligned along the eigenvector that makes up the projection operator with magnitude equal to the amplitude of $\ket \psi$ in that eigenvector. For example, $P_+ \ket \psi= (\ket + \bra + )\ket \psi = (\bra + \psi\rangle) \ket +$, where $\bra + \psi\rangle$ is the so called amplitude.
Also note that the projection operator acting on the eigenstate (from now, we will interchange the word eigenvector and eigenstate) that constructs it becomes the identity operator, i.e. $$P_+\ket + = (\ket + \bra +) \ket + = \ket + (\bra + +\rangle) = \ket +$$
And acting on a different eigenstate that does not construct the projection operator always give 0, i.e.
$$P_+\ket - = (\ket + \bra +) \ket - = \ket + (\bra + -\rangle) = 0$$
Measurement, again
Let’s return to the postulate that tells us the probability of measuring some eigenstate. For the sake of consistency, we will continue to denote things in the z axis spin states. $$\text{Probability for} \ket{+} = |\bra{+}\psi\rangle|^2$$ $$ = \bra + \psi \rangle ^* \bra + \psi \rangle $$ $$ = \bra \psi + \rangle \bra + \psi \rangle$$ $$ = \bra \psi P_+ \ket \psi$$ This shows us that we can restate the probability as the expectation value of the projection operator! In summary, we have seen that the projection operator acts on a state $\ket \psi$ and reorients it in the direction of the eigenstate that constructs the projection operator, with an amplitude that is given by the square root of the probability measuring that eigenstate. Mathematically: $$ \left|\psi^{\prime}\right\rangle=\frac{P_{+}|\psi\rangle}{\sqrt{\left\langle\psi\left|P_{+}\right| \psi\right\rangle}}=|+\rangle $$
Now we are ready to formulate the measurement postulate:
After a measurement of $\hat A$ that yields the result $a_n$, the quantum system is in a new state that is the normalized projection of the original system ket onto the ket (or kets) corresponding to the result of the measurement: $$ \left|\psi^{\prime}\right\rangle=\frac{P_{n}|\psi\rangle}{\sqrt{\left\langle\psi\left|P_{n}\right| \psi\right\rangle}} $$
Effectively, the old state collapsed into a new state, and hence the alternative name collapse postulate is sometimes used.