In general, the action of an observable operator on a quantum state is a projection into one of the operator's eigenstates. However, the action of an operator on a state is written $$ \hat A|\psi\rangle= \hat A\sum c_n|a_n\rangle=\sum c_na_n|a_n\rangle .$$

This has not executed the projection operation. Namely, a measurement of $\hat A$ should be $$ \hat A|\psi\rangle\to|a_k\rangle ,$$

so that if eigenvalue $a_k$ is obtained from the first measurement, any number of rapidly repeated measurements will also yield $a_k$. The operator $\hat A$ has projected $\psi$ into the 1D eigenspace spanned by $|a_k\rangle$. Unfortunately, I have not written an equation that shows that. I have only used the $\to$ symbol to say "and this where the magic happens." Is it required that after finding eigenvalue $a_k$, I must operate with $\hat P_k(\psi)=\frac{1}{c_k}|a_k\rangle\langle a_k|$: $$ \hat P_k(\psi)|\psi\rangle=|a_k\rangle ?$$

That extra step seems cumbersome and clunky but I am not sure which is the notation I am looking for.

Q: What formalism is standard for the mathematical statement that the act of measurement collapses a state to an eigenstate: $$ f\big(|\psi\rangle\big)=|a_k\rangle? $$

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    $\begingroup$ "the action of an observable operator on a quantum state is a projection into one of the operator's eigenstate". I mean, no, that's not what the action of the operator on the quantum state does. The action of the operator corresponding to the observable has basically nothing to do with the measurement of that observable. You have to define the projection operators onto the eigen-subspaces of the observable separately, and it is the action of those projection operators that makes up the measurement process. $\endgroup$
    – march
    Mar 11, 2022 at 5:42
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    $\begingroup$ suggest you try my answer to this question: physics.stackexchange.com/questions/457908/…; this should clear it up $\endgroup$ Mar 11, 2022 at 10:29

1 Answer 1


The most general formalism to describe the action of a quantum measurement is the POVM formalism (which stands for 'positive operator-valued measure'). In its most general form, a measurement is described by a measurement operator $M_k$ via

$$|\psi_k\rangle \rightarrow \frac{M_k |\psi\rangle}{\sqrt{\langle\psi|M^\dagger M|\psi\rangle}}$$

$M_k^\dagger M_k$ is called the POVM, which must satisfy the normalization condition

$$\sum_k M_k^\dagger M_k = I$$

The only difference between this and what you wrote is that the measurement operator does not contain the state-dependent normalization factor. For standard projective measurement of a Hermitian operator like your $\hat{A}$, we just take $M_k = |a_k\rangle\langle a_k|$, where $|a_k\rangle$ are the normalized eigenvectors of $\hat{A}$.

All of the above isn't really different from what you wrote at the end of the day, and there's no need to distinguish $M_k$ from $M_k^\dagger M_k$ for projective measurements. But it let's you handle measurements that can't be described as perfect projections, such as weak measurements. It's also nice that it describes time evolution by taking $M_0 = U$ where $U$ is unitary. So it unifies the two main postulates of quantum mechanics into one equation.


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