The statement

For simplicity, let's consider a finite-dimensional Hilbert space. (The question can probably be generalized, but I don't know enough about mathematical QM to properly do so.)

Let $A\colon H\to H$ be some observable (self-adjoint operator) with eigenvalues $\lambda_1,\ldots,\lambda_n$. Recall that $H$ is the direct sum of the eigenspaces: \begin{equation}\tag{1} H=\bigoplus_{i=1}^n H_i \end{equation} In other words, \begin{align} \Phi\colon H_1\times\cdots\times H_n&\to H\\ (\Psi_1,\ldots,\Psi_n)&\mapsto \Psi_1+\ldots+\Psi_n \end{align} is a bijection and we can consider the projection $$P_i\colon H\to H_i$$ for each $i=1,\ldots,n$.

Born's rule says that$$p_i:=\frac{\langle P_i\Psi|\Psi\rangle}{\langle\Psi|\Psi\rangle}=\frac{\langle \Psi|P_i\Psi\rangle}{\langle\Psi|\Psi\rangle}=\frac{\langle P_i\Psi|P_i\Psi\rangle}{\langle\Psi|\Psi\rangle}\in[0,1]$$ is the probability to measure $\lambda_i$ if our system is in the state $\Psi$.

My question

Would the formulation of Born's rule above typically be regarded as an axiom or as a result of some more fundamental assumptions?

  • $\begingroup$ Your expression is a probability of "finding" the system in the state $\Psi_i$, to be exact. $\endgroup$ Sep 11, 2021 at 13:02
  • $\begingroup$ @VladimirKalitvianski Isn't the system in the state $\Psi_i$ after the measurement? $\endgroup$
    – Filippo
    Sep 11, 2021 at 13:07
  • $\begingroup$ If the measurement is elastic, then yes. $\endgroup$ Sep 11, 2021 at 15:53
  • $\begingroup$ @VladimirKalitvianski What is an elastic measurement? $\endgroup$
    – Filippo
    Sep 11, 2021 at 16:29
  • $\begingroup$ Like an elastic scattering, when the target atom stays in the initial state after scattering. $\endgroup$ Sep 11, 2021 at 17:15

1 Answer 1


The correct statement is that the probability that a measurement of an observable represented by a Hermitian operator $A$ (with non-degenerate spectrum) over a state $\vert\psi\rangle$ would yield an eigenvalue $\lambda_i$ is given by

\begin{align}p_i=\frac{\langle\psi\vert\psi_i\rangle\langle\psi_i\vert\psi\rangle}{\langle\psi\vert\psi\rangle}\end{align} where $\vert\psi_i\rangle$ is the normalized eigenstate of the operator $A$ corresponding to the eigenvalue $\lambda_i$. However, this does not require that $\vert\psi\rangle=\sum_i\vert\psi_i\rangle$. The state vector $\vert\psi\rangle$ can be the most generic normalizable state and thus, would be represented, in general, as a generic linear combination $\vert\psi\rangle=\sum_ic_i\vert\psi_i\rangle$ where $c_i\in\mathbb{C}$.

This statement is called the Born rule.

It is needed to be supplied with a closely related axiom that goes by the name of the collapse postulate or the wavepacket reduction postulate to give a "complete" picture of what happens when you perform a measurement. It says that the aforementioned measurement evolves the state $\vert\psi\rangle$ to an eigenstate $\vert\psi_i\rangle$ corresponding to the outcome $\lambda_i$.

All of this can be made a bit more general to take care of measurements of operators with degenerate spectra using the projection operators, but the basic idea is already captured here. In the case of the measurement of an operator $A$ with distinct eigenvalues $\lambda_i$ such that $A=\sum_i\lambda_i\mathbb{P}_i$ where the $\mathbb{P}_i$s are the projection operators corresponding to the $i^\mathrm{th}$ eigensubspace, the probability of the outcome of the measurement yielding $\lambda_i$ is given by


The wavepacket reduction postulate now says that the aforementioned measurement evolves the state $\vert\psi\rangle$ to the state $\frac{\mathbb{P}_i\vert\psi_i\rangle}{\langle\psi\vert\mathbb{P}_i\vert\psi\rangle}$ corresponding to the measurement outcome being $\lambda_i$. Notice that the denominator here is needed to ensure that the resultant state is normalized.

In standard textbook quantum mechanics, both of these are always, as far as I know, taken to be basic axioms. One can formulate their quantum mechanics using a different mathematical formalism but they still have to provide some translation of these axioms as axioms in their framework as well -- as long they really are just another formulation of the standard textbook quantum mechanics in their physical content.

Having said that, there have been attempts, starting in 1957 and continuing to this day, to derive the Born rule. There have been mainly three approaches to attempt the derivation:

  • Measure-Theoretic/Frequentist Approaches

  • Symmetry-Based Approaches

    • The 2005 paper by Zurek derives the Born rule using an argument based on envariance which is an invariance that systems entangled with an environment exhibit.
    • The 2015 paper by Carroll and Sebens derives the Born rule in the context of many-worlds formulation of quantum mechanics. They use the "epistemic separability principle" which is just a weird/fancy way of saying that the probability of a measurement outcome shouldn't depend on the evolution of the environment that is decoupled and unentangled from the system.
  • Decision-Theoretic Approaches

    • I simply mention them for the sake of completeness and to invite a more informed reader to feel free to edit the answer and fill in the details.

Now, none of these attempts have been accepted, at least so far, by the community as true derivations of the Born rule. Basically, in standard quantum mechanics, there is no plausible way to do away with the wave-packet reduction axiom (which ought to accompany the Born rule for probabilities to make sense, otherwise there would simply be deterministic evolution according to the Schrodinger equation). So, even if one shows that the Born rule is the only consistent probability measure for the Hilbert spaces of quantum mechanics, it does not come in contact with the physical claims made by the standard axioms. Another approach, in particular, the papers by Carroll and Deutsch (the latter of whom has worked on decision-theoretic approaches) are in the framework of the many-words formulation. There, you can make sense of wavepacket reduction as the reduction of the relative state of a system with respect to an observer without violating underlying unitarity. However, it is conceptually difficult to derive the Born rule there. One reason is that the naive branch-counting leads to a contradiction with the Born rule. And the more sophisticated epistemic approaches have been criticized for either being circular or sloppy.

You can see the critiques of the derivations of the Born rule in papers by Adrian Kent, 1997 and 2014. I would also recommend having a look at this answer to my recent question by @ChiralAnomaly for some general comments on the derivations of the Born rule.

  • $\begingroup$ Thank you very much for the elaborate answer. Before reading further, I'd like to comment on the first paragraph, because I have the impression that there has been a misunderstanding. In my question, $\Psi_1,\ldots,\Psi_n$ is NOT an eigenbasis. The reason I used my formula is that it is basis-independent. In addition, I did not assume the eigenspaces are $1$-dimensional. $\endgroup$
    – Filippo
    Sep 11, 2021 at 11:17
  • $\begingroup$ I just compared our formulas. Are you sure that your formula for $p_i$ is correct? I read the Wikipedia article on Born's rule (thank you for mentioning the Born rule), and as far as I understand, the $\langle\Psi_i|\Psi_i\rangle$ in the denominator of your formula needs to be removed. But I might be wrong. $\endgroup$
    – Filippo
    Sep 11, 2021 at 11:18
  • $\begingroup$ @Filippo Yes, it is standard practice to normalize states such that their norm equals unity. The Wikipedia article works in the notation where the eigenstates are normalized but the state of the system might not be, and thus, they include the norm of the state vector in the denominator but not the norm of the eigenstates. I included both norms in the denominator just to allow for the convention where you don't normalize eigenstates (which would be a very bad practice in reality). $\endgroup$
    – ACat
    Sep 11, 2021 at 12:15
  • $\begingroup$ @Filippo Regarding your first comment, it does not make sense to talk about the probability of a certain measurement outcome without considering the inner product of the eigenstates with the state vector. It is already as basis independent as you can get, I am already working in the Dirac notation. No matter what basis you choose, an eigenstate of a given operator would remain an eigenstate of the given operator. [...] $\endgroup$
    – ACat
    Sep 11, 2021 at 12:19
  • $\begingroup$ [...] I am not sure what you are doing with the inner products of some arbitrary vectors $\Psi_i$ if they are not the eigenstates of the given operator. The only vectors at play in this discussion should be the eigenstates of the given operator and the state vector of the system. Regarding the dimensionality of the eigenspaces, you are correct, but as I mentioned, the case of degenerate eigen-subspaces can be straightforwardly handled using the corresponding projection operators. I will add a bit more explicit clarification on that point. $\endgroup$
    – ACat
    Sep 11, 2021 at 12:21

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