# Understanding the Born Rule

## Main Question or Discussion Point

We are taught that the probability of detecting a particle is <ψ*|ψ>. I understand it is a postulate of QM and required by the maths, but have often wondered as to why it seems to represent a joint probability. For a while it seemed to make some sense to me in the time symmetric QM interpretation, namely two halves making a whole so to speak. Today I read this on the Relativity Forum which suggests time has no particular significance and arrives a the Born probability another way.

http://uk.arxiv.org/pdf/gr-qc/0306059v1

My question is, can it be said that this formulation represents a joint probability and if so of what.

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stevendaryl
Staff Emeritus
We are taught that the probability of detecting a particle is <ψ*|ψ>. I understand it is a postulate of QM and required by the maths, but have often wondered as to why it seems to represent a joint probability. For a while it seemed to make some sense to me in the time symmetric QM interpretation, namely two halves making a whole so to speak. Today I read this on the Relativity Forum which suggests time has no particular significance and arrives a the Born probability another way.

http://uk.arxiv.org/pdf/gr-qc/0306059v1

My question is, can it be said that this formulation represents a joint probability and if so of what.
What do you mean by "a joint probability"?

The probability of A and B occurring. Sorry if that is not the right terminology.

vanhees71
Gold Member
2019 Award
Born's rule is one of the fundamental postulates of quantum theory and should be carefully stated. I don't know, why you quote the paper from the arXiv in your first posting. I'd not look into complicated issues like quantum theory in curved space time of General Relativity, before I've had a clear grasp of the foundations.

In quantum theory the (pure) states are represented by rays in Hilbert space, i.e., with each normalized vector $|\psi \rangle$ in Hilbert space (normalized: $\langle \psi|\psi \rangle=1$) any vector $|\psi' \rangle=\exp(\mathrm{i} \varphi) |\psi \rangle$ represents the same state.

Any observable $A$ is represented by a self-adjoint operator $\hat{A}$ of the Hilbert space, and the observable can take only values belonging to the spectrum (generalized eigenvalues) of $\hat{A}$. For simplicity let's assume the observable is not degenerate, i.e., for each (generalized) eigenvalue $a$ of $\hat{A}$ the eigenspace is one-dimensional. The (generalized) eigenvectors then build a complete set of orthogonal (generalized) vectors. For simplicity we assume that all eigenvectors are proper eigenvectors and thus the eigenvalues form a complete set. Then the completeness relation and orthonormality of the eigenbasis can be written as
$$\sum_{a} |a \rangle \langle a|=\hat{1}, \quad \langle a | a' \rangle=\delta_{a,a'}.$$

Then the Born rule can be formulated as follows: If a system is known to be in the quantum mechanical (pure) state, represented by a ray in Hilbert space with any normalized state vector $|\psi \rangle$ belonging to this ray, the probability to measure the value $a$ for the observable $A$ is given by
$$P(a|\psi)=|\langle a|\psi \rangle|^2.$$
Note that this probability is independent of the choice of the representing vector of the ray in Hilbert space (due to the modulus squared). The probability is also properly normalized to one, i.e., to measure some value for $A$ is trivially always certain:
$$\sum_a P(a|\psi)=\sum_a |\langle a|\psi \rangle|^2= \sum_a \langle \psi |a \rangle \langle a|\psi \rangle = \langle \psi|\psi \rangle=1.$$
Here we have used the completeness of the eigenbasis of the self-adjoint operator $\hat{A}$ and the normalization of the state vector to 1.

All this can be refined for the case of observables whose representing self-adjoint operators have a continuous (or both a descrete and a continuous) spectrum and that are degenerate. In the latter case you need to measure in addition other observables $B_1,B_2,\ldots, B_n$ that are compatible with $A$ in addition to completely label an eigenvector in the eigenspace of $\hat{A}$ for the (generalized) eigenvalue $a$. A complete set of compatible observables is represented by the corresponding set of self-adjoint operators that mutually commute, i.e.,
$$[\hat{A},\hat{B}_i]=[\hat{B}_i,\hat{B}_j]=0.$$
Then there exists a complete set of common eigenvectors $|a,b_1,\ldots,b_n \rangle$ that are orthonormalized (again assuming we have only true discrete eigenvalues and no continuous parts in the spectra of the operators):
$$\sum_{a,b_1,\ldots ,b_n} |a,b_1,\ldots , b_n \rangle \langle a,b_1,\ldots, b_n \rangle.$$
Then the probality to measure values $(a,b_1,\ldots, b_n)$ when jointly measuring the observables $A,B_1,\ldots, B_n$, supposed the system is prepared in the state represented by the normalized vector $|\psi \rangle$, is given by
$$P(a,b_1,\ldots, b_n|\psi)=|\langle a,b_1,\ldots, b_n|\psi \rangle|^2.$$
If you only measure $A$, the probability to find a possible value $a$ is then of course given by
$$P(a|\psi)=\sum_{b_1,\ldots, b_n} P(a,b_1,\ldots, b_n|\psi).$$

Thanks Vanhees71. I follow your post and am aware of the points you make. If you are not familiar with the paper I am referring to pages 12 and 13 which discusses how the variables of the extended configuration space can be interpreted as partial observables.

This is not my first post.

Hi to all. I searched the net and there is a lot of people confused about this rule or postulate. In feynmans lectures when he was explaining the quantum behaviour even for the water wave interference he took the same wave intensity postulate (wave function squared) and derive the interference probability. So in classical point of view i deduce that tthis is done to eliminate the negative amplitude of the waves. I hope i can clear myself.
Thanks.