# About Born's rule

I wanted to gain a better understanding of the Born rule to make my class on quantum mechanic feel less ad hoc. To do so I attempted to show that the version (1) given in my book is equivalent to the version (2) on Wikipedia.

1. The version in my book:

$P_x = \frac{\langle\psi_x\mid\psi_x\rangle}{\langle\psi\mid\psi\rangle}$ Where $P_x$ is the probability of getting a state $\psi_x$ when measuring.

2. Version on Wikipedia:

the probability of measuring a given eigenvalue $\lambda_i$ will equal $\langle\psi\mid P_i\mid\psi\rangle$, where $P_i$ is the projection onto the eigenspace of A corresponding to $P_i$.

There is a short explanation but I would be glad if someone could put it into better words. Specifically I don't understand what $P_i$ is. I do know about the Hermitian operator, its eigenvalues and eigenspaces. But for example, why is $P_i = \mid\psi_i\rangle\langle\psi_i\mid$ when the eigenspaces are one-dimensional. And how do I go from there to the form in my book?

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## 1 Answer

$P_i = \mid\psi_i\rangle\langle\psi_i\mid$ is the one-dimensional "projection" operator. By "one-dimensional" it means this projection operator projects $\psi$ onto a single dimension in Hilbert space.

Firstly, any wavefunction $\psi$ can be written as a linear combination of orthogonal components. That is, $\psi = \sum a_i\mid\psi_i\rangle$ where $a_i$ is some coefficient. If there are $n$ such non-zero coefficients, $\psi$ can be thought of as a vector in $n$ dimensions, having components in each direction of length $a_i$ in this $n-dimensional$ Hilbert space. $a_i$ is also the amplitude that the result $\psi_i$ will be obtained if the wave-function is measured in this basis. The probability is amplitude^2.

The projection measurement essentially "projects" the state $\psi$ onto one of these components.

It is easiest to demonstrate why $P_i = \mid\psi_i\rangle\langle\psi_i\mid$ by applying it to the state $\psi$.

$P_i\psi = P_i\sum a_k\mid\psi_k\rangle = \mid\psi_i\rangle\langle\psi_i\mid\sum a_k\mid\psi_k\rangle = \mid\psi_i\rangle\sum a_k\langle\psi_i\mid\psi_k\rangle = a_i\mid\psi_i\rangle$

Therefore, the operator $P_i$ acting on some arbitrary state $\psi$, projects $\psi$ onto its i-th component vector. (This is analogous to projecting a 2D vector onto say its x-component in Euclidean geometry. For instance if a vector $V = ax + by$, then the projection onto x-axis would yield $V_x = ax$)

So since, $P_i\mid\psi\rangle=a_i\mid\psi_i\rangle$ we can easily show that $\langle\psi\mid P_i\mid\psi\rangle=\sum \langle\psi_k\mid a_k^*a_i\mid\psi_i\rangle = \sum a_k^*a_i\langle\psi_k\mid\psi_i\rangle = a_ia_i^* = \mid a_i\mid^2$

Therefore we have shown that $\langle\psi\mid P_i\mid\psi\rangle$ gives the probability of the wavefunction being in the eigen-state $\psi_i$.

The next step is to show how the one in your book is also the probability. Note your book's use of $P_x$ is to represent probability and is not the projection operator. First consider the denominator $\langle\psi\mid\psi\rangle = \sum^j\sum^k\langle\psi_j\mid a_j^* a_k\mid\psi_k\rangle$. The only terms that survive is when i=j. Therefore we arrive at:

$\langle\psi\mid\psi\rangle = \sum^j\langle\psi_j\mid a_j^* a_j\mid\psi_j\rangle = \sum^j a_j^*a_j = \sum^j \mid a_j\mid ^2$ This is the total probability of any state which, if this is normalized, should be 1. Therefore $\langle\psi\mid\psi\rangle = 1$ for normalized states, otherwise it is the sum of all possible amplitudes^2.

Next we consider the numerator. This is the dot product of the x-components of the state, which will yield $a_x^* a_x = \mid a_x \mid ^2$ Therefore, numerator over denominator gives $\frac{\mid a_x \mid ^2}{\sum^j \mid a_j\mid ^2}$. This is the probability for a particular state x to occur divided by the probability that any of the possible states will occur (which should be 1 for normalized states).

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Great explanation, thank you! – user714 Sep 21 '12 at 6:55