The Born Rule
Claimed - Rehaan Naik 11/27/2022
The Born Rule is an important result of quantum mechanics that describes the probability density of a measured quantum system. In particular, it states that the square of the wavefunction is proportional to the probability density function. This result is a continuous form of the norm squared of the inner product being the probability that a known state takes the value of another.
Discrete Probability[math]\displaystyle{ P_{\psi}(A = a) = |\bra{\phi_{a}}\ket{\psi}|^2 }[/math]
Continuous Probability Density Function [math]\displaystyle{ P = \left | \Psi(x,t) \right |^2 }[/math]
The Wavefunction as a Continuous Representation of State
The wavefunction is a complex function. However, there is no physical interpretation of complex numbers. Converting a complex number to a real number involves taking the norm of the complex number. Now, see that the wavefunction [math]\displaystyle{ \psi(x) }[/math] is the continuous analogue of [math]\displaystyle{ \ket{\psi} }[/math] in the x basis. [math]\displaystyle{ \ket{\psi} }[/math] is a state vector for which the norm is finite, and the inner product is defined within the Hilbert Space. We know that the norm squared of the inner product of a state vector with itself is 1 since we have defined state vectors to have norm 1. Since the wavefunction is the analogue of the state vector, it makes sense that an analogue of the inner product is defined for it - the inner product of a state vector with itself is the multiplication of the the conjugate transpose of the state vector with itself. So, the norm squared of the wavefunction comes from the multiplication of the wavefunction with its complex conjugate.
[math]\displaystyle{ \left | \Psi(x,t) \right |^2 = \Psi^*(x,t)\Psi(x,t) }[/math]
Probability Density Function
Clearly, since we have mandated that the inner product of the two state vectors is finite, so must be the norm squared of the wavefunction. (The wavefunction belongs to the Hilbert space within which the inner product is defined).
In a discrete sense, the norm squared of the inner product gives the probability for an outcome upon measurement of a state for a given observable. However, for a wavefunction, we deal with an infinite Hilbert Space, and the spectrum of the observable is continuous - so its norm squared cannot give values of probability for a specific. Instead we interpret it as giving the probability density function of the state (as represented by the wavefunction) on the chosen observable.
Now, to get the probability over the entire observable, we integrate from [math]\displaystyle{ -\infty \rightarrow +\infty }[/math]. It makes sense that this integral would be 1, since the summation of probability over all the possible outcomes is necessarily 1. Since we are integrating the norm-squared of the wavefunction, and we have the following result:
[math]\displaystyle{ \int_{-\infty}^{\infty}{ \left | \Psi(x,t) \right |^2 dx}=1 }[/math],
as a given particle in question must be located somewhere between [math]\displaystyle{ -\infty \lt x \lt \infty }[/math]
Or more formally:
[math]\displaystyle{ P_{x\in[-\infty,\infty]}=1 }[/math]
We can say that [math]\displaystyle{ \Psi(x,t) }[/math] must be square integrable. (All this means is that [math]\displaystyle{ \int_{-\infty}^{\infty}{ \left | \Psi(x,t) \right |^2 dx} \lt \infty }[/math])
Calculating Probability
This is now straightforward since we have the probability density function.
The probability that the outcome of measurement of observable [math]\displaystyle{ A }[/math] lies between the values [math]\displaystyle{ a_1 }[/math] and [math]\displaystyle{ a_2 }[/math] for the state [math]\displaystyle{ \psi }[/math]is represented formally as
[math]\displaystyle{ Prob_\psi(a_1 \leqslant A \leqslant a_2) }[/math]
As expected from how probability distributions are defined:
[math]\displaystyle{ Prob_\psi(a_1 \leqslant A \leqslant a_2) = \int_{a_1}^{a_2}{\left | \Psi(a) \right |^2 da} }[/math]