Are there two aspects of Born's rule?

Question

I am having some problem understanding Born's rule. I am getting a little bit confused. Here it goes;

Let $f(x,t)$ be a solution of Schrodinger equation. Then Born's rule says that the square modulus of $f(x,t)$ gives the probability of locating a particle at position $x$ at time $t$.

Since quantum mechanics also says that a quantum state in superposition collapses to one of its eigenvalues upon observation, then I understood $x$ as an eigenvalue corresponding to the position operator and that Dirac-delta function is its eigenstate.

But then I do encounter another form of Born's rule which states that given that $g(x,t)$ is a linear combination of quantum states superimposed on each other, then the probability of $g(x,t)$ collapsing into one of its eigenfunctions, say $h(x,t)$ is the square modulus of the dot product of $g(x,t)$ and $h(x,t)$. Yes, I do understand how this dot product yields the coeficient of $h(x,t)$ but I don't understand well the connection between this latter rule and the former.

For instance, is it correct to say that the solution to Schrodinger equation, $y(x,t)$ is a linear combination of Gaussian bell functions that tends to Dirac-delta functions as we increase the certainty of the positions? If yes, is the square modulus of $y(x,t)$ equals to the square modulus of the dot product of $y(x,t)$ at the Dirac-delta function with eigenvalue of $x$ at time $t$?

Answer 1

Exactly, the former can be seen as a special instance of the more general latter rule and this would be done exactly along your lines. Consider your function $y(x)$ (please don't mind me dropping the $t$ parameter everywhere) and a measurement of position with the result of $x_0$. The generalized eigenstate would be represented by a Dirac delta function displaced to $x_0$, $$h(x) = \delta(x-x_0),$$ which in turn is a limit (in a quite special yet well-defined way) of $$h(x) = \lim_{\omega\to 0} \frac{1}{\sqrt{2\pi \omega^2}} e^{-\frac{(x-x_0)^2}{2\omega^2}}.$$ Let us compute the dot product of the two functions as a limit of dot products with the above Gaussians. $$(h(x),f(x)) = \int_\mathbb{R} h^\ast(x) f(x) \mathrm{d}x = \lim_{\omega\to 0} \int_\mathbb{R} \frac{1}{\sqrt{2\pi \omega^2}} e^{-\frac{(x-x_0)^2} {2\omega^2}} f(x) \mathrm{d}x.$$ A natural step is to substitute for $$y = \frac{x-x_0}{\sqrt{2\omega^2}}: \quad \mathrm{d}y = \frac1{\sqrt{2\omega^2}} \mathrm{d}x.$$ We don't need to know what $f(x)$ is, we just plug the inverse formula $$x = x_0 + \sqrt{2\omega^2} y$$ into it, yielding the integral $$(h(x),f(x)) = \lim_{\omega\to 0} \int_\mathbb{R} \frac{1}{\sqrt{\pi}} e^{-y^2} f(x_0 + \sqrt{2\omega^2} y) \mathrm{d}y.$$ Now it takes just some assumptions on the properties of $f(x)$ to move the limit operation inside the integral. Taking the limit there is simple: $$\lim_{\omega\to 0} \frac{1}{\sqrt{\pi}} e^{-y^2} f(x_0 + \sqrt{2\omega^2} y) = \frac{1}{\sqrt{\pi}} e^{-y^2} f(x_0).$$ And from here you only need to know that $$\int_\mathbb{R} e^{-y^2} \mathrm{d}y = \sqrt\pi$$ to derive that $$(h(x), f(x)) = f(x_0).$$ Now your square of the dot product indeed turns out to be $|f(x_0)|^2$.

Answer 2

You may write any solution of the Schrodinger equation as :

$$f(x,t) = \int dx_0 f(x_0,t) \delta (x-x_0) \tag{1}$$

Here $\delta(x-x_0)$ is a (not-normalized) eigenvector of the position operator $\hat x$ with the eigenvalue $x_0$.

The "dot product" of $\delta(x-x_0)$ and $f(x,t)$ and is simply : = $\int dx \delta (x-x_0) f(x) = f(x_0,t)$

So, there is no contradiction : $f(x,t)$ is the probability amplitude to find the particle at the position $x$, and $f(x_0,t)$ is also the coefficient of the decomposition of $f(x,t)$ in the basis $\delta(x-x_0)$ of the eingenvectors of the position operator $\hat x$.

Answer 3

Different people call different ideas "Born rule"; some of them think they are just variations of the same rule, some don't.

One meaning is that the solution $\psi(q_1,q_2,...q_{3N},t)$ to the time-dependent Schroedinger equation in $3N$-dimensional configuration space can be used to calculate probability that the system has configuration in some subset $c$ of the configuration space $C$:

$$ P(\mathbf q\in c)(t) = \int_c|\psi(q_1,...,q_{3N},t)|^2dq_1...dq_{3N} $$ So, $|\psi|^2$ gives probability density in the configuration space (not probability). The probability is an integral of square of $\psi$.

This only gives probability for configurations. For momenta, similar formula has been suggested: $$ P(\mathbf p\in c)(t) = \int_c|\tilde{\psi}(p_1,...,p_{3N},t)|^2dp_1...dp_{3N}. $$

where $\tilde{\psi}(p_1,...,p_N)$ is a Fourier transform of $\psi(q_1,...q_{3N})$ in all variables. Again, the probability is an integral of square of $\psi$.

Both these formulae have the advantage that they give definite answer directly from the solution to the Schroedinger equation; no special basis in the Hilbert space needs to be chosen.

However, there is another, different rule that is also called "Born rule" by some. This rule states that:

When a physical quantity is measured at time $t$, the result of the measurement can only be an eigenvalue of the corresponding self-adjoint operator $\hat{f}$ on the Hilbert space of admissible $\psi$'s. Which eigenvalue it will be cannot be predicted with certainty, but the probability of any eigenvalue (let's denote it $f'$) can be calculated from $\psi$ in the following way:

$$ P(f')(t) = \bigg|\int \phi'^*(q_1,...,q_{3N})\psi(q_1,...,q_{3N},t)dq_1dq_2...dq_{3N} \bigg|^2 $$

where $\phi'$ is the eigenfunction of the operator $\hat{f}$ (in non-degenerate case). In degenerate case one could think of additional summation over all the eigenfunctions corresponding to $f'$.

Clearly, this is very different rule: it gives probability, not probability density; it is specific for the quantity considered - its operator and eigenfunctions need to be chosen. This poses some problems in case the form of the operator is not definite but there are more possible choices (kinetic momentum in the presence of time-dependent electromagnetic field); and finally, the rule says we should calculate square of an integral, as opposed to the first rule , where we calculate integral of a square.