Does Born's rule guarantee component-wise probabilities? [closed]

Question

Suppose I have the wave function $\vec{\psi} = \begin{bmatrix} \psi_x \\ \psi_y \\ \psi_z \end{bmatrix}$.

If I understand Born's rule correctly, the equality $\vec{\psi}^* \cdot \vec{\psi} = f(x,y,z)$ (where $f(x,y,z)$ is the probability density of a given state) holds. For that same wave function, does it also hold that $\psi_x^* \psi_x = f(x)$?

My suspicion is that is doesn't hold based on analogizing that the square of a polynomial will be positive, but not every term in the expansion will be positive. But this is far from rigorous thinking.

Does Born's rule only imply the overall norm squared will be the probability density of a state, or will the products of the components similarly be probability densities of parts of the state?

Answer 1

Firstly, $c^* c$ will always be positive and real, because for any complex number $c=a+bi$ which is equal to $a^2+b^2\ge 0$.

Secondly, your notation looks a little confused! A vector $\vec\psi=(\psi_a,\psi_b,\psi_c)$ denotes a quantum system where, for example, one particle is either in state a, state b, or state c. More precisely, it's in a superposition of the three states. The Born rule probability to find the particle in state a would be $P(a)=|\psi_a|^2$, and it does not make sense to talk about the joint probability $P(a,b)$. The total probability is equal to one, $\vec\psi^*\cdot\vec\psi=|\psi_a|^2+|\psi_b|^2+|\psi_c|^2=1$.

If we're talking about a wavefunction in 3 dimensions, then $\psi(x,y,z)$ is a function with infinitely many states, and it does make sense to talk about the joint probability $P(x,y,z)=|\psi(x,y,z)|^2$. Then the statement that the total probability is 1 is: $\iiint dx dy dz |\psi(x,y,z)|^2=1$.

You can then do regular probability distribution stuff on this 3 dimensional probability distribution. For example, the marginal probability $P(x)=\iint dydz P(x,y,z)$, or the conditional probability $P(x|y,z)=P(x,y,z)/P(y,z)$. There are a few different aspects to Born's rule that could come up here, but it sounds like you might have more of a confusion on what the wavefunction is rather than how to apply Born's rule.