Question on Probability Amplitudes

Question

The Born rule implies that the probability density $\rho$ is defined as $$\rho(x,y,z)=|\psi(x,y,z,t_0)|^2$$ at time $t_0$. What is the difference in this probability density and the probability of a system being described by a particular eigenstate? I.e.,letting $|o_i\rangle$ be the $i$th eigenstate and letting $c_i$ be the coefficient of said eigenstate, the probability of finding the system in the state given by $|o_i\rangle$ is $$P(o_i)=|c_i|^2$$ My question is, what is the difference between these two? (Correct me if I'm wrong, but) Based on my understanding, they say essentially the same thing, so how can we show mathematically that they are equivalent?

Answer 1

If $\{|o_i\rangle\}$ forms a Complete Orthonormal Basis Set (CONS), then you can expand the state $|\psi\rangle$ as: $$|\psi\rangle = \sum_ic_i|o_i\rangle. $$

By probability conservation, you know that the norm of this state is equal to $1$: $$ 1 = \langle \psi | \psi \rangle = \left (\sum_j c_j^* \langle o_j| \right)\left (\sum_i c_i |o_i\rangle \right) = \sum_{ij} c_j^*c_i \langle o_j|o_i\rangle = \sum_i |c_i|^2,$$ where in the last step I used the orthonormality condition $ \langle o_j|o_i\rangle = \delta_{ij}.$

So, if the state vector contains a $c_i$ component of $|o_i\rangle$, the probability of measuring $|o_i\rangle$ is $P(o_i) = |c_i|^2$. The sum of all the outcomes, however, needs to be $1$. Your measurement gives at least something.

So let's now go from a state vector $|\psi\rangle$ in the abstract Hilbert space, so a wavefunction $\psi(x)$ in $L^2$: $$ \psi(x) = \langle x | \psi \rangle,$$ i.e. the wavefunction is just the probabilty amplitude of the state vector in the position basis. So that $|\psi\rangle = c_i |x_i\rangle$ or, since $x$ is a continuous variable, $|\psi = \int \mathrm{d}x\, c(x) |x\rangle.$

The identity is:

$$ \hat{1} = \int \mathrm{d}x\, |x\rangle \langle x|.$$

Let's insert in between $\langle \psi | \psi \rangle$: $$ 1 = \langle \psi |\hat{1}| \psi \rangle = \langle \psi |\int \mathrm{d}x\, |x\rangle \langle x| \psi \rangle = \int \mathrm{d}x\,\langle \psi |x\rangle \langle x| \psi \rangle = \int \mathrm{d}x\, \psi(x)^* \psi(x) = \int \mathrm{d}x\, |\psi(x)|^2, $$ where $|\psi(x)|^2$ is your $\rho(x)$ from the first equation.