Linear superposition. Measurments.

LagrangeEuler
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If in quantum physics some state is represented by
## \psi(x)=\sum_{k}C_k\psi_k(x)##
##C_m=\int \psi(x)\psi_m(x)dx##
Why probability to measure ##\psi_m(x)## is ##|C_m|^2##?
 
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We can actually express |\psi\rangle as a sum (integral) over all the position states:
|\psi\rangle = \int dx |x\rangle\langle x|\psi\rangle = \int dx |x\rangle \psi(x)

\psi(x) = \langle x|\psi\rangle is the component of |\psi\rangle that overlaps with the position basis vector |x\rangle.
It's an inner product, like with ordinary vectors. If you want to find the y-component of a vector \vec{v}, you take its inner product with the basis vector in the y-direction v_{y} = \vec{v}\cdot \hat{y}.


We can also express these components \langle x|\psi\rangle in other bases too. If we have some other (discrete) basis of states |k\rangle, we can express \langle x|\psi\rangle as:

\langle x|\psi\rangle = \sum_{k}\langle x|k\rangle\langle k|\psi\rangle = \sum_{k} \psi_{k}(x) C_{k}
where
C_{k} = \langle k|\psi\rangle

On the other hand, we can also represent C_{k} in terms of the position basis states |x\rangle, so that
C_{k} = \int dx\; \langle k|x\rangle\langle x|\psi\rangle = \int dx\; \psi_{k}^{*}(x)\psi(x)

The probability to measure \psi_{m}(x) is better thought of as just the probability of measuring k=m. This probability is just |\langle m|\psi\rangle|^{2} = |C_{m}|^{2}.
 
LagrangeEuler said:
If in quantum physics some state is represented by
## \psi(x)=\sum_{k}C_k\psi_k(x)##
##C_m=\int \psi(x)\psi_m(x)dx##
Why probability to measure ##\psi_m(x)## is ##|C_m|^2##?

I think that nobody knows from where Born rule comes. But we see how it goes to classical fields.
When a monochromatic source of particles is in front of a double slit particles hit the screen at a point x with probability p(x) and give it an energy e.
When there is a flux of particles this turns to be the intensity on the screen. For exemple with the electromagnetic field the density of intensity is E² + B²
 
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Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!

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