# Linear superposition. Measurments.

1. Mar 3, 2014

### LagrangeEuler

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 $|C_m|^2$?

2. Mar 3, 2014

3. Mar 3, 2014

### jfizzix

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 $|\langle m|\psi\rangle|^{2} = |C_{m}|^{2}$.

4. Mar 4, 2014

### naima

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

Last edited: Mar 4, 2014