# Why is the wave function not measurable alone?

1. Apr 1, 2012

### Toyona10

Hi,
why is the wavefunction not measurable as it is, but is measurable when the square of the absolute value is taken?

Thank you

2. Apr 1, 2012

### Mark M

Hi,

The wavefunction represents the probabilities of finding a particle in specific positions. So, let's say we have a particle that, upon observation, can express one of two traits, we'll call them Y and Z. Since it is in a superposition of both, we would have to write this as: $$\psi =\alpha (Y)+\beta (Z)$$Where $\alpha$ and $\beta$ are known as probability amplitudes. So, we write that the probability of finding the particle in state Y is $\left | \alpha \right |^{2}$

A normalized wavefunction, $\psi$ is a probability amplitude. If we are talking about a position x, this is also a probability amplitude. So, to write the probability of a wavefunction collapsing to the position x we must express it as $$\left | \psi (x) \right |^{2}$$ See here for more: http://en.wikipedia.org/wiki/Probability_amplitude
Hope I helped!

3. Apr 2, 2012

### ndung200790

I think that because of diffraction in QM,we must guest the wavefunction as probability amplitude but square of wavefunction is probability.Then the wavefunction is not measurable.