Why is the wave function not measurable alone?

[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

 

[Mark M]

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

Thank you

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!

 

[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.