Brushing Up: Understanding Why We Must Do |$\psi_i$|$^2$

[FrogPad]

I am brushing up on some basic quantum mechanics that we covered in the review for a course I just started. For some reason I cannot remember why the following is the case.

So let's say we have \psi representing a vector with components that represent a state.

Why do we have to do: | \psi_i |^2 to get the probability that we are in that given state?

I know I HAVE to, but I cannot remember WHY.

thanks

 

[Dick]

Because you are taking the \psi_i to be orthogonal and \psi to be normalized. So the probability to be in a given state is <psi_i|psi> which is what you said.

 

[FrogPad]

Because you are taking the \psi_i to be orthogonal and \psi to be normalized. So the probability to be in a given state is <psi_i|psi> which is what you said.

Would you mind expanding upon what you said? I cannot fully follow.

I have unfortunately never taken a proper quantum class. The two courses that I have even taken are a modern physics course (sophomore level -- talks about slit experiments and the likes), and a properties of solids (junior level - energy bands, probability amplitudes, etc...)

The course I am taken is a solid state device physics course. The review (first day) started with bra-ket notation; this is something I have never encountered before. Trying to review (more proper, catch up) has led me to a more thorough reading of the basics (including review of linear algebra).

 

[Dick]

If you know what a probability amplitude is then you are basically there. If you know the amplitude then you square it (in the complex number sense) to get the probability. If psi is split into a sum of orthogonal (so <psi_i|psi_j>=0 if i is not equal to j) components psi=sum(psi_i) and is normalized, so <psi|psi>=1. Then the probability of being in the state i is <psi_i|psi>=<psi_i|psi_i>. I don't think I'm explaining this very well, because it's really close to being one of the assumptions of quantum mechanics. Maybe somebody else can clarify.

 

[malawi_glenn]

This has to do with the amplitude as Dick said, if you look up the word probablity amplitude in your book om statistics, you will se parallells =)

I remember i struggeled a lot on this too in the begining, but then I compared that to the stuff I learned i statistics, and things became clearer.