# Normalization conditions in quantum mechanics

1. Dec 3, 2004

### tiger_striped_cat

I am familiar with the normalization

$$\int\left|\Psi\right|^{2}dx=1$$

Because we want to normalize the probability to 1. However if a state vector isn't in the x basis and is just a general vector in Hilbert space, we can take the normalization condition to be:

$$<\Psi|\Psi>=1$$

Correct? I guess this is making more and more sense to me. I guess I just never thought about it before. But since we can put a completeness relation in the middle of the latter equation then these two are basically equivalent. I guess I'm just having difficultly seeing how probability over all states is 1 in the second case.

Please tell me if this is incorrect.

Thanks!

2. Dec 3, 2004

### dextercioby

It is perfectly correct.Though the logic should contain the same ideas,but in a different order.From the general condition of normalization of vectors in a separable Hilbert space u have to make several assumptions (that is sentences which gradually make u lose generality) in order to obtain the integral over $$R^{3n}$$ in case of an "n" point-like particles quantum system.
Hopefully u're reading a "decent" ook which respects the logic mentioned above.

Both Schroedinger's wave mechanics and Heisenberg-Born-Jordan's matrix mechanics can be marvelously derived from the abstract theory by Dirac.A common example of showing that is the QM linear harmonic oscillator.

3. Dec 3, 2004

### Galileo

That's correct. Since every observable $Q$ has a complete set of orthonormal eigenvectors $|q_n\rangle$ we have:
$$\sum_n |q_n\rangle \langle q_n|= 1$$
and $|\Psi \rangle = \sum_n c_n|q_n \rangle$ with $c_n=\langle q_n|\Psi\rangle$.
so:
$$<\Psi|\Psi>=<\Psi|1|\Psi>=\sum_n \langle\Psi|q_n\rangle \langle q_n|\Psi \rangle=\sum_n |c_n|^2=1$$.

Last edited: Dec 3, 2004