# [Q]Is physical quantity depending on coordinate on which we solve problem?

1. Oct 16, 2008

### good_phy

Hi.

I have a question whether eigenvalue(corresponding physical quantity) is up to coordinate

we establish to solve certain problem.

For example, I have state function combined by momentum eigen states such as $\Psi = \sqrt{\frac{1}{2}}ae^{ikx} + \sqrt{\frac{1}{2}}ae^{-ikx}$

a is normalization factor of each eigenstate and $\sqrt{\frac{1}{2}}$ means

probability each eigenstate is chosen after momentum measurement is the same.

I think expactation value of momentum of that state is 0 because momentum we can get

from each eigenstate is the same with opposite sign and probability is the same, according

to $$<a> = \sum p_{i}a_{i}$$, $p_{i}$ is probability $a_{i}$ can be obtained.

I attempted to get expectation value with firsthand calculation with $<\Psi|\stackrel{{^}}{p}\Psi >$

I establish integral range from 0 to a, [0,a] and i get fianl integral form $$\sqrt{\frac{1}{2}}\hbar ka\int_0^{a}2isin(2kx)dx [/itex] but its calculation is not zero except for case in which integral range is [-a/2,a/2] or $[-\infty,\infty]$ i can not accept physical quantity is up to on what coordinate we solve problem. and i have a very important question, orthogonality [tex] <\varphi_{k} | \varphi_{k^{'}} >=0$$ only if integral range is $[-\infty,\infty]$ ?

You can calculate expectation value by $<\Psi|\stackrel{{^}}{p}\Psi >$ only
if the wavefunction is normalized. In your case the function is not normalized no matter what normalization constant you choose, because $<\Psi|\Psi >$=infinity.