Linear algebra solution to quantum mechanics; state collapse

[aswin]

In quantum mechanics only parameters( observables) equal to eigen values of the operator used, are measurable and once measured the system state collapses to the corresponding eigen state( vector). Just tried to give a mathematical emphasis for the above postulate.

Present state is arbitrary

Present state = linear comb of eigen vectors of the hermitian operator which forms the basis of the hilbert space in which each vector is a state of the system.

I m stuck!

Or is there some error in my calculation...

 

[topsquark]

I'm not sure what your question is but I'll take a guess. You initially have an indeterminate state [math] | x > = \sum_n \beta _n | E_n > [/math] and when you make a measurement the state collapses to [math] | E_k >[/math]. If you apply the energy operator we have [math] H | x > = \sum_n \alpha _n \beta _n | E_n > [/math] and when the state is collapsed [math]H | x> = \alpha _k | E _k >[/math].

We have to renormalize to [math] \sum_n | \alpha _n \beta _n |^2 = 1[/math] but we don't need to have [math] |\alpha _k | ^2 = 1[/math] for the collapsed state. (The renormalized state for the collapsed wavefunction is [math] | \alpha _k |^2 = 1[/math]. The renormalization is different for the indeterminate state.)

-Dan