Can we rewrite Schrodinger equation using observable variable?

[phdphysics]

We know that in Schrodinger equation, Ψ is called wave function, which is not observable, while Ψ·Ψ* is the probability, which is observable.
can we rewirte the Schrodinger equation to a form without Ψ but only Ψ·Ψ*?

because I think, in this way can I figure out all conservations in the equation. Although I can make it with present Schrodinger equation, it's obvious that the Schrodinger equation will change if I make t→-t transformation.

Thanks~

 

[vanhees71]

Write down the equation of motion for [itex]\psi^*[/itex] given that [itex]\psi[/itex] fulfills the usual Schrödinger equation. Then change [itex]t \rightarrow -t[/itex]!

 

[phdphysics]

Write down the equation of motion for [itex]\psi^*[/itex] given that [itex]\psi[/itex] fulfills the usual Schrödinger equation. Then change [itex]t \rightarrow -t[/itex]!

you mean, this equation set(containing two equation, Ψ and Ψ*) does not change?
by the way, could you tell me, can Schrodinger equation be rewrited to Ψ·Ψ* mathemetically?
thanks

 

[tom.stoer]

can we rewirte the Schrodinger equation to a form without Ψ but only Ψ·Ψ*?

No, we can't.

Define

##\psi = R \, e^{iS}##

with two real variables R and S.

Then introduce

##\rho = \psi^\ast \psi = R^2##

Now we see that an equation in R or ρ is an equation in one single real variable, whereas the original equation was an equation in two independent real variables R and S.

 

[gadong]

We know that in Schrodinger equation, Ψ is called wave function, which is not observable, while Ψ·Ψ* is the probability, which is observable.
can we rewirte the Schrodinger equation to a form without Ψ but only Ψ·Ψ*?


Thanks~

That's essentially the idea behind density functional theory (DFT).

 

[tom.stoer]

That's essentially the idea behind density functional theory (DFT).

But that's an approximation.

 

[gadong]

But that's an approximation.

If you mean that the DF theory itself is an approximation - no, it's exact. Practical implementations are approximations, however.

 

[tom.stoer]

If you mean that the DF theory itself is an approximation - no, it's exact. Practical implementations are approximations, however.

But DFT works only for the ground state, whereas the SG works for all states including all bound and scattering states.

 

[gadong]

But DFT works only for the ground state, whereas the SG works for all states including all bound and scattering states.

DFT works for the lowest states of a given symmetry, of which there might be several. The calculation of excited state properties (e.g. absorption spectra) can be carried out using time-dependent DFT. More information here: http://en.wikipedia.org/wiki/Time-dependent_density_functional_theory.

 

[tom.stoer]

That's essentially the idea behind density functional theory (DFT).

I still think that this answer is missleading in our context. Otherwise you would have to prove that DFT is fully equivalent to the SG including the complete set of states (bound plus scattering states) in Hilbert space plus all derived phenomena like superpositions, interference, entanglement etc.