Can we rewrite Schrodinger equation using observable variable?

In summary: So that it would be fully equivalent to SG. Otherwise I could claim that also a wavefunction is an exact solution to the SG because it is correct for the 1e⁻ case.In summary, the Schrodinger equation can be rewritten using the concept of density functional theory (DFT) where the wave function Ψ is replaced with the product of its complex conjugate Ψ* and itself, Ψ·Ψ*. DFT is an exact theory, but practical implementations are approximations. It works for the lowest states of a given symmetry and can also be used for calculating excited state properties through time-dependent DFT. However, DFT is not fully equivalent to the Schrodinger equation as it only works for the ground
  • #1
phdphysics
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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~
 
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  • #2
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]!
 
  • #3
vanhees71 said:
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
 
  • #4
phdphysics said:
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.
 
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  • #5
phdphysics said:
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).
 
  • #6
gadong said:
That's essentially the idea behind density functional theory (DFT).
But that's an approximation.
 
  • #7
tom.stoer said:
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.
 
  • #8
gadong said:
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.
 
  • #9
tom.stoer said:
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.
 
  • #10
gadong said:
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.
 

FAQ: Can we rewrite Schrodinger equation using observable variable?

Can we rewrite the Schrodinger equation using observable variables?

Yes, it is possible to rewrite the Schrodinger equation using observable variables. This is known as the Heisenberg representation, where the evolution of the system is described by the change in the observable values rather than the wave function.

Why would we want to rewrite the Schrodinger equation using observable variables?

Rewriting the Schrodinger equation using observable variables can provide a more intuitive understanding of the physical properties of the system. It also allows for a more direct connection to experimental observations.

What is the difference between the Schrodinger and Heisenberg representations?

The Schrodinger representation describes the evolution of a quantum system in terms of the wave function, while the Heisenberg representation describes the evolution in terms of observable variables. In the Schrodinger representation, the wave function evolves in time, while in the Heisenberg representation, the observable values are time-dependent.

How do we convert between the Schrodinger and Heisenberg representations?

The conversion between the Schrodinger and Heisenberg representations is achieved through the use of mathematical operators known as Heisenberg operators, which relate the wave function to the observable variables. These operators can be derived from the Schrodinger equation.

Are there any limitations to using the Heisenberg representation?

While the Heisenberg representation can provide a more intuitive understanding of quantum systems, it is not always the most convenient or practical representation to use. In some cases, the Schrodinger representation may be more suitable, such as when dealing with systems that have time-independent Hamiltonians.

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