State vectors and Eigenvalues?

[kq6up]

If I define a state ket in the traditional way, Say:

$$|\Psi \rangle =\sum _{ i }^{ }{ a_{ i }|\varphi _{ i }\rangle \quad } $$

Where $$a_i$$ is the probability amplitude.

How does:

$$\hat {H } |\Psi \rangle =E|\Psi \rangle $$ if the states of $$\Psi$$ could possibly represent states that have different energy levels, and one would not be able to factor the energy eigenvalue out of the summation. I know I am missing something big here. Could someone point it out?

Edit: Does $$ \hat {H} $$ collapse psi to one state phi, and so render only energy eigenvalue of the one phi that remains?

Thanks,
Chris Maness

 

[atyy]

If I define a state ket in the traditional way, Say:

$$|\Psi \rangle =\sum _{ i }^{ }{ a_{ i }|\varphi _{ i }\rangle \quad } $$

Where $$a_i$$ is the probability amplitude.

How does:

$$\hat {H } |\Psi \rangle =E|\Psi \rangle $$ if the states of $$|\Psi \rangle$$ could possibly represent states that have different energy levels, and one would not be able to factor the energy eigenvalue out of the summation. I know I am missing something big here. Could someone point it out?

Thanks,
Chris Maness

When you write $\hat {H } |\Psi \rangle =E|\Psi \rangle$, $|\Psi \rangle$ cannot represent a state with more than one energy. Each possible $|\Psi \rangle$ that is a solution to that equation is an energy eigenstate with a definite energy.

That equation has many solutions, each corresponding to a different energy. An arbitrary state will be a superposition of energy eigenstates with different energies.

The equation $\hat {H } |\Psi \rangle =E|\Psi \rangle$ is the time-independent Schroedinger equation or the the energy eigenstate equation. It is only an intermediate step in the solution of the full time-dependent Schroedinger equation.

 

[kq6up]

Ok, so $$ \Psi $$ can't be a superposition anymore.

Does this work for a generalized statement?

$$ \hat { B } \hat { A } |\Psi \rangle =ab|\Psi \rangle $$ Where if, psi is not collapsed, it is at least degenerate, and where A & B commute.

Thanks,
Chris Maness

 

[Matterwave]

Ok, so $$ \Psi $$ can't be a superposition anymore.

Does this work for a generalized statement?

$$ \hat { B } \hat { A } |\Psi \rangle =ab|\Psi \rangle $$ Where if, psi is not collapsed, it is at least degenerate, and where A & B commute.

Thanks,
Chris Maness

Statements such as these are eigenvalue problems. In these cases, by definition, $\Psi$ must be an eigenvector of the operator.

The time-dependent Schroedinger equation is NOT an eigenvalue problem:

$$H\left|\Psi\right>=i\hbar\frac{\partial}{\partial t}\left|\Psi\right>$$

And so ACTUAL wave functions do NOT have to be eigen-vectors of any operator. We just use the eigenvalue problems to help us solve the full problem.

 

[Jilang]

It's the phis that have the different energy levels!

 

[atyy]

Ok, so $$ \Psi $$ can't be a superposition anymore.

Does this work for a generalized statement?

$$ \hat { B } \hat { A } |\Psi \rangle =ab|\Psi \rangle $$ Where if, psi is not collapsed, it is at least degenerate, and where A & B commute.

Thanks,
Chris Maness

Yes, in the example you give, if $|\Psi \rangle$ is an eigenstate of A and an eigenstate of B, then it is an eigenstate of AB or BA. This does not mean that $|\Psi \rangle$ is not a superposition. It is a superposition of eigenstates of other operators that do not commute with A or B.

http://en.wikipedia.org/wiki/Complete_set_of_commuting_observables
http://www2.ph.ed.ac.uk/~ldeldebb/docs/QM/lect2.pdf (Section 2.7)