General solution of Schrodinger eq. proof

[dingo_d]

Homework Statement

Let $$|\psi\rangle$$ and $$|\psi '\rangle$$ be solutions to the same Schrodinger equation. Show than, that $$c|\psi\rangle+c'|\psi '\rangle$$ is the solution, where c and c' are arbitrary complex coefficients, for which holds: $$|c|^2+|c'|^2=1$$

The Attempt at a Solution

Now this follows from linearity of the Schrodinger equation (that every linear combination is the solution). But how to prove it directly?

I've started with:

$$i\hbar \frac{\partial}{\partial t}|\psi\rangle=\hat{H}|\psi\rangle$$
$$i\hbar \frac{\partial}{\partial t}|\psi'\rangle=\hat{H}|\psi'\rangle$$

And added them up:

$$i\hbar\left( \frac{\partial}{\partial t}|\psi\rangle+\frac{\partial}{\partial t}|\psi'\rangle\right)=\hat{H}(|\psi\rangle+|\psi'\rangle)$$

And... now I don't know what to do next :\

I think I'm taking the superposition principle way to lightly :\

 

[arkajad]

Call $$|\psi''\rangle=c|\psi\rangle+c'|\psi'\rangle$$

Show that $$|\psi''\rangle$$ solves the Schrodinger equation. You will se that conditions on c and c' are irrelevant for this question.

 

[dingo_d]

Ok... but how? :\

 

[arkajad]

Calculate

$$i\hbar \frac{\partial}{\partial t}|\psi''\rangle$$

remembering that c and c' are constants, and see if it the same as $$\hat{H}|\psi''\rangle$$

 

[dingo_d]

Ok so,

$$i\hbar \frac{\partial}{\partial t}|\psi ''\rangle=\hat{H}|\psi ''\rangle$$

$$i\hbar \frac{\partial}{\partial t}|\psi ''\rangle=i\hbar\frac{\partial}{\partial t}\left(c|\psi\rangle+c'|\psi'\rangle\right)=c\underbrace{i\hbar\frac{\partial}{\partial t}|\psi\rangle}+c'\underbrace{i\hbar\frac{\partial}{\partial t}|\psi'\rangle}=c\hat{H}|\psi\rangle+c'\hat{H}|\psi'\rangle=\hat{H}(c|\psi\rangle+c'|\psi'\rangle)=\hat{H}|\psi''\rangle$$

And that's it?

 

[arkajad]

And that's it!

 

[dingo_d]

W00t! Thanks a lot :D