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

<br /> i\hbar \frac{\partial}{\partial t}|\psi&#039;&#039;\rangle<br />

remembering that c and c' are constants, and see if it the same as \hat{H}|\psi&#039;&#039;\rangle

 

[dingo_d]

Ok so,

<br /> i\hbar \frac{\partial}{\partial t}|\psi &#039;&#039;\rangle=\hat{H}|\psi &#039;&#039;\rangle<br />

<br /> i\hbar \frac{\partial}{\partial t}|\psi &#039;&#039;\rangle=i\hbar\frac{\partial}{\partial t}\left(c|\psi\rangle+c&#039;|\psi&#039;\rangle\right)=c\underbrace{i\hbar\frac{\partial}{\partial t}|\psi\rangle}+c&#039;\underbrace{i\hbar\frac{\partial}{\partial t}|\psi&#039;\rangle}=c\hat{H}|\psi\rangle+c&#039;\hat{H}|\psi&#039;\rangle=\hat{H}(c|\psi\rangle+c&#039;|\psi&#039;\rangle)=\hat{H}|\psi&#039;&#039;\rangle<br />

And that's it?

 

[arkajad]

And that's it!

 

[dingo_d]

W00t! Thanks a lot :D