Linearity of Schrodinger Equation

[dainylee]

Could someone please address this question?
How do you algebraically demonstrate the superposition principle revealed by the Schrodinger equation (ie. If Psi1(x,t) and Psi2(x,t) are both solutions then Psi(x,t)= Psi1(x,t)+Psi2(x,t) is also a solution.)?

[George Jones]

What do you get when you substitute Psi(x,t) into the left side of the Schrodinger equation?

[dainylee]

I am not quite sure what you are asking me to do...Psi(x,t) is already in there?

[Rach3]

No, \psi(x,t) is not in there. However, you do know that \psi_1(x,t) is a solution -

-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi_1(x,t)+V(x) \psi_1(x,t)=i \hbar \frac{d}{dt}\psi_1(x,t)


And \psi_2(x,t) is also in there:

-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi_2(x,t)+V(x) \psi_2(x,t)=i \hbar \frac{d}{dt}\psi_2(x,t)


However, you have to show that \psi=\psi_1+\psi_2 also satisfies this dynamic equation.

And then you can extend your result to a\psi_1 + b\psi_2 for any coefficents a,b.

[George Jones]

I am not quite sure what you are asking me to do...Psi(x,t) is already in there?

You have to show that


-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \Psi(x,t) + V(x) \Psi(x,t) = i \hbar \frac{\partial}{\partial t} \Psi (x,t).


In the left, substitute \Psi = \Psi_1 + \Psi_2, and, using that \Psi_1 and \Psi_2 both statisfy Schrodinger's equation, work your way to the right side of the equation you must show true.