Schrodinger equation for a free particle in 3d space

[GeolPhysics]

I've got the solution to the question but I just need more detail. I can't work out the first step of the solution to the second step.

That should read, I don't know what they multiplied ih-bar by to make it (i/h-bar)^2?

 

[DrClaude]

I'm confused as to what you donate understand. Is it the appearance of $(i/\hbar)^2$ or it going away on the next line?

 

[GeolPhysics]

I'm confused as to what you donate understand. Is it the appearance of $(i/\hbar)^2$ or it going away on the next line?

Yeah, it's the appearance of $(i/\hbar)^2$

 

[DrClaude]

It comes from the derivative of the exponential.

$$
\frac{\partial^2}{\partial x^2} e^{ax} = a^2 e^{ax}
$$

 

[GeolPhysics]

It comes from the derivative of the exponential.

$$
\frac{\partial^2}{\partial x^2} e^{ax} = a^2 e^{ax}
$$

I only managed to get this. Can you show me the exact steps on how you ended up with the second line?

And what are the values of a, e and x?

 

[Nugatory]

And what are the values of a, e and x?

Did you intend to ask what $e$ is? It’s Euler’s number, and $exp(a)$ is another notation for $e^a$. If you not already familiar with its properties you’re going to have to put some time into first-year differential and integral calculus before you can take on Schrodinger’s equation.

If that was just a slip of the typing fingers, go back to the $\psi$ suggested in the text, then rewrite the exponential of sums as a product of exponentials. Now you’ll be able to take the second derivative with respect to $x$ to get the $(i/\hbar)^2$ factor.

 

[DrClaude]

To add to what @Nugatory said, $a \in \mathbb{C}$ is a constant and $x$ is the independent variable, as in the differentiation.