# Deriving the solution for a free particle from the TISE

1. Nov 26, 2012

### CAF123

1. The problem statement, all variables and given/known data
In my physics course, we have been given 'plausibility' arguments for the solutions of the time dependent/time independent SE. However, have done a Calculus course alongside, I feel I should really derive these solutions, since I have learnt about the techniques of second order diff eqns.

3. The attempt at a solution
Considering the simple case first: $V = 0$. In such a region, the TISE reduces to $$\frac{-\hbar^2}{2m} \frac{d^2}{dx^2} \psi(x) - E\psi(x) = 0.$$ Take the auxiliary eqn of this to get $-\frac{\hbar^2}{2m}r^2 - E = 0 => r = ±\frac{\sqrt{2mE}}{\hbar}i$ and so the general solution is $$\psi(x) = A\cos(\frac{\sqrt{2mE}}{\hbar}) + B\sin(\frac{\sqrt{2mE}}{\hbar})$$
First question:I recognise my expression for $r$ as the expression for $k$, the wavenumber. Can I simply let $r = k$ to recover this expression? If so, why?
Second question:I know the general soln in this case is $\psi(x) = Ae^{ikx} + Be^{-ikx}.$How does this follow from what I have? I thought about using the method of reduction of order to perhaps find this other term but then I noticed that the expression that I got did not have a term $i\sin$, so what I have does not conform with the answer anyway.
Many thanks.

2. Nov 26, 2012

### dextercioby

You can set r=k on dimensionality arguments.
Use Euler's formula linking sin/cos and the complex exponential.

3. Nov 26, 2012

### CAF123

Are you talking about $$e^{ikx} = \cos(kx) + i\sin(kx)?$$
In my case, I have $Ae^{ikx} + Be^{-ikx} = A[\cos(kx) + i\sin(kx)] + B[\cos(kx) -i\sin(kx)] = \cos(kx)[A+B] +i\sin(kx)[A-B]$. In my expression I have no i. The only thing I can see of doing is saying $A+B =$some constant, $C$ and $(A-B)i =$some constant, $D$, right?

4. Nov 27, 2012

### haruspex

What if you allow that A and B could be complex? Aren't the two solution formats then equivalent?

5. Nov 27, 2012

### CAF123

Is that not what I tried? Instead of A and B, I used C and D, so that in the end I get $$\psi(x) = C\cos(\frac{\sqrt{2mE}}{\hbar}) + D\sin(\frac{\sqrt{2mE}}{\hbar}) = (A+B)\cos(\frac{\sqrt{2mE}}{\hbar}) + (A-B)i\sin(\frac{\sqrt{2mE}}{\hbar})$$ Simplifying gives $\psi(x) = Ae^{ikx} + Be^{-ikx}$.

6. Nov 27, 2012

### haruspex

OK. It wasn't clear to me that you had come to understand that the source of your difficulty was the assumption that A and B were real.