# Time-independent Schrodinger Equation

1. Sep 12, 2011

### ralqs

Why is it that we assume that the solutions to the time-independent Schrodinger Equation are real? Why can't they be complex?

2. Sep 13, 2011

### tom.stoer

We do not assume that; in general they will be complex.

3. Sep 13, 2011

### Bill_K

If ψ is a solution then ψ* is also a solution with the same energy eigenvalue. The trivial cases are ψ = ψ*, or ψ and ψ* differ by a constant phase. If ψ and ψ* are linearly independent then the energy level is degenerate. This often happens, for example, in three dimensions when ψ ~ Ylm(θ, φ) and ψ* ~ Yl,-m(θ, φ)

4. Sep 13, 2011

### ralqs

I'll give you an example from my text:
For the infinite potential well, the TISE is
$$\frac{d^2 \psi}{dx^2} = -\frac{2mE}{\hbar^2}\psi = -k^2 \psi$$
The solution to this is given as $A \sin{kx} + B \cos{kx}$, which is real.

5. Sep 13, 2011

### zymurge

But $A$ and $B$ are in general complex, and you can write
$$\cos kx = \frac{e^{ikx}+e^{-ikx}}{2},~\sin kx = \frac{e^{ikx}-e^{-ikx}}{2i}.$$

6. Sep 13, 2011

### Bill_K

Let me expand on this further. If ψ is a solution of the time-independent Schrodinger equation then so is ψ*, with the same energy E. This is a general consequence of time reversal invariance. And if ψ and ψ* are linearly independent, you may always form real combinations ψ + ψ* and i(ψ - ψ*) and obtain two real solutions.

However there may be reasons for not wanting to do this. For example you may want to use ψ's that are also the eigenfunctions of some other variable besides the energy. In the three-dimensional example I gave, you could form linear combinations of Ylm(θ, φ) and Yl,-m(θ, φ), but they would not be eigenfunctions of m.

As another example, consider a finite square well in one dimension. For E > 0 the eigenstates are the scattering states, and twofold degenerate. A natural choice is to use traveling waves to the left and right, which are complex and contain factors e±ikx. You could, if you wanted, use standing wave solutions by forming symmetric and antisymmetric combinations of these, but the disadvantage is that they would not be eigenstates of momentum.

7. Sep 13, 2011

### ralqs

Okay sure, but then won't you have to prove that ψ + ψ* and i(ψ - ψ*) are linearly independent?

8. Sep 14, 2011

### tom.stoer

One should try via Reductio ad absurdum: Suppose there IS a complex eigenfunction. How can we show that we cannot transform it into a real one in general?

Or make an ansatz like ψ = R*exp(iS) and show that in general S cannot be a constant.

Last edited: Sep 14, 2011