# [Basic QM] Potential step & particle flow, my answer is inconsistent ?

1. Jan 6, 2009

### Nick89

1. The problem statement, all variables and given/known data
A flow of particles collides with a potential step. The potential is zero for x < 0 and a certain value $$V_0$$ for x > 0.
The flow of particles comes in from the left (x<0 area).

One of the questions is to give the general form of the wavefunction in the area x > 0 and to find an expression for the wavenumber k.

2. Relevant equations
$$-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} + V_0 \psi = E \psi$$
(For x > 0)

3. The attempt at a solution

I first write the schrodinger eq. into a more general form:
$$-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} + (V_0 - E) \psi = 0$$
$$- \frac{d^2 \psi}{dx^2} + \frac{2m (V_0-E)}{\hbar^2} \psi = 0$$
$$\frac{d^2 \psi}{dx^2} - \frac{2m (V_0-E)}{\hbar^2} \psi = 0$$
$$\frac{d^2 \psi}{dx^2} - k^2 \psi = 0$$

So
$$k^2 = \frac{2m (V_0-E)}{\hbar^2}$$

And the solution are complex exponents due to the minus-sign (they would be real exponents if it was a plus sign):
$$\psi (x) = A e^{ikx} + B e^{-ikx}$$

Now, when I look at the answer, I noticed that they also use complex exponents, but their k was different:
$$k^2 = \frac{2m (E-V_0)}{\hbar^2}$$

This is a pretty important error because:

In my case: If E > V, the particle is 'above the potential step' and thus it should be a complex exp. (sin/cos). But, if E > V then k becomes complex, and the exponents become real!
If E < V, the particle is 'inside the potential step' and thus it should be a real exponent (as usual with penetration in classically forbidden area). But, if E < V then k is real and the exponents stay complex!...

Obviously I made an error with a minus sign somewhere, but I honestly can't find it..!?

I also tried switching the minus sign for a plus sign (and swapping the E and V again obviously) to get:
$$\frac{d^2 \psi}{dx^2} + \frac{2m (E-V_0)}{\hbar^2} \psi = 0$$
So now, my k is in accordance with the answer (E-V), but because of the + I get REAL exponents, and if E > V, k is real and the exponents stay real (wrong), and if E < V, k is complex and the exponents become complex (wrong)...???

I can't figure it out...

2. Jan 6, 2009

### daschaich

This is where things went backwards. The solution to
$$\frac{d^2 \psi}{dx^2} = k^2 \psi$$
is
$$\psi (x) = A e^{kx} + B e^{-kx}$$
with no $$i$$ in the exponents. In order to get the other form we should define
$$k^2 = \frac{2m (E-V_0)}{\hbar^2}$$
so that
$$\frac{d^2 \psi}{dx^2} = -k^2 \psi.$$

This is the answer "they" get (and has the right behavior for $$E > V_0$$, $$E < V_0$$), but you can easily check the signs just by substituting $$\psi$$ back into the differential equation.

3. Jan 6, 2009

### Nick89

Omg... You're completely right of course. I checked and re-checked that, and I was actually under the impression that the solution to
$$y'' - k^2y = 0$$
had complex exponents and the solution to
$$y'' + k^2y = 0$$
had real exponents... It's the other way around of course. Can't believe I mixed them up, must have been the holidays hehe...