# Real and Complex Parts of a Wave Function

1. Sep 21, 2007

### Rahmuss

1. The problem statement, all variables and given/known data

Just a snipit of one of my homework problems. I'm trying to find out what $$\Psi \frac{\partial \Psi^{*}}{\partial x}$$ equals to help me find out what the probability current for a given free particle is.

2. Relevant equations
$$\Psi = Ae^{i(kx-\frac{\hbar k^{2}t}{2m})}$$

3. The attempt at a solution

I view $$\Psi^{*}$$ as the complex part of the given wave function; but in this case there is no real part, it's all complex. Does that mean the real part is zero? If so then $$\Psi \frac{\partial \Psi^{*}}{\partial x} = 0$$. If $$\Psi = \Psi^{*}$$, then the larger equation I'm trying to calculate comes out to be zero because it's:

$$\Psi \frac{\partial \Psi^{*}}{\partial x} - \Psi^{*} \frac{\partial \Psi}{\partial x}$$

So what am I missing here? Does it actually have a zero probability current because it's a "free particle" (whatever that really means)?

2. Sep 21, 2007

### Kurdt

Staff Emeritus
I think you've been misinformed somewhere. The starred notation means the complex conjugate (certainly if this is quantum mechanics HW) of the function. Basically to find the complex conjugate of a complex function you reverse the sign in front of any i. For example:

If we have a general complex number z = a + ib then it has a complex conjugate of z* = a - ib.

http://mathworld.wolfram.com/ComplexConjugate.html

Last edited: Sep 21, 2007
3. Sep 21, 2007

### Rahmuss

Kurdt - Ah, brilliant. Ok, yes, I do remember it being the conjugate. I think in my mind I had the idea that it dealt with something complex (I don't mean complicated; but $$i$$ ), and so I must have mentally given it the value of the complex portion of the wave function. Thanks for that clarification.

So, if I understand correctly, then in the case listed I would have

$$\Psi = Ae^{i(kx - \frac{\hbar k^{2}t}{2m})}$$
and
$$\Psi^{*} = Ae^{-i(kx - \frac{\hbar k^{2}t}{2m})}$$​

If that's correct, then I think I can figure it out. Thanks for the help. I'll jump back on later if I need more help; but for now it's off to class.

4. Sep 21, 2007

### Kurdt

Staff Emeritus
Yeah thats it basically.