Real and Complex Parts of a Wave Function

Rahmuss
Messages
222
Reaction score
0

Homework Statement



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.

Homework Equations


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


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)?
 
Physics news on Phys.org
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.

There is more information about complex conjugates here:

http://mathworld.wolfram.com/ComplexConjugate.html
 
Last edited:
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.
 
Rahmuss said:
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.

Yeah that's it basically.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top