Probability Current for Free Particle Wave Function

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Homework Help Overview

The discussion revolves around finding the probability current, J, for a free particle wave function, specifically examining the expression for J and its implications regarding the direction of flow. The subject area is quantum mechanics, focusing on wave functions and probability currents.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive the probability current from the given wave function and questions the correctness of their complex conjugate calculation. They also inquire about determining the direction of the probability current.

Discussion Status

Some participants confirm the original poster's understanding of the direction of the probability current based on its sign. There is a discussion about the assumptions regarding the wave function's amplitude and its implications for the calculation.

Contextual Notes

Participants note that the wave function is a plane wave and discuss the implications of assuming the amplitude A is real versus complex. There is mention of a potential misunderstanding regarding the representation of the amplitude in the probability current formula.

singular
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[SOLVED] Probability Current for Free Particle Wave Function

Homework Statement


Find the probability current, J for the free particle wave function. Which direction does the probability current flow?

Homework Equations


[tex]J(x,t) = \frac{ih}{4\pi m}\left(\Psi \frac{\partial \Psi^{*}}{\partial x} - \Psi^{*} \frac{\partial \Psi}{\partial x}\right)[/tex]

[tex]\Psi_{k}\left(x, t\right) = Ae^{i\left(kx - \frac{hk^{2}}{4\pi m}t}\right)[/tex]

The Attempt at a Solution


I won't take the time to put my math into Latex, but I come up with

[tex]J(x,t) = \frac{A^{2}hk}{2\pi m}[/tex]

Is this correct or did I do the complex conjugate wrong?
How would I find the probability current flow direction?
 
Last edited:
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I just read that the direction is simply the sign of J(x,t) ( - corresponds to left and + corresponds to right). If this is so, that would be great. Can anyone confirm? (it wasnt exactly a textbook source)
 
singular said:
I just read that the direction is simply the sign of J(x,t) ( - corresponds to left and + corresponds to right). If this is so, that would be great. Can anyone confirm? (it wasnt exactly a textbook source)

Yes, that's correct (and you can tell that the wavefunction you have is a plane wave traveling to the right since the sign of the x and t terms in the exponential have opposite signs). Your current looks good if A is assumed real (you should really have |A|^2 there, not A^2 since a gets complex conjugated).
 
kdv said:
Yes, that's correct (and you can tell that the wavefunction you have is a plane wave traveling to the right since the sign of the x and t terms in the exponential have opposite signs). Your current looks good if A is assumed real (you should really have |A|^2 there, not A^2 since a gets complex conjugated).

Great, thank you very much.
 

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