Probability Density Current

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Homework Statement



Calculate the probability current density vector [itex]\vec{j}[/itex]for the wave function [itex]\psi = Ae^{-(wt-kx)}[/itex].

Homework Equations


From my very poor and beginner's understanding of probability density current it is :

[itex]\frac{d(\psi \psi^{*})}{dt}=\frac{i\hbar}{2m}[\frac{d\psi}{dx}\psi^{*}-\frac{d\psi^{*}}{dx}\psi][/itex]


The Attempt at a Solution


By applying the RHS of the above equation :

[itex]\frac{i\hbar}{2m}[-A^{2}ikxe^{-i(ωt-kx)}e^{i(ωt-kx)}-A^{2}ikxe^{i(ωt-kx)}e^{-i(ωt-kx)}][/itex]

This gives :

[itex]\frac{-2iA^{2}ik\hbar}{2m}=\frac{k \hbar A^{2}}{m}[/itex]

This is not the correct answer. :( What have I done wrong ?
 
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Answers and Replies

  • #2
TSny
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[itex]\frac{d(\psi \psi^{*})}{dt}=\frac{i\hbar}{2m}[\frac{d\psi}{dx}\psi^{*}-\frac{d\psi^{*}}{dx}\psi][/itex]

This equation isn't correct. But the right hand side does represent the current density.

The Attempt at a Solution


By applying the RHS of the above equation :

[itex]\frac{i\hbar}{2m}[-A^{2}ikxe^{-i(ωt-kx)}e^{i(ωt-kx)}-A^{2}ikxe^{i(ωt-kx)}e^{-i(ωt-kx)}][/itex]

This gives :

[itex]\frac{-2iA^{2}ik\hbar}{2m}=\frac{k \hbar A^{2}}{m}[/itex]

This is not the correct answer. :( What have I done wrong ?

This looks correct to me for the magnitude of the current density.
Maybe you need to express your answer as a vector.

Or, maybe you are supposed to allow A to be a complex number. If so, then A2 in your answer would need to be modified.
 
  • #3
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There seems to be a sign mistake - I think.
 
  • #4
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This equation isn't correct. But the right hand side does represent the current density.



This looks correct to me for the magnitude of the current density.
Maybe you need to express your answer as a vector.

Or, maybe you are supposed to allow A to be a complex number. If so, then A2 in your answer would need to be modified.

No, letting A be complex doesn't change anything.
 
  • #5
TSny
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No, letting A be complex doesn't change anything.

The probability current must be real. If A is complex (i.e., nonzero imaginary part), then A2 is also complex.
 
  • #6
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This equation isn't correct. But the right hand side does represent the current density.



This looks correct to me for the magnitude of the current density.
Maybe you need to express your answer as a vector.

Or, maybe you are supposed to allow A to be a complex number. If so, then A2 in your answer would need to be modified.

What is the correct form of the LHS of the equation ? I'm assuming the RHS is completely correct ?

I realised that [itex]\hbar k[/itex] = mv = p can be substituted into the last step of my workings to obtain j = A2v which is the correct answer.




There seems to be a sign mistake - I think.

Where exactly ?
 
  • #7
TSny
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What is the correct form of the LHS of the equation ? I'm assuming the RHS is completely correct ?

The left hand side represents the rate of change of the probability current at some point. This should equal the negative of the divergence of the probability current vector at that point. In one dimension the gradient is just the derivative with respect to x. So, to make the equation correct, you would need to apply -d/dx to the RHS.

I realised that [itex]\hbar k[/itex] = mv = p can be substituted into the last step of my workings to obtain j = A2v which is the correct answer.

Good.
 
  • #8
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The probability current must be real. If A is complex (i.e., nonzero imaginary part), then A2 is also complex.

No, A2 is not necessarily equal to A x A. It is a fairly common notation to define A2 = AA* which is real.
 
  • #9
TSny
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No, A2 is not necessarily equal to A x A. It is a fairly common notation to define A2 = AA* which is real.

That seems confusing to me. There are times when we really do want the square of a complex number z (i.e., z2) rather than the square of the magnitude (i.e., |z|2 = z*z).

But, anyway, at least we agree on the substance of the answer to the question.
 

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