Probability Current of a Wave Function

1. Nov 19, 2012

leonmate

Find the probability current of Ae^i(kx - ωt) + Be^-i(kx+ωt)

Ok, to my understanding the probability current is the probability that you will find a certain particle as it moves with time, thus the probability of finding it changes with time. Quantum physics is a tricky one to grasp, I've got to say.

I have an equation for probability current:

J = (i*hbar / 2*m)( ψ∇ψ* - ψ*∇ψ)

So, I have figured that ψ = Ae^i(kx - ωt) + Be^-i(kx+ωt)
Thus ψ* = Ae^-i(kx - ωt) + Be^i(kx+ωt)

Not too sure if I have done the complex conjugate correctly here, I'm used to simple uses of it.. z = a + bi, z* = a - bi ... for example

Next, I'm assuming ∇ is del. I've used this function before, it works out as d/dx + d/dy + d/dz .. otherwise known as gradψ for this example.

I've tried differentiating it but ψ∇ψ* is some horrible maths, should i expect that? Wolfram alpha shows that d/dx of ψ is:

Ake^i(kx - ωt) * i'(kx - ωt) - Bke^-i(kx - ωt) * i'(kx - ωt)

I dont really know what i' is? and multiplying that out looks horrific and makes me think I've gone down the wrong path...

2. Nov 19, 2012

leonmate

Could someone confirm for me that I have the complex conjugate correct?

if ψ = Ae^i(kx - ωt) + Be^-i(kx+ωt)
ψ* = Ae^-i(kx - ωt) + Be^i(kx+ωt)

When I work it through with the equation above it looks very close to having lots of values cancel out and leave me with just [i*hbar / 2*m][ikA^2 + ikB^2]

But with the complex conjugate I have at the moment I'm a couple of negative signs away from that

3. Nov 19, 2012

TSny

You complex conjugate looks correct. What do you get for the derivative with respect to x of ψ*?

4. Nov 19, 2012

leonmate

Alright, to my delight it does cancel down somewhat but not quite enough.

The derivatives I found were:

∇ψ = ikAe^i(kx-wt) - ikBe^-i(kx+wt)
∇ψ* = -ikA*e^-i(kx-wt) + ikB*e^i(kx+wt)

where the complex conjugate of A is A* and likewise for B*
Not sure if this is a necessary thing to do, but I found it somewhere and it seemed correct.

Anyway, I wont type it all out because it will take forever but basically I found that:

ψ∇ψ* - ψ*∇ψ = 2[ψ∇ψ*] = 2[-ikAA* + ikBB* +ikAB*e^2kxi - A*Be^-2kxi]

Now I've put it into my equation for J = (i*hbar / 2*m)( ψ∇ψ* - ψ*∇ψ)
The i and 2 cancel out and I'm left with:

J = hbar/m[ψ∇ψ*]

Don't really know where to go from here, some of the people on my course have told me that it should eventually result in zero.

5. Nov 19, 2012

TSny

That looks correct.
I don't think that's right. Certain terms in ψ∇ψ* will cancel certain terms in ψ*∇ψ when you combine ψ∇ψ* - ψ*∇ψ.
The term Ae^i(kx-wt) represents a wave traveling to the right with amplitude A. Be^-i(kx+wt) travels to the left with amplitude B. If |A| ≠ |B| would you expect the current to be zero?

6. Nov 20, 2012

leonmate

When i calculate ψ∇ψ* - ψ*∇ψ i wind up with:

ψ∇ψ* = [Ae^i(a-b) + Be^-i(a+b)][-ikA*e^-i(a-b) + ikB*e^i(a+b)]
where a = kx , b = wt (for simplicity)

ψ∇ψ* = -ikAA* + ikBB* + ikAB*e^[i(a-b)+i(a+b)] - ikA*Be^[-i(a-b)-i(a+b)]

i(a-b)+i(a+b) = 2ai
-i(a-b)-i(a+b) = -2ai

ψ∇ψ* = -ikAA* + ikBB* + ikAB*e^2ai - ikA*Be^-2ai

Next,

ψ*∇ψ = [A*e^-i(a-b) + B*e^i(a+b)][ikAe^i(a-b) - ikBe^-i(a+b)]
= ikAA* - ikBB* - ikA*Be^[-i(a-b)-i(a+b)] + ikAB*e^[i(a+b)+i(a-b)]
= ikAA* - ikBB* - ikA*Be^-2ai + ikAB*e^2ai

So, when these values are subtracted:

[-ikAA* + ikBB* + ikAB*e^2ai - ikA*B^-2ai] - [ikAA* - ikBB* - ikA*Be^-2ai + ikAB*e^2ai]

This looks to me like it results in 2[-ikAA* + ikBB* + ikAB*e^2ai - ikA*Be^-2ai] = 2[ψ∇ψ*]

I follow your logic, however this is where the maths leads me. Perhaps I've made a mistake somewhere in this?

7. Nov 20, 2012

andrien

There is an easy way out for calculating probability current by using the eqn.
∇.j=-∂ρ/∂t,since every thing is dependent on x you can write it,
∂j/∂x=-∂ρ/∂t,where ρ=|ψ|2,and then integrate with respect to x.see if it helps.

8. Nov 20, 2012

TSny

Do the corresponding colored terms cancel? Maybe I'm overlooking a sign.

Last edited: Nov 20, 2012
9. Nov 20, 2012

leonmate

Feel like a fool for missing that, yeah it does cancel down nicely!
Thanks a lot for your help, got a few more tricky quantum questions so I will probably post if/when I get stuck on those haha