# Two different ways? (probability denstities and currents)

1. Jun 10, 2015

### AMB

Hi, I would appreciate some help with this issue: I want to calculate the probability denstities and currents from the Schrödinger and Klein-Gordon equations, and I've found 2 ways so far, the one that gives the "standard" result (the one I've seen on my course, or wikipedia) but I don't understand, and the one that is obvious, but something I don't like about it (more later).

The first solution is the one in Bjorken's book, and here: http://goo.gl/D04Xlo, one just multiplies the conjugate wave function to the left of the equation, or the wave function to the left of the conjugate equation, and substracts both, but I don't understand why, for example, ψ*(∂ψ/∂t)+ψ(∂ψ*/∂t)=∂(ψ*ψ)/∂t (I think they are implying that), or even why the m^2 terms in K-G dissapear, the same with the ∇ relations, to me those things do not commute, or is there something else happening?

The second solution can be found here https://goo.gl/0Rq9nx and in many other sites, the difference is that one wave function is multiplied to the RIGHT of the equations, so the operations are straightforward, the m^2 terms obviously cancel out, but the formulas for the probability denstities and currents are not the same as before.

So my questions are: what I am missing in the first case? Are the solutions on the second case somewhat equivalent to the first ones? why can I call both probability denstities and currents?

Thanks!!

2. Jun 10, 2015

### stevendaryl

Staff Emeritus
The slides don't actually give a derivation of the current and density, but really count as a guess.

$\rho = i (\psi^* \dfrac{\partial \psi}{\partial t} - \psi \dfrac{\partial \psi^*}{\partial t})$

1. $\dfrac{\partial \rho}{\partial t} = i (\dfrac{\partial \psi^*}{\partial t} \dfrac{\partial \psi}{\partial t} + \psi^* \dfrac{\partial^2 \psi}{\partial t^2} - \dfrac{\partial \psi}{\partial t} \dfrac{\partial \psi^*}{\partial t} - \psi \dfrac{\partial^2 \psi^*}{\partial t^2})$
2. $\dfrac{\partial \rho}{\partial t}= i (\psi^* \dfrac{\partial^2 \psi}{\partial t^2} - \psi \dfrac{\partial^2 \psi^*}{\partial t^2})$ (the first and third terms on the right cancel)
3. $\dfrac{\partial \rho}{\partial t}= i (\psi^* (\nabla^2 - m^2) \psi - \psi (\nabla^2 - m^2) \psi^*$ (by using the Klein-Gordon equation to replace $\frac{\partial^2}{\partial t^2}$)
4. $\dfrac{\partial \rho}{\partial t}= i (\psi^* \nabla^2 \psi - \psi \nabla^2 \psi^*$ (the $m^2$ terms on the right cancel out)
With $\vec{j} = i (\psi^* \nabla \psi - \psi \nabla \psi^*)$, you have:
1. $\nabla \cdot \vec{j} = i ((\nabla \psi^*) \cdot (\nabla \psi) + \psi^* \nabla^2 \psi - (\nabla \psi) \cdot (\nabla \psi^*) - \psi \nabla^2 \psi^*)$
2. $\nabla \cdot \vec{j} = i (\psi^* \nabla^2 \psi - \psi \nabla^2 \psi^*)$ (because the first and third terms cancel)
So with that choice of $\rho$ and $\vec{j}$, you get the continuity equation.

3. Jun 10, 2015

### stevendaryl

Staff Emeritus
Hmm. Actually, there might be a sign error somewhere. It seems to me that I get

$\dfrac{\partial \rho}{\partial t} = + \nabla \cdot \vec{j}$

$\dfrac{\partial \rho}{\partial t} = - \nabla \cdot \vec{j}$

4. Jun 10, 2015

### AMB

Thanks for your answer, but my doubts are just the same, I don't see why "the first and third terms on the right cancel", or why "the m2 terms on the right cancel out", since I assume ψ and ψ* do not commute... or if the solutions on the second case I mentioned are valid, and why.

5. Jun 10, 2015

### stevendaryl

Staff Emeritus
For the Klein-Gordon equation, $\psi$ is a real number field, it commutes with any other real number field (such as $\psi^*$)

6. Jun 10, 2015

### stevendaryl

Staff Emeritus
I tried that link, and I could not access the page. Some permissions thing.

7. Jun 10, 2015

### AMB

Thanks, I was long suspecting that at this stage, but I still can't find where to read about that with more detail. I hope that for the Schrödinger equation is just the same.

It happened to me once but it worked after trying again, it's page 93 of An Introduction to Advanced Quantum Physics, by Hans Paar (it's also here http://goo.gl/WRLFQy [Broken]), anyway if ψ and ψ* commute the solutions are the same.

I need to convince myself about some things, I was studying fermions and I tought I read ψ and ψ* do not commute, now I'm not sure when that happens or not.

Last edited by a moderator: May 7, 2017
8. Jun 10, 2015

### stevendaryl

Staff Emeritus
Well, there are two different things that look very similar: wave functions and field operators. Field operators don't commute, in general.

Last edited by a moderator: May 7, 2017
9. Jun 10, 2015

### AMB

I know what was confusing me: in first quantization ψ it's not an operator, it doesn't have sense to speak about commutation, while in second quantization it is, and I was reading the later before the former...

Thanks!