# Two different ways? (probability denstities and currents)

• AMB
In summary, the conversation discusses two solutions for calculating probability densities and currents from the Schrödinger and Klein-Gordon equations. The first solution involves multiplication of the conjugate wave function to the left of the equation, while the second solution involves multiplication to the right. The conversation also touches on the issue of commutation and the validity of the second solution. It is mentioned that for the Klein-Gordon equation, the wave function is a real number field and thus commutes with any other real number field. However, there may be confusion between wave functions and field operators, as field operators do not necessarily commute.
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!

AMB said:
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 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.

stevendaryl said:
So with that choice of $\rho$ and $\vec{j}$, you get the continuity equation.

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

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

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

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.

AMB said:
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.

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

AMB said:
... or if the solutions on the second case I mentioned are valid, and why.

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

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

stevendaryl said:
I tried that link, and I could not access the page. Some permissions thing.
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 ), 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:
AMB said:
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 ), 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.

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:
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!

## What is the difference between probability densities and currents?

Probability densities and currents are two different ways of representing the distribution of a certain quantity. Probability density refers to the likelihood of a particular value occurring within a given range, while current is the flow or rate of change of that quantity. Probability densities are typically used in statistics and probability theory, while currents are commonly used in physics and engineering.

## How are probability densities and currents calculated?

Probability densities are calculated by dividing the probability of a certain value occurring by the total number of possible outcomes. Currents, on the other hand, are calculated by taking the derivative of the probability density function with respect to time. In simpler terms, probability densities are a measure of the likelihood of a certain value, while currents indicate the rate of change of that value.

## What are some real-world applications of probability densities and currents?

Probability densities are commonly used in fields such as statistics, finance, and risk assessment to analyze and predict the likelihood of certain events or outcomes. Currents, on the other hand, have applications in physics and engineering, such as in the study of electromagnetism and fluid dynamics. They can also be used to analyze and predict the behavior of financial markets and economic systems.

## Can probability densities and currents be used together?

Yes, probability densities and currents can be used together to gain a more comprehensive understanding of a particular system or phenomenon. For example, in physics, the probability density of an electron can be used to calculate its current density, which is essential in understanding the behavior of electronic devices.

## What are the limitations of using probability densities and currents?

Probability densities and currents are mathematical models that are based on assumptions and simplifications of real-world systems. Therefore, their predictions may not always accurately reflect the true behavior of a system. Additionally, these models may not be applicable in all situations, and their calculations may be affected by external factors or uncertainties.

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