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## Homework Statement

I was just studying the Klein Gordon equation with fields. In particular I was reviewing the continuity equation. In the derivation for it, the usual approach is to take the klein-gordon equation (I'm using 4-vector covariant notation), multuply by the complex conjugate of the wave function by the left, then take the complex conjugate of this equation, then substract one from the other, and working one gets the divergence of the 4-current. I've tryied something else, I just wanted to see what I've obtained if instead of taking the complex conjugate of the equation, I've just stated the same equation for the complex conjugate of the wave function, then multiplied by the left by the wave function, and substracted this from the klein-gordon with fields equation for the wave function multiplied by the left by the complex conjugate. I hoped to get nothing at all, perhaps an identity of the form 0=0, and did this just as a practice. I wanted to discuss the result I get, so here I am.

## Homework Equations

The klein gordon equation with fields reads as follows:

##\displaystyle \left ( \hat p^{\mu}-\frac{e}{c}A^{\mu} \right ) \left ( \hat p_{\mu}-\frac{e}{c}A_{\mu} \right ) \psi=m_0^2 c^2 \psi##

##\displaystyle \left [ g^{\mu \nu} \left ( \frac{\partial}{\partial x^{\nu}} - \frac{e}{c}A_{\nu} \right ) \left ( \frac{\partial}{\partial x^{\mu}}-\frac{e}{c}A_{\mu} \right ) \right ] \psi=\frac{m_0^2 c^2}{\hbar^2} \psi## (1)

Here p is the four momentum operator, A is the four vector for the electromagnetic field.

##\displaystyle A^{\mu}=(A_0,\vec A)=g^{\mu \nu}A_{\nu}##

##\displaystyle \hat p^{\mu}=i\hbar \frac{\partial}{\partial x_{\mu}}=i\hbar \nabla^{\mu}=\left ( i\hbar\frac{\partial}{\partial (ct)},-i\hbar \vec \nabla \right ) ##

3. The Attempt at a Solution [/B]

So this is what I did. I took the Klein gordon equation in the form (1), and then multiplied it by the c.c. of the wave function, to obtain:

##\displaystyle \psi^* \left [ g^{\mu \nu} \left ( \frac{\partial}{\partial x^{\nu}} - \frac{e}{c}A_{\nu} \right ) \left ( \frac{\partial}{\partial x^{\mu}}-\frac{e}{c}A_{\mu} \right ) \right ] \psi=\psi^* \frac{m_0^2 c^2}{\hbar^2} \psi=\frac{m_0^2 c^2}{\hbar^2} \left | \psi \right |^2 ## (2)

In the same way I've considered the equation for the complex conjugate of the wave function, and multiplied it by the wave function by the left to obtain:

##\displaystyle \psi \left [ g^{\mu \nu} \left ( \frac{\partial}{\partial x^{\nu}} - \frac{e}{c}A_{\nu} \right ) \left ( \frac{\partial}{\partial x^{\mu}} -\frac{e}{c}A_{\mu} \right ) \right ] \psi^*=\frac{m_0^2 c^2}{\hbar^2} \left | \psi \right |^2## (3)

Then I made the substraction (2)-(3), and after some algebra I've arrived to:

##\displaystyle -g^{\mu \nu} \left [ \frac{\partial}{ \partial x^{\mu} } \left ( \psi^* \frac{ \partial \psi}{\partial x^{\nu}} - \psi \frac{ \partial \psi^* }{ \partial x^{\nu} } \right ) + \frac{ie}{\hbar c} \left ( \frac{\partial \psi }{\partial x^{\nu}} A_{\mu} \psi^* - \frac{\partial \psi^*}{\partial x^{\nu}} A_{\mu} \psi \right ) \right ]=0##

I don't know if this result is right, I could made some mistake, I have eliminated lots of products of operators and things like that. I wanted to know if someone had tried this before, and if this relation has any sense at all.

Thanks in advance.

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