# Homework Help: Klein-Gordon equation with fields

1. Aug 7, 2015

### Telemachus

1. The problem statement, all variables and given/known data
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.

2. Relevant 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

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.

Last edited: Aug 7, 2015
2. Aug 8, 2015

### Telemachus

I'll give in here furhter details on the calculation I've carried on, so I can get some corrections in the procedure, and in the way make a review and perhaps detecting some mistakes by my self. First of all there was a mistake in eq. (1) in a sign, but that was when I wrote this here, in my calculations I did that right.

So (1) should be:

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

And similarly for (2) and (3)

After the substraction (2)-(3) I get:

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

Expanding the terms:

$\displaystyle -g^{\mu \nu} \left \{ \psi^* \left [ \frac{\partial}{\partial x^{\nu}} \frac{\partial}{\partial x^{\mu}} + \frac{\partial}{\partial x^{\nu}} \frac{ie}{\hbar c}A_{\mu} + \frac{ie}{\hbar c}A_{\nu} \frac{\partial}{\partial x^{\mu}} -\frac{e^2}{\hbar^2 c^2 } A_{\nu}A_{\mu} \right ] \psi - \psi \left [ \frac{\partial}{\partial x^{\nu}} \frac{\partial}{\partial x^{\mu}} + \frac{\partial}{\partial x^{\nu}} \frac{ie}{\hbar c}A_{\mu} + \frac{ie}{\hbar c}A_{\nu} \frac{\partial}{\partial x^{\mu}} -\frac{e^2}{\hbar^2 c^2 } A_{\nu}A_{\mu} \right ] \psi^* \right \}=0$

Then:
$\displaystyle -g^{\mu \nu} \left \{ \psi^* \frac{\partial}{\partial x^{\nu}} \frac{\partial}{\partial x^{\mu}} \psi - \psi \frac{\partial}{\partial x^{\nu}} \frac{\partial}{\partial x^{\mu}} \psi^* + \frac{ie}{\hbar c} \left ( \psi^* \frac{\partial}{\partial x^{\nu}} A_{\mu} \psi - \psi \frac{\partial}{\partial x^{\nu}} A_{\mu} \psi^* + \psi^* A_{\nu} \frac{\partial}{\partial x^{\mu}} \psi - \psi A_{\nu} \frac{\partial}{\partial x^{\mu}} \psi^* \right ) \\ -\frac{e^2}{\hbar^2 c^2 } \left ( \psi^* A_{\nu}A_{\mu} \psi - \psi A_{\nu}A_{\mu}\psi^* \right ) \right \}=0$

The last term in parenthesis equals zero.

For the next step, I've used that:

1) $\displaystyle \frac{\partial}{\partial x^{\nu}}\left ( \psi^* \frac{\partial}{\partial x^{\mu}}\psi \right )= \frac{\partial \psi^*}{\partial x^{\nu}}\frac{\partial \psi }{\partial x^{\mu}}+\psi^* \frac{\partial}{\partial x^{\nu}}\frac{\partial}{\partial x^{\mu}}\psi$

2) $\displaystyle \frac{\partial}{\partial x^{\nu}}\left ( \psi^* A_{\mu} \psi \right )= \frac{\partial \psi^*}{\partial x^{\nu}} A_{\mu} \psi+\psi^* \frac{\partial}{\partial x^{\nu}}A_{\mu}\psi$

3) $\displaystyle g^{\mu \nu} \left ( \frac{\partial \psi^*}{\partial x^{\nu}}\frac{\partial \psi }{\partial x^{\mu}} - \frac{\partial \psi}{\partial x^{\nu}}\frac{\partial \psi^* }{\partial x^{\mu}} \right ) = 0$

And after making all that substitutions I get the result I've posted at the beggining of this topic (I have one correction for it, that I've noted while typing this):

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

Last edited: Aug 8, 2015
3. Aug 11, 2015

### sgd37

I think there is a sign error somewhere because all you're showing is that $$\partial (\psi^*\partial\psi)+\frac{ie}{\hbar c}\partial\psi A \psi$$ is real i.e. $$z-z^*=2Im(z)$$ which is not a trivial statement for complex fields

4. Aug 11, 2015

### Telemachus

Yes, I think you are right. But for a different reason. I actually didn't substract the c.c. of the equation (what I did was substracting the k-g eq. for the . But anyway, the first term is equal to the one that one obtains by doing exactly what you said, that is the procedure the book follows to obtain the continuity equation (I'm following Greiner's relativistic quantum mechanics btw). So the second term must be equal to the one obtained in that way (because it must be equal to the first term) or it could be zero. I think it actually should be zero, so what I would obtain si the continuity equation without fields, but I should demonstrate that

$\displaystyle \frac{\partial \psi }{\partial x^{\nu}} A_{\mu} \psi^*= \frac{\partial \psi^*}{\partial x^{\nu}} A_{\mu} \psi$