Similarly to the probability current in non-relativistic quantum mechanics, there is a conserved current for the Klein Gordon equation, however a different one. I'm trying to calculate that.

The KG equation reads $$(\partial^2 + m^2)\psi. $$ From this I want to show that when $\psi$ satisfies the KG equation, then the following is also satisfied:

$$\psi^* \partial^2 \psi - \psi \partial^2 \psi^* = 0.$$

Which would imply the current conservation law

$$\partial_{\mu}[\psi^* \partial^{\mu} \psi - \psi \partial^{\mu}\psi^*] = 0.$$

However I can't seem to find a way to get rid of the mass term by simply slapping the complex conjugated on the left side or something like that.

What am I missing?

  • $\begingroup$ the mass is a real parameter, so it doesn't change when you take the c.c. of the KG equation $\endgroup$ – Phoenix87 Jan 13 '15 at 11:38

Consider the KG equation for a complex scalar field $\phi(x) \in \mathbb{C}$ $$ \tag{1} ( \square + m^2 ) \phi(x) = 0,$$ and its complex conjugate $$ \tag{2} ( \square + m^2 ) \phi^*(x) = 0.$$ Multiplying at the left (1) by $\phi^*(x)$ and (2) by $\phi(x)$ you have respectively $$ \tag{3} \phi^*(x) (\square + m^2 ) \phi(x) = 0$$ and $$ \tag{4} \phi(x) ( \square + m^2 ) \phi^*(x) = 0.$$ Subtracting now (4) from (3) you have $$ \tag{5} \phi^*(x) (\square + m^2 ) \phi(x) - \phi(x) ( \square + m^2 ) \phi^*(x) = 0 $$ $$ \tag{5-1} \Longrightarrow (\phi^* \square \phi - \phi \square \phi^*) + m^2 (\phi^*\phi - \phi \phi^*) = 0$$ where in the last step we used the fact that $m \in \mathbb{R}$ and hence $\phi(x) m = m \phi(x)$ (and omitted the $x$ argument for lazyness).

Because also $\phi(x), \phi^*(x) \in \mathbb{C}$, i.e. they are just complex number (as opposite to operator fields in QFT), you have $\phi(x)\phi^*(x)=\phi^*(x)\phi(x)$ and hence the conclusion: $$ \phi^* \square \phi - \phi \square \phi^* = 0 \Longleftrightarrow \partial_\mu ( \phi^*(x) \partial^\mu \phi(x) - \phi(x) \partial^\mu \phi^*(x) ) = 0, $$ i.e. $$ \partial_\mu J^\mu(x) = 0, \qquad J^\mu(x) \equiv \phi^*(x)\partial^\mu\phi(x) - \phi(x)\partial^\mu \phi^*(x). $$

Note: I'm using the notation $$ \square \equiv \partial^2 \equiv \partial_\mu \partial^\mu, $$ and my $\phi(x)$ is your $\psi(x)$.


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