I Klein-Gordon equation and continuity equation

[dyn]

Hi
I am using the textbook "Modern Particle Physics" by Thomson. Working from the K-G equation and comparing with the continuity equation he states that the probability density is given by

ρ = i ( ψ^*(∂ψ/∂t) - ψ(∂ψ^*/∂t) )

He then states that the factor of i is included to ensure that the probability density is real. My question is - why does the factor of i make this real. This implies the quantity inside the bracket is pure imaginary. Why is that true ? ψ could be real , complex or pure imaginary. It is just a general wavefunction
Thanks

 

[PeterDonis]

This implies the quantity inside the bracket is pure imaginary.

Yes. You can verify this easily by writing $\psi = \psi_R + i \psi_I$, where $\psi_R$ and $\psi_I$ are real, and computing the quantity inside the bracket. You should find that all of the real parts cancel.

ψ could be real , complex or pure imaginary.

Yes, but the combination of terms inside the bracket is not $\psi$ and, as above, can be shown to be pure imaginary for a general $\psi$.

 

[dyn]

Thank you very much for your help

 

[vanhees71]

A real KG field describes an uncharged particle, and thus for this case $\rho=0$.

In the case of charged fields, $\rho$ is the time-component of the Noether charge from invariance under $\psi \rightarrow \exp(-\mathrm{i} \alpha) \psi, \quad \psi^* \rightarrow \exp(+\mathrm{i} \alpha) \psi^*$,
$$j_{\mu} = \mathrm{i} (\psi^* \partial_{\mu} \psi - \psi \partial_{\mu} \psi^*).$$
That's a real vector field, which is immediately clear when taking the complex conjugate of this definition.

That it's conserved follows from the (free) KG equation,
$$\partial_{\mu} j^{\mu} = \mathrm{i} (\psi^* \Box \psi-\psi \Box \psi^*) = -\mathrm{i} m^2 (\psi^* \psi-\psi \psi^*)=0.$$

 

[Hill]

ψ could be real

Could it be purely real? I don't think Schrödinger equation has purely real solutions.

 

[vanhees71]

Sure, the KG equation is
$$(\Box+m^2) \psi=0$$
and thus has real solutions.

The time-dependent Schrödinger equation has a complex coefficient and thus has no real solutions.

 

[Hill]

Sure, the KG equation is
$$(\Box+m^2) \psi=0$$
and thus has real solutions.

The time-dependent Schrödinger equation has a complex coefficient and thus has no real solutions.

Ah, I see. So, when $\psi$ is real, then $j_{\mu} = \mathrm{i} (\psi^* \partial_{\mu} \psi - \psi \partial_{\mu} \psi^*) = 0$ and, in the OP, $\rho=0$.