Klein-Gordon equation and continuity equation

  • #1
dyn
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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
 
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  • #2
dyn said:
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.

dyn said:
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##.
 
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  • #3
Thank you very much for your help
 
  • #4
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,
$$\partial_{\mu} j^{\mu} = \mathrm{i} (\psi^* \Box \psi-\psi \Box \psi^*) = -\mathrm{i} m^2 (\psi^* \psi-\psi \psi^*)=0.$$
 
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  • #5
dyn said:
ψ could be real
Could it be purely real? I don't think Schrödinger equation has purely real solutions.
 
  • #6
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.
 
  • #7
vanhees71 said:
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##.
 
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