# Klein-Gordon, gauge transformation [closed]

It must be really simple, but I cannot get why can we add an $i e \frac{\partial \Lambda}{\partial x}$ in the second row below.

The propagation of a charged scalar particle, along the x-axis and in an electromagnetic field is described by the equation

$$\left[\left(\frac{\partial}{\partial x} − ieA_x\right)^2 − \frac1{c^2}\left(\frac{∂}{∂t} - ieV\right)^2 − \frac{m^2c^2}{\hbar^2}\right] \phi(x, t) = 0$$

Show that this equation is not changed under the gauge transformation

$$V \rightarrow V−\frac{\partial \Lambda}{\partial t}$$ $$A_x \rightarrow A_x+\frac{∂Λ}{∂x}$$ $$\phi \rightarrow \phi \exp(ie\Lambda),$$

which shows that the scalar field must be complex, in order to respect gauge invariance.

• There is actually a sign problem in your covariant derivative, which should read $\partial_{t}+ieV$. Then the shift given by the time derivative of the exponential cancels with your choice of the covariant transformation of the gauge potential. May 11, 2014 at 7:46
• oh thank you, it would have taken me years to figure that out!
– mtgt
May 11, 2014 at 10:04

The term $i e \frac{\partial \Lambda}{\partial x}$ simply comes from taking the derivative of $e^{i e \Lambda}$. $$\partial_x\left(\phi e^{ie\Lambda}\right)=e^{ie\Lambda}\left(\partial_x+i e \partial_x\Lambda\right)\phi$$
• $ie\Lambda$ should actually be $i e \partial_x \Lambda$? May 11, 2014 at 1:39