How to solve the Klein Gordon Complex Field?

[Ben Niehoff]

The point to remember is that a complex field actually has two degrees of freedom: a real part and an imaginary part. Alternatively, we can use a more convenient basis by using linear combinations:

z = x + iy
\bar z = x - iy

and the inverse transformation

x = \frac12 z + \frac12 \bar z
y = -i \frac12 z + i \frac12 \bar z

and so, instead of using x and y, we can just as easily use z and z* as independent variables. This is why we treat phi and its conjugate as independent fields.

 

[sadness]

The point to remember is that a complex field actually has two degrees of freedom: a real part and an imaginary part. Alternatively, we can use a more convenient basis by using linear combinations:

z = x + iy
\bar z = x - iy

and the inverse transformation

x = \frac12 z + \frac12 \bar z
y = -i \frac12 z + i \frac12 \bar z

and so, instead of using x and y, we can just as easily use z and z* as independent variables. This is why we treat phi and its conjugate as independent fields.

but how do you define the derivative with respect to z*?

 

[Ben Niehoff]

Like I said, you just treat z and z* as independent variables. You can call them z and w if it makes you feel better. Then just apply the formula

w = z*

as a constraint equation when you're done.Or, here's yet another way you can look at it, using the definitions above:

\frac{\partial}{\partial z} = \frac{\partial x}{\partial z} \frac{\partial}{\partial x} + \frac{\partial y}{\partial z} \frac{\partial}{\partial y} = (\frac12) \frac{\partial}{\partial x} + (-i \frac12) \frac{\partial}{\partial y}

and therefore

\frac{\partial \bar z}{\partial z} = \frac{\partial}{\partial z} (x - iy) = \frac12 \frac{\partial}{\partial x} (x - iy) - i \frac12 \frac{\partial}{\partial y} (x - iy) = \frac12 - i \frac12 (-i) = \frac12 (1 + i^2) = 0

 

[sadness]

Like I said, you just treat z and z* as independent variables. You can call them z and w if it makes you feel better. Then just apply the formula

w = z*

as a constraint equation when you're done.Or, here's yet another way you can look at it, using the definition z = x + iy:

\frac{\partial}{\partial z} = \frac{\partial x}{\partial z} \frac{\partial}{\partial x} + \frac{\partial y}{\partial z} \frac{\partial}{\partial y} = (1) \frac{\partial}{\partial x} + (-i) \frac{\partial}{\partial y}

and therefore

\frac{\partial \bar z}{\partial z} = \frac{\partial}{\partial z} (x - iy) = \frac{\partial}{\partial x} (x - iy) - i \frac{\partial}{\partial y} (x - iy) = 1 - i (-i) = 1 + i^2 = 0

Okay. One can somewhat define it formally, but it's not a very decent derivative I think. May I ask that what's your opinion on the point I raised in the post #30?

 

[Ben Niehoff]

Okay. One can somewhat define it formally, but it's not a very decent derivative I think. May I ask that what's your opinion on the point I raised in the post #32?

In what sense is it not "decent"? Note, I forgot some factors of 1/2, went and fixed them.

The first expression you have for the Hamiltonian is correct (with both \pi \partial_t \phi and \pi^* \partial_t \phi^*). Try working the problem in detail to see why.

 

[sadness]

In what sense is it not "decent"? Note, I forgot some factors of 1/2, went and fixed them.

The first expression you have for the Hamiltonian is correct (with both \pi \partial_t \phi and \pi^* \partial_t \phi^*). Try working the problem in detail to see why.

I mean, you can not write the derivative to z* without the help of splitting into x and y. I see your point now. Thanks for pointing out my mistake.

 

[Ben Niehoff]

Or alternatively, you cannot write a derivative w.r.t. x without splitting into z and z*. It's a matter of using different coordinate systems to represent the same thing.

\frac{\partial}{\partial x} = \frac{\partial z}{\partial x} \frac{\partial}{\partial z} + \frac{\partial \bar z}{\partial x} \frac{\partial}{\partial \bar z}