Treating phi and phi-dagger as independent

In summary: Basically, the fact that the symmetries are isomorphic means that we can write the complex scalar field in terms of two real scalar fields (and vice versa) and the physics will be the same. This is because the transformation between the real and imaginary parts (SO(2) transformation) is equivalent to the transformation between the modulus and phase (U(1) transformation) of the complex scalar field. Therefore, the symmetries are equivalent and the fields are equivalent. In summary, treating the complex scalar field in QFT, the real and imaginary parts are treated as independent variables and the symmetries SO(2) and U(1) are isomorphic, meaning the fields are equivalent.
  • #1
gamma5772
22
0
In treating the complex scalar field in QFT, [itex]\phi[/itex] and [itex]\phi^\dagger[/itex] are treated as independent variables. I'd like to make sure I understand what this actually means:

Is this the same idea as treating [itex]\phi[/itex] and [itex]\partial \phi[/itex] as independent variables in the Lagrangian? Formally, we view the Lagrangian as a function of two variables, and we can differentiate with respect to the two slots as we please, but at the end of the day, we will plug [itex]\phi (t)[/itex] and [itex] \partial \phi (t) [/itex] into the slots. Physicists abbreviate this procedure as treating [itex]\phi[/itex] and [itex]\partial \phi[/itex] as independent variables.

Is this correct?
 
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  • #2
No. A complex scalar field is actually two real scalar fields that obey SO(2) symmetry. Since SO(2) ~ U(1), these two fields are equivalent to one complex scalar field. You should think of the real and imaginary parts of the complex scalar field as independent variables.
 
  • #3
Dickfore said:
No. A complex scalar field is actually two real scalar fields that obey SO(2) symmetry. Since SO(2) ~ U(1), these two fields are equivalent to one complex scalar field. You should think of the real and imaginary parts of the complex scalar field as independent variables.

A couple questions:
What precisely does it mean to "obey a symmetry?"

QFT in a Nutshell and Peskin and Schroeder (If I recall correctly -- I don't have the latter on hand right now) adopt the view that [itex]\phi[/itex] and [itex]\phi ^\dagger[/itex] are independent variables, writing expressions such as [itex] \phi^\dagger \phi [/itex] in the lagrangian and differentiating with respect to them. What do you make of, for example, differentiating with respect to [itex]\phi^\dagger[/itex], as this is seemingly incompatible with your assertion.
 
  • #4
Well, we know that:

[tex]
\begin{array}{rcl}
\phi & = & \frac{\phi_{R} + i \, \phi_{I}}{\sqrt{2}} \\

\phi^{\dagger} & = & \frac{ \phi_{R} - i \, \phi_{I}}{\sqrt{2}}
\end{array} \Rightarrow
\begin{array}{rcl}
\phi_{R} & = & \frac{\phi + \phi^{\dagger}}{\sqrt{2}} \\

\phi_{I} & = & \frac{\phi - \phi^{\dagger}}{\sqrt{2} \, i} \\
\end{array}
[/tex]
where [itex]\phi_{R}, \phi_{I}[/itex] are real scalar fields and the factor [itex]2^{-1/2}[/itex] is introduced to make the Jacobian of the transformation have modulo 1.

Then, one can write the expression [itex]\phi^{\dagger} \, \phi[/itex] as:
[tex]
\frac{1}{2} \, \left(\phi^{2}_{R} + \phi^{2}_{I} + i \, [\phi_{R}, \phi_{I}]\right)
[/tex]
and the last term drops out assumng the two real fields commute between each other, which is the same as:
[tex]
[\phi_{R}, \phi_{I}] = \left[\frac{\phi + \phi^{\dagger}}{\sqrt{2}}, \frac{\phi - \phi^{\dagger}}{\sqrt{2} \, i}\right] = i [\phi, \phi^{\dagger}] = 0
[/tex]
Then, following the chain rule for partial differentiation:
[tex]
\frac{\partial}{\partial \phi^{\dagger}} = \frac{\partial \phi_{R}}{\partial \phi^{\dagger}} \, \frac{\partial}{\partial \phi_{R}} + \frac{\partial \phi_{I}}{\partial \phi^{\dagger}} \, \frac{\partial}{\partial \phi_{I}} = \frac{1}{\sqrt{2}} \, \left(\frac{\partial}{\partial \phi_{R}} + i \, \frac{\partial}{\partial \phi_{I}}\right)
[/tex]
The partial derivatives w.r.t. to the real and imaginary part are:
[tex]
\frac{\partial}{\partial \phi_{R}} \, \left(\phi^{\dagger} \, \phi\right) = \frac{\partial}{\partial \phi_{R}} \, \left[\frac{1}{2} \, \left(\phi^{2}_{R} + \phi^{2}_{I}\right)\right] = \phi_{R}
[/tex]
[tex]
\frac{\partial}{\partial \phi_{I}} \, \left(\phi^{\dagger} \, \phi\right) = \frac{\partial}{\partial \phi_{I}} \, \left[\frac{1}{2} \, \left(\phi^{2}_{R} + \phi^{2}_{I}\right)\right] = \phi_{I}
[/tex]
and we finally have:
[tex]
\frac{\partial}{\partial \phi^{\dagger}} \, \left(\phi^{\dagger} \, \phi\right) = \frac{\phi_{R} + i \, \phi_{I}}{\sqrt{2}} = \phi
[/tex]
which is exactly as it should be.
 
  • #5
The symmetry SO(2) means that we have a liberty of choosing what we call [itex]\phi_{R}[/itex] and [itex]\phi_{I}[/itex]. Namely, we can perform a 'rotation':

[tex]
\left(\begin{array}{c}
\phi'_{R} \\

\phi'_{I}\end{array}\right) = \left(\begin{array}{cc}
\cos{\alpha} & \sin{\alpha} \\

-\sin{\alpha} & \cos{\alpha}
\end{array}\right) \cdot \left(\begin{array}{c}
\phi_{R} \\

\phi_{I}\end{array}\right)
[/tex]
The matrix that achieves this rotation is the general form of a 2x2 matrix that is orthogonal (meaning [itex]O \cdot O^{T} = 1[/itex]), hence the term [itex]O(2)[/itex] and has a determinant one (hence the prefix special).

If one translates what this means for the complex scalar field, one gets:
[tex]
\phi' = \frac{\phi'_{R} + i \, \phi'_{I}}{\sqrt{2}} = \frac{1}{\sqrt{2}} \, \left[(\cos{\alpha} \, \phi_{R} + \sin{\alpha} \, \phi_{I}) + i \, (-\sin{\alpha} \, \phi_{R} + \cos{\alpha} \, \phi_{I})\right] = (\cos{\alpha} - i \, \sin{\alpha}) \, \frac{\phi_{R} + i \, \phi_{I}}{\sqrt{2}} = e^{-i \, \alpha} \, \phi
[/tex]
and similarly:
[tex]
\phi'^{\dagger} = \frac{\phi'_{R} - i \, \phi'_{I}}{\sqrt{2}} = \frac{1}{\sqrt{2}} \, \left[(\cos{\alpha} \, \phi_{R} + \sin{\alpha} \, \phi_{I}) - i \, (-\sin{\alpha} \, \phi_{R} + \cos{\alpha} \, \phi_{I})\right] = (\cos{\alpha} + i \, \sin{\alpha}) \, \frac{\phi_{R} - i \, \phi_{I}}{\sqrt{2}} = e^{i \, \alpha} \, \phi^{\dagger}
[/tex]
This is what is called [itex]U(1)[/itex] transfromation meaning a unitary 1x1 matrix (i.e. a complex number with modulo 1).

The term [itex]\phi^{\dagger} \, \phi[/itex] remains unchanged under the [itex]U(1)[/itex] transformation and so are the other terms in the lagrangian. Hence, the Lagrangian has [itex]U(1)[/itex] symmetry. You can rewrite everything in term of the real and imaginary parts and see that the Lagrangian has [itex]SO(2)[/itex] symmetry. Hence, these two symmetries are isomorphic.
 
Last edited:
  • #6
Thank you! This was helpful.

I am still confused about the symmetry business: you claim that because SO(2) ~ U(1) that two real scalar fields (endowed with SO(2) symmetry) behave as a single complex field. I understand what you mean by possession of a symmetry now, but I don't understand why isomorphic symmetries implies equivalent fields.
 

1. What is the significance of treating phi and phi-dagger as independent?

Treating phi and phi-dagger as independent is a common approach used in quantum mechanics. It allows us to mathematically describe the behavior of a quantum system, particularly in terms of creation and annihilation operators.

2. How does treating phi and phi-dagger as independent affect the calculations in quantum mechanics?

By treating phi and phi-dagger as independent, we can simplify and streamline calculations in quantum mechanics, making it easier to solve complex problems and derive important equations.

3. Is treating phi and phi-dagger as independent always accurate?

No, treating phi and phi-dagger as independent is an approximation that is valid in certain cases, such as when the system is in a coherent state or in the limit of large occupation numbers.

4. Can treating phi and phi-dagger as independent be applied to all quantum systems?

No, treating phi and phi-dagger as independent is most commonly used in systems with a large number of particles, such as in quantum field theory. It may not be applicable to single-particle systems or systems with a small number of particles.

5. Are there any limitations or drawbacks to treating phi and phi-dagger as independent?

One limitation is that treating phi and phi-dagger as independent does not account for interactions between particles, which can have significant effects on the behavior of a quantum system. Additionally, it may not accurately describe systems with a small number of particles or in certain extreme conditions.

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