Ryder's SU(2) Example in Quantum Field Theory

In summary: Note that you can't just replace the transformed quantities in the expression for the covariant derivative because that would result in a different expression for the covariant derivative, which is not what we want.
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ShayanJ
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In section 3.5 of his textbook Quantum Field Theory, Ryder discusses an example of a non-Abelian gauge theory. He considers a 3D internal space and rotations in this space.

At first he shows that the fields in this internal space transform like ##\delta \vec \phi=-\vec \Lambda \times \vec \phi ## under a rotation ## \vec \Lambda ## in the internal space. Then he shows that ## \delta(\partial_\mu \vec \phi)=-\vec \Lambda \times (\partial_\mu \vec \phi)-(\partial_\mu \vec \Lambda)\times \vec \phi##.

He then introduces the covariant derivative ## D_\mu \vec \phi=\partial_\mu \vec \phi+g\vec W_\mu \times \vec \phi ## and demands that ## \delta(D_\mu \vec \phi)=-\vec \Lambda \times (D_\mu \vec \phi) ## which gives him the transformation rule for ## \vec W_\mu## which is ## \delta \vec W_\mu=-\vec \Lambda \times \vec W_\mu+\frac 1 g \partial_\mu \vec \Lambda ##.

Now here is where my confusion begins. He writes ##\delta(D_\mu \vec \phi)=\delta(\partial_\mu \vec \phi)+g(\delta \vec W_\mu)\times \vec \phi+g\vec W_\mu \times (\delta \vec \phi) ## and then using the above expressions, he verifies that ## \delta(D_\mu \vec \phi)=-\vec \Lambda \times (D_\mu \vec \phi) ##.
But it seems to me that without using ##\delta(D_\mu \vec \phi)=\delta(\partial_\mu \vec \phi)+g(\delta \vec W_\mu)\times \vec \phi+g\vec W_\mu \times (\delta \vec \phi) ##, I still should be able to get the same conclusion with just replacing the transformed quantities in the expression for the covariant derivative, so I write:
## (D_\mu \vec \phi)_{transformed}=\partial_\mu \vec \phi-\vec \Lambda \times (\partial_\mu \vec \phi)-(\partial_\mu \vec \Lambda)\times \vec \phi+g\left( \vec W_\mu-\vec\Lambda\times \vec W_\mu+\frac 1 g \partial_\mu \vec \Lambda \right) \times \left( \vec\phi-\vec\Lambda\times \vec \phi \right)##.

But when I do the calculations, the result is far from what Ryder gets(which should already be obvious from my expression).

What's wrong with this method?
Thanks
 
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  • #2
Shyan said:
In section 3.5 of his textbook Quantum Field Theory, Ryder discusses an example of a non-Abelian gauge theory. He considers a 3D internal space and rotations in this space.

At first he shows that the fields in this internal space transform like ##\delta \vec \phi=-\vec \Lambda \times \vec \phi ## under a rotation ## \vec \Lambda ## in the internal space. Then he shows that ## \delta(\partial_\mu \vec \phi)=-\vec \Lambda \times (\partial_\mu \vec \phi)-(\partial_\mu \vec \Lambda)\times \vec \phi##.

He then introduces the covariant derivative ## D_\mu \vec \phi=\partial_\mu \vec \phi+g\vec W_\mu \times \vec \phi ## and demands that ## \delta(D_\mu \vec \phi)=-\vec \Lambda \times (D_\mu \vec \phi) ## which gives him the transformation rule for ## \vec W_\mu## which is ## \delta \vec W_\mu=-\vec \Lambda \times \vec W_\mu+\frac 1 g \partial_\mu \vec \Lambda ##.

Now here is where my confusion begins. He writes ##\delta(D_\mu \vec \phi)=\delta(\partial_\mu \vec \phi)+g(\delta \vec W_\mu)\times \vec \phi+g\vec W_\mu \times (\delta \vec \phi) ## and then using the above expressions, he verifies that ## \delta(D_\mu \vec \phi)=-\vec \Lambda \times (D_\mu \vec \phi) ##.
But it seems to me that without using ##\delta(D_\mu \vec \phi)=\delta(\partial_\mu \vec \phi)+g(\delta \vec W_\mu)\times \vec \phi+g\vec W_\mu \times (\delta \vec \phi) ##, I still should be able to get the same conclusion with just replacing the transformed quantities in the expression for the covariant derivative, so I write:
## (D_\mu \vec \phi)_{transformed}=\partial_\mu \vec \phi-\vec \Lambda \times (\partial_\mu \vec \phi)-(\partial_\mu \vec \Lambda)\times \vec \phi+g\left( \vec W_\mu-\vec\Lambda\times \vec W_\mu+\frac 1 g \partial_\mu \vec \Lambda \right) \times \left( \vec\phi-\vec\Lambda\times \vec \phi \right)##.

But when I do the calculations, the result is far from what Ryder gets(which should already be obvious from my expression).

What's wrong with this method?
Thanks
Nothing is wrong with your method if you keep in mind that we are supposed to work in first order of the transformation parameter ##\Lambda##. If you drop all your terms with two powers of ##\Lambda## and then look at the difference between your expression and ## D_\mu \phi ##(because you want the variation ## \delta D_\mu \phi ##) then you should get his expression.
 
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Related to Ryder's SU(2) Example in Quantum Field Theory

1. What is Ryder's SU(2) Example in Quantum Field Theory?

Ryder's SU(2) Example in Quantum Field Theory is an example used to illustrate the concept of symmetry breaking within the framework of quantum field theory. It involves a system with a symmetry described by the special unitary group SU(2) and shows how this symmetry can be spontaneously broken, leading to the emergence of mass for certain particles.

2. How does the SU(2) symmetry work in this example?

The SU(2) symmetry in this example is related to the symmetry between particles and antiparticles, known as charge conjugation. This symmetry implies that particles and antiparticles have equal masses and other properties. However, when the symmetry is spontaneously broken, particles and antiparticles can have different masses, leading to a violation of this symmetry.

3. What is the significance of Ryder's SU(2) Example in Quantum Field Theory?

Ryder's SU(2) Example is significant because it illustrates an important concept in quantum field theory, namely the spontaneous breaking of symmetries. This phenomenon has important implications for the behavior of particles and the emergence of mass, and is an important aspect of the Standard Model of particle physics.

4. How does Ryder's SU(2) Example relate to other examples of symmetry breaking in physics?

Ryder's SU(2) Example is just one specific example of symmetry breaking, which is a common phenomenon in physics. Other examples include the breaking of rotational symmetry in magnets and the breaking of chiral symmetry in strong interactions. However, the underlying mechanism of spontaneous symmetry breaking is the same in all of these examples.

5. What are some real-world applications of Ryder's SU(2) Example in Quantum Field Theory?

While Ryder's SU(2) Example is primarily used as a theoretical tool to understand the behavior of particles and symmetries in quantum field theory, it has also found practical applications in areas such as condensed matter physics and cosmology. For example, the Higgs mechanism, which is based on the ideas of symmetry breaking, is used to explain the origin of mass in the Standard Model. Additionally, the concept of spontaneous symmetry breaking has implications for the formation of structures in the early universe.

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