What is the point of regularization?

In summary, dimensional regularization is used to isolate the divergent terms in QFT integrals and express them in a way that respects gauge symmetry. The verb "regularize" means to make finite. The conditions for using the LSZ formula can take care of divergences, making renormalization less artificial. One can argue that counterterms are introduced to satisfy the LSZ conditions and are not meant to cancel divergences, but end up doing so miraculously. This is similar to the shift between terms in the Lagrangian and their physical counterparts in QM.
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AndrewGRQTF
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Take for example dimensional regularization. Is it correct to say that the main point of the dimensional regularization of divergent momentum integrals in QFT is to express the divergence of these integrals in such a way that they can be absorbed into the counterterms? Can someone tell me what the definition of the verb "regularize" is?

Also, is it true that the conditions required to be able to use the LSZ formula, like the pole of the exact propagator being at the physical mass and it having residue one, take care of the divergences, so that renormalization is not completely artificial? Can one argue the point of view that counterterms are introduced to satisfy the LSZ conditions, and that they are not meant to cancel any divergences, but end up doing so miraculously?
 
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AndrewGRQTF said:
Take for example dimensional regularization. Is it correct to say that the main point of the dimensional regularization of divergent momentum integrals in QFT is to express the divergence of these integrals in such a way that they can be absorbed into the counterterms?
Yes, it allows you to isolate the divergent terms in a way that respects gauge symmetry.

AndrewGRQTF said:
Can someone tell me what the definition of the verb "regularize" is?
Basically "make finite".

AndrewGRQTF said:
Also, is it true that the conditions required to be able to use the LSZ formula, like the pole of the exact propagator being at the physical mass and it having residue one, take care of the divergences, so that renormalization is not completely artificial?
As far as we can tell yes.

AndrewGRQTF said:
Can one argue the point of view that counterterms are introduced to satisfy the LSZ conditions, and that they are not meant to cancel any divergences, but end up doing so miraculously?
Yes. There would be a shift between terms in the Lagrangian and their physical counterparts regardless, i.e. ##e_0## in the QED Lagrangian would not be the same as ##e## the physical charge. This occurs even in QM for the anharmonic oscillator with Lagrangian:
$$\mathcal{L} = \frac{m\dot{q}^{2}}{2} - \frac{kq^{2}}{2} - \frac{\lambda_0 q^{4}}{4!}$$
Where the physical ##\lambda## is not the same as the ##\lambda_0## in the Lagrangian and one needs to renormalize to attain the latter. It just so happens this also cures divergences in many QFTs.
 
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FAQ: What is the point of regularization?

What is regularization and why is it important in science?

Regularization is a technique used in science to prevent overfitting in statistical models. It involves adding a penalty term to the cost function, which helps to reduce the complexity of the model and prevent it from fitting too closely to the training data. This is important because overfitting can lead to poor generalization and inaccurate predictions on new data.

How does regularization work?

Regularization works by adding a penalty term, such as the L1 or L2 norm, to the cost function of a model. This penalty term penalizes large weights, thereby reducing the complexity of the model and preventing overfitting. The optimal value of the penalty term is determined through techniques such as cross-validation.

What is the difference between L1 and L2 regularization?

L1 and L2 regularization are two common types of regularization techniques. L1 regularization, also known as Lasso regression, adds the absolute value of the weights to the cost function, while L2 regularization, also known as Ridge regression, adds the squared values of the weights. L1 regularization tends to result in sparse solutions, while L2 regularization tends to shrink the weights towards zero.

When should I use regularization in my research?

Regularization should be used when working with complex models that have a large number of features, as these models are more prone to overfitting. It is also useful when dealing with limited or noisy data, as it helps to prevent the model from fitting too closely to the training data and producing inaccurate predictions on new data.

Are there any downsides to using regularization?

While regularization can help prevent overfitting and improve the generalization of a model, it can also lead to a trade-off between bias and variance. By reducing the complexity of the model, regularization may also introduce some bias, which can result in less accurate predictions. Additionally, the optimal value of the penalty term may be difficult to determine and can vary depending on the dataset and problem at hand.

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