
What is the Perturbation Theory in quantum physics ?
Answer
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Hint: In quantum mechanics, perturbation theory is a series of approximation schemes for explaining a complex quantum system in terms of a simpler one that are directly related to mathematical perturbation. The idea is to take a simple system with a known mathematical solution and add a "perturbing" Hamiltonian that represents a weak disruption to the system. The various physical quantities associated with the perturbed system (e.g., its energy levels and eigenstates) can be expressed as "corrections" to those of the simple system if the disruption is not too high.
Complete answer:
If the problem at hand cannot be solved exactly, but can be formulated by adding a "thin" word to the mathematical definition of the exactly solvable problem, perturbation theory can be used. For example, tiny changes in the spectral lines of hydrogen induced by the presence of an electric field (the Stark effect) can be measured by applying a perturbative electric potential to the quantum mechanical model of the hydrogen atom. Since the sum of a Coulomb potential and a linear potential is unstable (has no true bound states), even though the tunnelling time (decay rate) is very long, this is just an estimate.
Although the expressions produced by perturbation theory are not exact, they can lead to accurate results if the expansion parameter, for example$\alpha $, is kept small. Typically, the results are expressed in terms of finite power series in that, when summed to higher orders, appear to converge to exact values.
However, since the series are normally divergent, the results get worse after a certain order\[n{\text{ }}\~{\text{ }}1/\alpha \]. The vibrational approach is the most effective way to transform them into convergent sequences, which can be tested for large-expansion parameters. Also convergent perturbations can produce incorrect results, and divergent perturbation expansions can produce positive results at lower orders.
Note: Precisely in some cases, perturbation theory is the wrong path to take. When the system we want to describe can't be described by a small perturbation imposed on a simple system, this happens. The interaction of quarks with the gluon field, for example, cannot be viewed perturbatively at low energies in quantum chromodynamics because the coupling constant (the expansion parameter) becomes too big.
Complete answer:
If the problem at hand cannot be solved exactly, but can be formulated by adding a "thin" word to the mathematical definition of the exactly solvable problem, perturbation theory can be used. For example, tiny changes in the spectral lines of hydrogen induced by the presence of an electric field (the Stark effect) can be measured by applying a perturbative electric potential to the quantum mechanical model of the hydrogen atom. Since the sum of a Coulomb potential and a linear potential is unstable (has no true bound states), even though the tunnelling time (decay rate) is very long, this is just an estimate.
Although the expressions produced by perturbation theory are not exact, they can lead to accurate results if the expansion parameter, for example$\alpha $, is kept small. Typically, the results are expressed in terms of finite power series in that, when summed to higher orders, appear to converge to exact values.
However, since the series are normally divergent, the results get worse after a certain order\[n{\text{ }}\~{\text{ }}1/\alpha \]. The vibrational approach is the most effective way to transform them into convergent sequences, which can be tested for large-expansion parameters. Also convergent perturbations can produce incorrect results, and divergent perturbation expansions can produce positive results at lower orders.
Note: Precisely in some cases, perturbation theory is the wrong path to take. When the system we want to describe can't be described by a small perturbation imposed on a simple system, this happens. The interaction of quarks with the gluon field, for example, cannot be viewed perturbatively at low energies in quantum chromodynamics because the coupling constant (the expansion parameter) becomes too big.
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