# More doubts in perturbation theory

• I
• Kashmir
In summary, the conversation discusses the challenges of using perturbation theory for degenerate systems in quantum mechanics. Specifically, it explores the issue of choosing the correct basis for the perturbation series to converge. The necessity of choosing the "right" basis only arises in perturbation theory because it cannot approximate jump-discontinuities. The solution to this problem is known as "degenerate perturbation theory" and is discussed in textbooks.
Kashmir
Townsend, quantum mechanics
" In our earlier derivation we assumed that each unperturbed eigenstate ##\left|\varphi_{n}^{(0)}\right\rangle## turns smoothly into the exact eigenstate ##\left|\psi_{n}\right\rangle## as we turn on the perturbing Hamiltonian. However, if there are ##N## states
##
\left|\varphi_{n, i}^{(0)}\right\rangle \quad i=1,2, \ldots, N
## all with the same energy, it isn't clear which are the right linear combinations of the unperturbed states that become the exact eigenstates. For example, in the case of two-fold degeneracy, is it
##
\left|\varphi_{n, 1}^{(0)}\right\rangle \text { and }\left|\varphi_{n, 2}^{(0)}\right\rangle
##
or
##
\frac{1}{\sqrt{2}}\left(\left|\varphi_{n, 1}^{(0)}\right\rangle+\left|\varphi_{n, 2}^{(0)}\right\rangle\right) \quad \text { and } \frac{1}{\sqrt{2}}\left(\left|\varphi_{n, 1}^{(0)}\right\rangle-\left|\varphi_{n, 2}^{(0)}\right\rangle\right)
##
or some other of the infinite number of linear combinations that we can construct from these two states? If we choose the wrong linear combination of unperturbed states as a starting point, even the small change in the Hamiltonian generated by turning on the perturbation with an infinitesimal ##\lambda## must produce a large change in the state
1) We find the eigenstate of the total Hamiltonian using the below series ##\begin{aligned}\left|\psi_{n}\right\rangle &=\left|\varphi_{n}^{(0)}\right\rangle+\lambda\left|\varphi_{n}^{(1)}\right\rangle+\lambda^{2}\left|\varphi_{n}^{(2)}\right\rangle+\cdots \\ E_{n} &=E_{n}^{(0)}+\lambda E_{n}^{(1)}+\lambda^{2} E_{n}^{(2)}+\cdots \end{aligned}## What happens if I use the above series expansion for the wrong states? 2) Is it that the states which change abruptly aren't "the exact eigenstates" of the total perturbed Hamilton ?

1) You end up with meaningless expressions since the ##(E_1 - E_2)^{-1}## factors diverge.
2) No, the necessity to choose the "right" basis only arises in perturbation theory because we're looking for a power series in terms of ##\lambda## which cannot approximate jump-discontinuities.

Kashmir
HomogenousCow said:
1) You end up with meaningless expressions since the ##(E_1 - E_2)^{-1}## factors diverge.
2) No, the necessity to choose the "right" basis only arises in perturbation theory because we're looking for a power series in terms of ##\lambda## which cannot approximate jump-discontinuities.
So it is not possible to write the series expansion using"wrong" states ?

This case is known as "degenerate perturbation theory" and is discussed in any textbook. Specifically, you first have to diagonalize the perturbation V in the set of degenerate zeroth order states.

vanhees71
DrDu said:
This case is known as "degenerate perturbation theory" and is discussed in any textbook. Specifically, you first have to diagonalize the perturbation V in the set of degenerate zeroth order states.
I'm having confusion reading my textbook 'McIntyre'

So try to find another book which fits your way of thinking better :-)

Kashmir
I like the treatment in Sakurai, Modern Quantum Mechanics.

Kashmir said:
So it is not possible to write the series expansion using"wrong" states ?
Anyone??

Maybe you can think of it like this. This degeneracy can be lifted in several ways. For example, in a spin 1/2 problem, you can add either a sigma_z or a sigma_x term to lift the degeneracy. Assume you take the degenerate case as the limit of a small sigma_z potential tending to 0. Now if you add a sigma_x potential as a real perturbation, the convergence of the series will break down as soon as the term containing sigma_x becomes larger than the term containing sigma_z. In the limit of vanishing sigma_z, the perturbation series will not converge at all.

## 1. What is perturbation theory?

Perturbation theory is a mathematical method used in physics and other sciences to approximate solutions to complex problems that cannot be solved exactly. It involves breaking down a problem into simpler, solvable parts and then adding small adjustments to the solution to account for the more complex aspects of the problem.

## 2. How is perturbation theory used in science?

Perturbation theory is used in a wide range of scientific fields, including quantum mechanics, astrophysics, and fluid dynamics. It is particularly useful for problems that involve small changes or disturbances in a system, allowing scientists to make accurate predictions and calculations.

## 3. What are the limitations of perturbation theory?

One limitation of perturbation theory is that it relies on small perturbations, and therefore may not accurately describe systems with large changes or disturbances. It also assumes that the perturbations are independent of each other, which may not always be the case.

## 4. How does perturbation theory relate to quantum mechanics?

In quantum mechanics, perturbation theory is used to approximate the behavior of a system when it is subjected to small changes or disturbances. This allows scientists to make predictions about the behavior of quantum systems, such as atoms and subatomic particles, which are often too complex to solve exactly.

## 5. Can perturbation theory be applied to nonlinear systems?

Yes, perturbation theory can be applied to nonlinear systems, but it becomes more complex and less accurate. Nonlinear systems involve interactions between different variables, making it difficult to accurately predict the behavior of the system using perturbation theory. In these cases, other mathematical methods may be more suitable.

• Quantum Physics
Replies
10
Views
1K
• Quantum Physics
Replies
3
Views
998
• Quantum Physics
Replies
3
Views
828
• Quantum Physics
Replies
6
Views
1K
• Quantum Physics
Replies
2
Views
716
• Quantum Physics
Replies
5
Views
345
• Quantum Physics
Replies
2
Views
1K
• Quantum Physics
Replies
9
Views
1K
• Quantum Physics
Replies
21
Views
2K
• Quantum Physics
Replies
14
Views
2K