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I am, in full generality, confused about perturbation theory in quantum mechanics.

My textbook and Wikipedia have the same general approach to explaining it: given some Hamiltonian $H=H^{(0)} + H^\prime$, we can break down each eigenfunction $\left\vert n \right\rangle$ into a power series in an invented constant $\lambda$ and the eigenenergies likewise:

$\left\vert n \right\rangle = \sum\lambda^i\left\vert n^{(i)}\right\rangle$

$E_n = \sum \lambda^i E_n^{(i)}$

$\left(H^{(0)} + H^\prime\right) \left(\left\vert n^{(0)}\right\rangle + \lambda \left\vert n^{(0)}\right\rangle + \cdots \right) = \left(E^{(0)}+ \lambda E^{(1)} + \cdots\right) \left(\left\vert n^{(0)}\right\rangle + \lambda \left\vert n^{(1)}\right\rangle + \cdots \right)$

... and then they take $\lambda\to1$.

My question is - what's the logic here? Where did this come from? What purpose does $\lambda$ serve, given that the actual size of each contribution will be determined by the $E^{(i)}$'s and $\left\vert n^{(i)}\right\rangle$'s?

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up vote 4 down vote accepted

Firstly, I refer you to Prof. Binney's textbook (see below) which covers perturbation theory in quantum mechanics in explicit detail. When doing perturbation theory, we perturb the Hamiltonian $H^{(0)}$ of a system which has been solved analytically, i.e. the eigenstates and eigenvalues are known. Specifically,

$$H^{(0)}\to H^{(0)} + \lambda H'$$

where $H'$ is the perturbation, and $\lambda$ is a coupling constant. Why include such a constant? As Binney says, it provides us a 'slider' which when gradually increased to unity increases the strength of the perturbation. When $\lambda = 0$, the system is unperturbed, and when $\lambda=1$ we 'fully perturb the system.'

Introducing a coupling constant $\lambda$ also provides us with a manner to refer to a particular order of perturbation theory; $\mathcal{O}(\lambda)$ is first order, $\mathcal{O}(\lambda^2)$ is second order, etc. As we increase in powers of the coupling constant, we hope the corrections decrease. (The series may not even converge.)

A caveat: the demand that a coupling $\lambda \ll1$ may not be sufficient or correct to ensure that the coupling is small; this is only the case when the coupling is dimensionless. For example, if the coupling, in units where $c=\hbar=1$, had a mass (or equivalently energy) dimension of $+1$, then to ensure a weak coupling we would need to demand, $\lambda/E \ll 1$, where $E$ had dimensions of energy. Such couplings are known as relevant as at low energies they are high, and at high energies the coupling is low.

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Which textbook is this? – linkhyrule5 Mar 26 '14 at 11:15
A free PDF of the book is provided by Binney at: – JamalS Mar 26 '14 at 11:16
For easy reference, can you link that in the question? – Kvothe Mar 26 '14 at 12:55
@Kvothe: Certainly. – JamalS Mar 26 '14 at 12:55

As far as I understand, the logic behind this is the following.

We write down the Hamiltonian for the perturbed system as the Hamiltonian for the unperturbed one plus some perturbation \begin{equation} H = H^{(0)} + H' \, . \end{equation}

Assuming that the perturbation is applied gradually we then introduce $H(\lambda)$ operator \begin{equation} H(\lambda) = H^{(0)} + \lambda H' \, , \end{equation} which is identical to $H^{(0)}$ when $\lambda = 0$ and is identical to $H$ when $\lambda = 1$, thus giving a continuous change from the unperturbed to the perturbed system.

We finally assume that the time-independent Schrödinger equation holds for all $\lambda \in [0, 1]$ \begin{equation} H(\lambda) |n(\lambda)\rangle = E(\lambda) |n(\lambda)\rangle \, , \end{equation} and we introduce the power series expansions for $|n(\lambda)\rangle$ and $E(\lambda)$ you mentioned.

Lately we often set $\lambda$ equal to 1 if we are interested in the fully perturbed system.

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But what does a "first-order-approximation" really mean in this case, then? Since $\lambda \in [0,1]$, there's no particular reason why a higher-order term should be smaller than a lower-order one that I can see... – linkhyrule5 Mar 26 '14 at 10:50
First-order approximation to energy (state vector) is the coefficient of the first power of $\lambda$ in the expansions of energy (state vector) in powers of $\lambda$, i.e. $E^{(1)}$ ($n^{(1)}$) in the notation you used. – Wildcat Mar 26 '14 at 10:55
Right, but ... I mean, technically it is indeed an approximation in the first order of $\lambda$; what I'm asking is, what makes it necessarily more accurate than the third-order terms (alone)? What makes the contributions of higher-order terms less significant, once you take $\lambda\to1$? – linkhyrule5 Mar 26 '14 at 10:56
@linkhyrule5 hmmm... the higher-order terms are not less significant. Where did you get that? Perturbative series are not even guaranteed to converge. – Wildcat Mar 26 '14 at 11:01
@linkhyrule5 well, you can successively calculate first-order corrections, second-order corrections, and so forth and check the convergence. If perturbative series do converge, you can safely use the theory, but if not, then you are in a trouble. – Wildcat Mar 26 '14 at 11:16

The point of introducing the coupling constant $\lambda$ is that the perturbation series in $\lambda$ might not have radius of convergence $\geq 1$, i.e. the power series might not be convergent at $\lambda=1$, and hence that it might not make sense to substitute $\lambda=1$. In fact, that's typically the case.

Nevertheless, a divergent series still make sense as a formal power series if we have a free parameter $\lambda$. (One may think of $\lambda$ as a convenient bookkeeping device, which keeps track of the perturbative order.) Of course, a formal power series is of limited use if we don't know how to sum it.

However, a divergent formal power series may in turn be an asymptotic series. If we are granted that the system makes sense non-perturbatively (so that we can talk about the correct result), it might still be the case that the first few terms of the perturbative power series expansion in $\lambda$ may constitute an excellent approximation, even if the full perturbation series in $\lambda$ is divergent.

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In case $H'$ is small in some sense wrt. $H_0$ one usually writes $$H(\lambda)=H_0+{\lambda}H'.$$ If the eigenvalues of $H_0$ are known one then obtains a perturbation series expressing the eigenvalues and eigenvectors of $H$ in terms of those of $H_0$. $\lambda$ is primarily introduced to keep track of terms.

Complications occur in case an eigenvalue of $H_0$ is degenerate and if it is continuum-embedded.

There is a vast literature about this matter. In the volumes of Reed and Simon you can find much about the mathematical background.

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