Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
add comment

4 Answers 4

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.

share|improve this answer
    
Which textbook is this? –  linkhyrule5 Mar 26 at 11:15
    
A free PDF of the book is provided by Binney at: www-thphys.physics.ox.ac.uk/people/JamesBinney/QBhome.htm –  JamalS Mar 26 at 11:16
    
For easy reference, can you link that in the question? –  Kvothe Mar 26 at 12:55
    
@Kvothe: Certainly. –  JamalS Mar 26 at 12:55
add comment

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.

share|improve this answer
    
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 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 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 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 at 11:01
1  
@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 at 11:16
show 1 more comment

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.