In perturbation theory for non-relativistic quantum mechanics, you begin with a Hamiltonian of the form $$H=H_0+\lambda H'$$ and assume that the perturbed eigenstates and eigenvalues can be written as power series in $\lambda$:


$$E_n=E_n^0+\lambda E_n^1+\lambda^2 E_n^2+\dots.$$

Then you plug these expansions into the eigenvalue equation, $H\left|\psi\right>=\lambda\left|\psi\right>$, and set the coefficients of like powers of $\lambda$ equal to each other.

At least three concerns arise:

  1. Why are we justified in assuming these Taylor expansions exist, i.e., have nonzero radii of convergence?

  2. Why are we justified in Taylor expanding $\left|\psi\right>$, as it is an abstract vector in Hilbert space and not a scalar-valued function?

  3. Why are we justified in setting like coefficients of $\lambda$ equal to each other?

For #3, I have seen some mathematically precise arguments for why, if $P(x)=Q(x)$ for all $x$, where $P$ and $Q$ are finite polynomials, then their coefficients must be equal. But we now have infinite Taylor series, as well as coefficients of $\lambda$ that are not scalars but rather abstract vectors in infinite-dimensional Hilbert space—should it still be obvious that we can do this?

EDIT: To clarify question 1, by "convergence" it don't just mean convergence at all, but even stronger: convergence to the correct value. There are non-analytic functions, like $f(x) = \{0$ if $x=0, e^{-1/x^2}$ otherwise$\}$ which have Taylor expansions that converge (to some value) everywhere, but to the correct value only at the single point of expansion (at zero, for this example).

  • $\begingroup$ The bit I find unclear is the relationship between the Hilbert space itself on the left hand side of the eigenstate expansion with the old one on the right hand side. $\endgroup$
    – Keith
    Commented Feb 24, 2019 at 22:33
  • $\begingroup$ About #1; even if the Taylor expansion has a zero radius of convergence, there are way of extracting information from that series, which actually corresponds to physical results. $\endgroup$
    – Our
    Commented Feb 25, 2019 at 15:49

2 Answers 2

  1. By the definition of a Hilbert space, any element of a Hilbert space can be represented as a sum $\sum_n c_n \lvert \psi_n\rangle$ with the sequence $(c_n)_{n\in\mathbb{N}}$ being square summable and the $\lvert \psi_n\rangle$ being normalized and orthogonal. So if $\lambda$ is small enough that the sequence $(\lambda^n)_{n\in\mathbb{N}}$ is square summable, then these series converge. This series is a geometric series and converges if and only if $\lambda < 1$.

    However, there is no generic reason for the vectors appearing in the perturbative expansion to be orthogonal, and neither can you expect the series to converge generically. An argument usually attributed to Dyson says, for example, that the standard perturbation series of interacting quantum field theories in terms of the coupling constant $\lambda$ can never converge around $\lambda = 0$ because the theory is unstable for $\lambda < 0$ (as it has no ground state and would produce infinite particles from the vacuum). Many perturbative series that are non-convergent in this sense are nevertheless asymptotic around small $\lambda$.

    By their very nature, perturbative expansions cannot capture non-perturbative effects, where "non-perturbative" is basically synonymous with "not captured by the Taylor series". Famously, e.g. instanton effects are non-perturbative. The perturbative series does not necessarily capture all physics involved.

  2. Taylor's theorem holds in arbitrary Banach spaces, cf. e.g. this random search result for "Taylor series in Banach spaces". A Hilbert space is a Banach space.

  3. A power series is zero on an open subset if and only if all its coefficients are zero, cf. e.g. this math.SE post, this is also the underpinning of the holomorphic identity theorem. So if two power series agree on an open interval of $\lambda$ (not just a point), their coefficients are equal.

  • $\begingroup$ Thanks for these responses, though #1 is not exactly what I was getting at—please see my update to the question. Also for #3, that fact about power series is true for power series with scalar coefficients, but does it hold true for those with vector coefficients? $\endgroup$
    – WillG
    Commented Feb 24, 2019 at 21:46
  • $\begingroup$ @WillG For #1, see my edit. For #3, a power series with vector coefficients is just a vector of power series of scalar coefficients, so theorems about scalar series carry over to those with vector coefficients. $\endgroup$
    – ACuriousMind
    Commented Feb 24, 2019 at 21:54
  • 1
    $\begingroup$ Issue 1 can be quite subtle since the perturbation expansion may diverge but still Borel summable to the full function as in the harmonic oscillator case. An old survey of such results is the article by Barry simon: math.caltech.edu/SimonPapers/R23.pdf $\endgroup$ Commented Feb 25, 2019 at 20:39

In the anharmonic oscillator (quartic perturbation) the series does not converge, although it can be shown to hold asymptotically as $\lambda \to 0$. The reason is that for $\lambda \lt 0$ the Hamiltonian is unbounded below so the eigenvalue does not exist.

Kato discusses perturbations in finite dimensional spaces, where convergenge can be proven. This can be extended to infinite dimensional spaces, and under some conditions to unbounded operators.


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