# Eigenkets belonging to a range of eigenvalues

• I
When one wants to represent a general ket in a basis consisting of eigenkets each attributed to an eigenvalue in a range, say from a to b, why does one take the integral of said kets from a to b w.r.t. the eigenvalues? I understand that the integral here plays a role analogous to a sum in the case where a general ket is expressed in terms of eigenkets belonging to discrete eigenvalues, but I don't understand why each vector is multiplied by an infinitesimal change near the eigenvalue it belongs to. Interpreting this integral as the limit of a sum we get:
[P>=Σ[ξ>Δξ (lim.Δξ→0)
where I do not understand the role of Δξ.

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stevendaryl
Staff Emeritus
I looked over the thread about Rigged Hilbert Spaces, and I'm not sure that it completely explained the relationship between continuous and discrete bases.

In a non-rigorous way, you can think of the continuous basis as a limiting case of the discrete basis. However, in going from discrete to continuous, the normalization convention for basis elements changes.

Let me illustrate. Suppose you have an operator ##\Lambda## with discrete eigenvalues ##\lambda_j##. I think in order for the continuum limit to make sense int the most straightforward way, you need to assume that ##\lambda_{j+1} > \lambda_j##, and that there are infinitely many ##\lambda_j##, and that the corresponding eigenstates ##|n\rangle## form a complete orthonormal basis. That means that
1. If ##n \neq m##, then ##\langle n|m\rangle = 0##
2. ##\langle n|n\rangle = 1##
3. If ##|\psi\rangle## is a properly normalized state, then ##|\psi\rangle = \sum_n \langle n|\psi\rangle |n\rangle##
4. ##\sum_n |\langle n|\psi\rangle|^2 = 1##
Now, if the coefficients ##\langle n|\psi\rangle## change slowly with ##n## (and maybe we also have to assume that ##(\Delta \lambda)_n \equiv \lambda_{n+1} - \lambda_n## remains bounded? I'm not sure...) then we can define a new ket with a different normalization:

##|\lambda_n\rangle \equiv \frac{1}{\sqrt{(\Delta \lambda)_n}} |n\rangle##

In terms of the ##|\lambda_n\rangle##, we have:

##|\psi\rangle = \sum_n (\Delta \lambda)_n \langle \lambda_n |\psi\rangle |\lambda_n\rangle##

The kets ##|\lambda_n\rangle## have a different normalization:

• ##\langle \lambda_n | \lambda_m \rangle = 0## (if ##m \neq n##)
• ##\langle \lambda_n | \lambda_n \rangle = \frac{1}{(\Delta \lambda)_n}##
If the states ##|\lambda_n\rangle## change smoothly with ##n##, then this can be approximated by an integral:

##|\psi\rangle = \int d\lambda \langle \lambda |\psi\rangle |\lambda\rangle##

Last edited:
So dλ is introduced to make the product between each eigenket and dλ finite, since the eigenkets will be of "infinite length" in the sense of lim.Δλ→0[1/Δλ]. That makes sense. In that case, will the coefficients <λIΨ> be infinitesimal? Otherwise the integral would diverge, even over a finite range.

bhobba
Mentor
So dλ is introduced to make the product between each eigenket and dλ finite, since the eigenkets will be of "infinite length" in the sense of lim.Δλ→0[1/Δλ]. That makes sense. In that case, will the coefficients <λIΨ> be infinitesimal? Otherwise the integral would diverge, even over a finite range.

Yes - but unless you want to get into non-standard analysis infinitesimals are a load of the proverbial, although used every now and then when speaking informally. I do it but shouldn't really.

There is no way to understand it properly unless you study the references in my link.