0
$\begingroup$

The propagator function for discrete eigenstates is $$u(t)=\sum_{n=1}^{\infty}|E_n\rangle\langle E_n|e^{-iE_nt/ \hbar } \tag{1}\ .$$

But when we have continuous eigenstates, (like for the case of free particle) then: $$u(t)=\int_{- \infty}^{\infty} |E\rangle\langle E|e^{-iEt/\hbar}\,dE \tag{2}$$

My question is, how do we translate from the discrete eigenstates to continuous eigenstates? And what is the proof that (2) is the generalization of (1)?

This doubt also arose while defining the inner product for 2 continuous vectors. We said that for continuous states, the inner product will tend towards infinity and thus we introduce a normalizing factor. But, shouldn't this be the case when the sum is of discrete eigenstates but infinite?

For discrete inner products: $$\langle f,g\rangle=\sum_i f_i(x) g_i(x)$$ and for continuous inner product, magically: $$\langle f,g\rangle=\int f(x)^*g(x)\,dx.$$

$\endgroup$
  • 1
    $\begingroup$ I'm not sure that you can prove that $(2)$ is a generalization of $(1)$. All you can do is check that it has the same properties, or that one reduces to the other in some limit. $\endgroup$ – Javier Feb 27 '15 at 17:28
1
$\begingroup$

You do not obtain the rules for the infinite-dimensional case by "proving" them from the finite-dimensional rules. Rather, you know that you need to have a Hilbert space, which is a complex vector space with an inner product, essentially.

If you now search for infinite-dimensional Hilbert spaces that could possibly be used in quantum mechanics, you find that all separable Hilbert spaces are isomorphic to the complex valued square-integrable functions $L^2(\mathbb{R})$ with inner product

$$ (f,g) = \int_\mathbb{R} f^*g$$

Now, you implement quantum mechanics by choosing a self-adjoint Hamiltonian operator $H$ on the space, and just impose the Schrödinger equation

$$ Hf = \mathrm{i}\hbar \partial_t f$$

Now, as yuggib write, the spectral theorem gives you that the self-adjoint Hamiltonian has an orthonormal eigenbasis, and the form of the propagator follows naturally.

$\endgroup$
1
$\begingroup$

The correct formulation of the result is called spectral theorem. It encompasses all type of spectra for self-adjoint operators, and thus both (1) and (2).

Given a self adjoint operator $A$ with only discrete spectrum (you have such situations for compact operators, and operators with compact resolvent) the spectral theorem has a nice and simple form: $$A=\sum_{i}\lambda_i P_i\; ,$$ where the sum ranges over all possible eigenvalues $\lambda_i$ and $P_i$ is the projection on the eigensubspace corresponding to $\lambda_i$ (that if the eigenvalue has multiplicity one has the form $\lvert \psi_i\rangle\langle \psi_i\rvert$ where $\psi_i$ is the normalized eigenvector). The spectral theorem gives also the so-called functional calculus (I will not be completely precise mathematically but to give the idea will be enough): given a suitable function $f:\mathbb{R}\to \mathbb{C}$, the operator $f(A)$ is defined by $$f(A)=\sum_i f(\lambda_i)P_i\; .$$ The particular case of $e^{-itA}$ is the one that gives you quantum evolution.

Now if $B$ is an operator with general spectrum (i.e. a discrete part and a continuum one) the theorem still holds, but instead of having a simple discrete family of projections $P_i$ we have what is called a spectral family $\{P_\Omega\}_{\Omega\in \{X, X\subseteq \mathbb{R}\}}$ of projections. This family is uncountable, so you cannot take the sum, but it defines a projection valued measure $dP_\lambda$ (a function from sets to self-adjoint projections that satisfies the axioms of a measure) and it is possible to integrate functions with respect to it. So the spectral theorem for (measurable) functions $f$ of $B$ becomes: $$B=\int_{\mathbb{R}}\lambda dP_\lambda\; ,$$ $$f(B)=\int_{\mathbb{R}}f(\lambda) dP_\lambda\; .$$ Or, in the case of the evolution operator $e^{-itB}$ applied to $u\in \mathscr{H}$ (the Hilbert space): $$u(t)=e^{-itB}u=\int_{\mathbb{R}}e^{-it\lambda} dP_\lambda u \; .$$

Observe that if $B$ has only discrete spectrum, the corresponding measure is just the sum of delta measures pointed at each discrete eigenvalue $\lambda_i$, so it can be written $dP_\lambda = \sum_i \delta(\lambda-\lambda_i)P_\lambda d\lambda$. Therefore the discrete spectrum is just a special case of the spectral theorem.

Anyways, what you are writing is an operator on the right hand side, and a function on the left hand side, so it is not correct as written ;-)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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