# About completeness relation from discrete to continuous limit

In quantum mechanics, the completeness relation for discrete and continuous basis are \begin{align} \sum_n \lvert n \rangle \langle n\rvert &= 1 \tag{1} \\ \int \lvert x \rangle \langle x \rvert \mathrm{d}x = 1 \tag{2} \end{align} respectively. The integral form can be written as $$\lim_{n \rightarrow \infty} \sum_n \lvert x_n \rangle \langle x_n\rvert \Delta x_n=1 \tag{3}$$ such that $$\langle \psi \lvert \psi \rangle= \int \langle \psi \lvert x \rangle \langle x \lvert \psi \rangle \mathrm{d}x = \lim_{n \rightarrow \infty} \sum_n \langle \psi \lvert x_n \rangle \langle x_n\lvert \psi \rangle \Delta x_n \tag{4}$$ like the definition of definite integral.

My question is, suppose space is discrete, Eq. (2) may be written as $$\sum_n \lvert x_n \rangle \langle x_n \rvert =1 \tag{6}$$ If I approach the continuous limit (2) from discrete basis (6), how the $\Delta x_n$ in Eq. (3) should appear?

## 2 Answers

To add a slightly different angle to PhotonicBoom's sound answer, the link between the two entities - discrete sum and integral - is the concept of measure, not of limit. You can think of your sum as a Lebesgue integral if you choose a discrete measure for the real line with the measure's "anchors" at a countable set of "allowed values". Discrete and continuous measures are highly analogous insofar that they both have all the "Real MacCoy" properties of measures: non-negativity, positivity and countable ($\sigma$-) additivity. Ultimately, though, the two are as different as are $\aleph_0$ and $\aleph_1$, dramatically illustrated by the Cantor Slash argument: a quantum observable which can in principle yield any real number in an interval as a measurement and one which can only have discrete values as measurements are very different beasts.

In quantum mechanics, or at least all the QM I've seen(see footnote), one makes an assumption of a separable or a first countable Hilbert space for the state space. This means, in effect, that there exists a countable basis for the space of states: for example, the quantum harmonic oscillator's state can be expressed as a superposition of the countable set of energy eigenstates. So in "normal QM", there is always a co-ordinate transformation which will turn an integral completeness relation into a discrete one, although, at the same time, you are changing the observable whose eigenstates span the state space.

Footnote: Separability is part of the Wightman axioms. But sometimes the quantum field theorists drop even this assumption, although I understand that they still assume a separable subspace containing physical fields embedded in a non-separable state space of potential fields.

The integral form is not an alternative way of writing the completeness relation. The integral form is the definition of the completeness relation ONLY when a continuous variable is involved.

What you have included in your question, equation (2) only shows how to compute the completeness relation when the number of states are infinite i.e the spatial configuration x can be infinitely divisible therefore a sum is useless and integration is the solution to the problem.

Answering your actual question now. If space is discrete, the it has a discrete basis and (6) IS the definition of the completeness theorem.

• My question is, is there any way to approach the continuous limit (3) from the discrete basis (6) and vice versa? If the answer is yes, then how should $\Delta x$ come? – user26143 Oct 14 '14 at 3:33
• The answer is no. (3) is only the formal definition, or derivation if you like, of (2). Therefore only the discrete basis can be used. – Constandinos Damalas Oct 14 '14 at 3:35