# Why is $\int_{\mathbb{R}^3} |p\rangle \langle p| d\lambda(p)=id$?

As I have written in the headline, I am curious how the relation $\int_{\mathbb{R}^3} |p \rangle \langle p| d\lambda(p)=id$ that physicists use, where $|p\rangle$ is the eigenfunction to the eigenvalue $p$ of the momentum operator can be mathematically understood? How is this integral defined and is there a better way to write this operator more rigorously?

• Please don't use < and > instead of \langle and \rangle, it looks terrible. – Najib Idrissi Oct 14 '14 at 8:28
• alright, also edited the title. – user66906 Oct 14 '14 at 8:30
• The object $\lvert p \rangle \langle p \rvert d\lambda(p)$ can be interpreted completely rigorously and fairly straightforwardly as a projection-valued measure; the link also gives details on how to rigorously integrate functions against a projection-valued measure to get operators. This machinery really does come directly from the spectral theorem for general unbounded self-adjoint operators, which may well have continuous spectrum. – Branimir Ćaćić Oct 14 '14 at 8:47
• excuse me, could you turn your comment into an answer and maybe even explain, how you integrate against this measure in this particular example? it would be really help to me and probably also for other physics students :-) – user66906 Oct 14 '14 at 8:51
• I just did as you asked. – Branimir Ćaćić Oct 14 '14 at 13:27

A projection-valued measure on $\mathbb{R}$ is an assignment $\Pi$ that maps a Borel measurable subset $U$ of $\mathbb{R}$ (e.g., an interval) to an orthogonal projection $\Pi(U)$, such that
1. $\Pi(\emptyset) = 0$ and $\Pi(\mathbb{R}) = I$,
2. if $\{U_k\}$ is a countable collection of Borel measurable subsets such that $U_i \cap U_j = \emptyset$ for $i \neq j$, then $\Pi_{\cup_k U_k} = \sum_k \Pi_{U_k}$, where the sum on the right converges strongly, i.e., $\Pi_{\cup_k U_k}\xi = \sum_k \Pi_{U_k}\xi$ for all vectors $\xi$ in your Hilbert space $H$,
or equivalently, such that for any $\xi$, $\eta \in H$, $\mu_{\xi,\eta} : U \mapsto \langle \xi \lvert \Pi(U) \rvert \eta \rangle$ defines a complex measure on $\mathbb{R}$. In particular, for any unit vector (i.e., pure state) $\xi \in H$, $\mu_{\xi,\xi} : U \mapsto \langle \xi \lvert \Pi(U) \rvert \xi \rangle$ defines a probability measure on $\mathbb{R}$. Hence, for any Borel measurable function $f : \mathbb{R} \to \mathbb{C}$, one can define an unbounded operator $\int_{-\infty}^\infty f(x) d\Pi(x)$ on $H$ with dense domain $D_f := \{\xi \in H \mid \int_{-\infty}^\infty \lvert f(x) \rvert^2 d\mu_{\xi,\xi}(x) < \infty \}$ by setting $$\forall \xi,\; \eta \in D_f, \quad \left\langle \xi \left\lvert \int_{-\infty}^\infty f(x) d\Pi(x) \right\rvert \eta \right\rangle := \int_{-\infty}^\infty f(x) d\mu_{\xi,\eta}.$$
Now, what does all this have to do with observables? Let $A$ be a quantum observable, i.e., a possibly unbounded self-adjoint operator. The spectral theorem then tells you that there exists a unique spectral valued measure $\Pi$ on $\mathbb{R}$ such that $A = \int_{-\infty}^\infty a d\Pi(a)$; for example, in the case of momentum $\hat{p} = i\tfrac{d}{dx}$, $\Pi(U) = \int_U \lvert p \rangle\langle p \rvert dp$ in physics notation, so that $\lvert p \rangle\langle p \rvert dp$ can really be interpreted as physics notation for $d\Pi(p)$ wherever it appears. In particular, for any Borel subset $U$ of $\mathbb{R}$, $$\Pi(U) = \int_{-\infty}^\infty \chi_U(a)d\Pi(a) = \int_U d\Pi(a)"$$ is the orthogonal projection onto the subspace of pure states where $A$ is observed to take its value in $U$, and if $\xi \in H$ is a unit vector, and thus a pure state, then $\langle \xi \lvert \Pi(U) \rvert \eta \rangle = \int_U d\mu_{\xi,\xi}$ is the probability that the observed value of $A$ in the pure state $\xi$ lies in $U$. Still more is true: if $a$ denotes, by abuse of notation, the classical observable corresponding to $A = \hat{a}$, then for any measurable function $f : \mathbb{R} \to \mathbb{R}$, $f(A) = \int_{-\infty}^\infty f(a) d\Pi(a)$ is the quantum observable corresponding to the classical observable $f(a)$, and its expectation value in any pure state (unit vector) $\xi \in H$ is $\langle \xi \lvert f(A) \rvert \xi \rangle = \int_{-\infty}^\infty f(a) d\mu_{\xi,\xi}$.
Finally, suppose that $A$ is a self-adjoint operator with discrete spectrum $\{a_k\}$, let $H_k := \ker(A - a_k I)$ be the eigenspace corresponding to the eigenvalue $a_k$, and let $P_k$ denote the orthogonal projection onto $H_k$. Then the corresponding projection-valued measure $\Pi$ is given by $\Pi(U) := \sum_{a_k \in U} P_k,$ where the right hand side converges strongly, i.e., $\Pi(U)\xi = \sum_{a_k \in U} P_k\xi$ for all $\xi \in H$, so that for any measurable function $f : \mathbb{R} \to \mathbb{C}$, $$f(A) = \int_{-\infty}^\infty f(a) d\Pi(a) = \sum_k f(a_k) P_k,$$ where the sum on the right hand side, again, converges strongly on the domain of $A$; in particular, the spectral theorem reads $$A = \int_{-\infty}^\infty a d\Pi(a) = \sum_k a_k P_k,$$ which, up to convergence issues, is precisely the spectral theorem of finite-dimensional linear algebra. From a symbolic standpoint, you can symbolically write $$d\Pi(a) = \sum_k P_k \delta(a-a_k)da,$$ where $\delta(a-a_k)da$ is the Dirac delta supported at $a_k$.