I'm currently reading Sakurai's book Modern QM and I've been stuck after the introduction of observables with continuous eigenvalues. The thing that bothers me are expressions such as $$ \int d\xi |\xi\rangle \langle \xi| $$ and $$ \int_{-\infty}^\infty dx |x\rangle \langle x| \alpha \rangle. $$

So far I've been able to make sense of the whole bra-ket notation by imagining the kets being members of Hilbert space $\mathcal{H}$ and the bras being the continuous elements of $\mathcal{H}^*$. If we consider the natural isomorphism $$\tau : \mathcal{H} \otimes \mathcal{H}^* \longrightarrow \mathcal{L}(\mathcal{H},\mathcal{H}) $$ which sends $ |v\rangle \otimes \langle u| :=v\otimes\phi_u \longmapsto A$, such that $A(h) = \phi_u(h)v = \langle u,h \rangle v$ for every $h \in \mathcal{H}$. Then we can view $|\xi \rangle \langle \xi |$ as $$ |\xi \rangle \langle \xi | = \tau(|\xi \rangle \otimes \langle \xi |). $$

With this point of view the second integral takes value in $\mathcal{L}(\mathcal{H},\mathcal{H})$. and the first in $\mathcal{H}$ both of which are infinite dimensional. I don't see how I can make sense of such integral. I can make sense of vector valued integrals in finite dimensional space but I've never seen definition of integral that justifies the above expressions.

So my question is what am I missing and how exactly do I interpret those symbols? Is there any math that I need to study in order to rigorously be able to make sense of such expression?

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    $\begingroup$ The integral of operator-valued functions is rigorously defined using the so-called Bochner integral, at least as long as the aforementioned function has values in the Banach space of bounded operators and it is Borel measurable (the preimage of Borel (open) sets in the space of bounded operators is measurable). $\endgroup$ – yuggib Sep 18 '17 at 9:32
  • $\begingroup$ However, the expression involving "eigenfunctions" of the position operator is only formal, and thus does not make precise mathematical sense. The most common way to handle rigorously such generalized eigenfunctions and in general self-adjoint operators with continuous spectrum is by means of the spectral theorem, and the associated projection-valued measures. $\endgroup$ – yuggib Sep 18 '17 at 9:32

I'll start with your first expression: As you correctly noticed, $| \xi \rangle\langle \xi |$ can be seen as an element of $\mathcal L(\mathcal H)$, therefore you are integrating an $\mathcal L(\mathcal H)$-valued function which gives a result $$ \int | \xi \rangle\langle \xi | \, \mathrm d\xi \in \mathcal L(\mathcal H) . $$

Your second expression is interesting: It can be understood in two ways. These two ways are equivalent because the tensor product is associative.

  1. You can see it as $$ \left( \int \mathrm dx | x \rangle\langle x | \right) | \alpha \rangle , $$ which is obviously an element of $\mathcal H$ according to what I said about your first expression.

  2. Alternatively, you can see it as an integral of a $\mathcal H$-valued function: $$ \int | x \rangle\langle x | \alpha \rangle \, \mathrm dx = \int \langle x | \alpha \rangle \, | x \rangle \, \mathrm dx . $$

Finally, I should mention that these integrals are still not well defined from a strict mathematical point of view, as soon as there is continuous spectrum. The correct way to go about is to view the whole expression $| x \rangle \langle x | \, \mathrm dx$ as a projection-valued measure (Wikipedia). That subject is too complex to explain in just one post here, though.

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  • $\begingroup$ Your final remark about projective-valued measures was what I was looking for. $\endgroup$ – gcc-6.0 Sep 18 '17 at 10:18

I'm personally unsure if Sakurai gives the relevant examples for discrete eigenvalues before the continuous, but for completeness' sake I will start there to answer this question.

Suppose we're talking about a spin 1/2 state. Then our two states are $|+\rangle$ and $|-\rangle$, meaning up and down respectively. Let's consider the following expression:

$$\sum_s|s\rangle\langle s| = |+\rangle\langle +| + |-\rangle\langle -|.$$

A good way to try and figure out an expression like this is to try and work with it: suppose we have a state $|\psi\rangle = a|+\rangle + b|-\rangle$. Then,

\begin{align} \left(\sum_s|s\rangle\langle s|\right)|\psi\rangle = |+\rangle\langle +|\psi\rangle + |-\rangle\langle -|\psi\rangle = \sum_s|s\rangle\langle s|\psi\rangle = a|+\rangle\langle +|+\rangle + b|+\rangle\langle +|-\rangle \\+ a|-\rangle\langle -|+\rangle + b|-\rangle\langle -|-\rangle \end{align} and since $\langle s|s'\rangle = \delta_{ss'}$, we have that

$$\left(\sum_s|s\rangle\langle s|\right)|\psi\rangle = a|+\rangle + b|-\rangle = |\psi\rangle.$$

Now since this was for any vector $|\psi\rangle$ expressed in this basis, we must have that

$$\sum_s|s\rangle\langle s| = \mathbf{Id},$$

the identity operator.

All of these integrals are expressing the exact same thing, just for bases of an infinite dimensional Hilbert space. In particular, this means your interpretations of the two integrals you've shown are flipped: $\int d\xi \ |\xi\rangle\langle\xi|$ is the identity operator and so is an element of $\mathcal{L}(\mathcal{H},\mathcal{H})$ and $\int_{-\infty}^{\infty}dx \ |x\rangle\langle x|\alpha\rangle$ is an integral of a scalar times an element of the Hilbert space, so it is an element of the Hilbert space $\mathcal{H}$.

Now for making sense of these integrals, you do the same thing you would do in the finite-dimensional case: the first integral is just the sum of infinitely many products of kets and bras, which makes it an infinite sum of matrices (operators) to make a new matrix (operator). The second integral is actually just the expansion of the state $|\alpha\rangle$ in the $|x\rangle$ basis, just as if you wanted to express a state $|\beta\rangle$ in the spin 1/2 basis above, you would write it as

$$\sum_{s}|s\rangle\langle s|\beta\rangle = |+\rangle\langle +|\beta\rangle + |-\rangle\langle - |\beta\rangle = a|+\rangle + b|-\rangle.$$

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