In quantum mechanics we know that if $q$ corresponds to a complete set of parameters characterizing a quantum system, then the state vectors $|q\rangle$ satisfy the following identity: $$\int |q\rangle\langle q| d\tau_q = \Bbb 1$$ where $|q\rangle\langle q|$ is a projection operator and $\Bbb 1$ is the identity operator. I know how to use this identity, but my question is what does it mean mathematically to integrate an operator?
4
-
$\begingroup$ What's the measure $d\tau_q$? $\endgroup$– Silvia GhinassiDec 11, 2015 at 0:44
-
$\begingroup$ I don't actually know any measure theory, but $d\tau_q$ will change depending on what we're integrating. For instance if $|q\rangle=|x\rangle$ are the eigenstates of the position operator $\hat x$, then we'd integrate over all position space: $\int |x\rangle\langle x| dx = \Bbb 1$. If they were the eigenstates of the momentum operator $\hat p$ then we'd integrate over momentum space, making the identity $\int |p\rangle\langle p| dp = \Bbb 1$. $\endgroup$– user297767Dec 11, 2015 at 0:50
-
$\begingroup$ have a look at this article: en.wikipedia.org/wiki/Projection-valued_measure $\endgroup$– user159517Dec 11, 2015 at 1:19
-
$\begingroup$ you can also try to understand these integrals in terms of riemann sums, with the role of absolute value played by the hilbert space norm. Personally, I don't find this approach very helpful. $\endgroup$– user159517Dec 11, 2015 at 1:26
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
This equation is not intended to mean the integral of an operator is the identity operator. The meaning is that if we take a ket, for example $|\psi\rangle$, and act on it with the operator $|x\rangle\langle x|$ and integrate over all values of $x$, then we get back $|\psi\rangle$, the same as operating on $|\psi\rangle$ with the identity operator $\Bbb 1$. In other words, the operator equation is just a short way to say that $$\int |x\rangle\langle x|\psi\rangle dx = |\psi\rangle$$ for any $|\psi\rangle$. And, of course, we could act on any bra $\langle \phi|$ with the projection operator, integrate, and get back $\langle \phi|$.
$\begingroup$
$\endgroup$
The concept of POVM (projector operator valued measure) does not necessarily arise in infinite dimensional Hilbert space. A common example are the so called spin coherent states. For example you can check that
$$ \int_{S_2} P(\Omega) \mu (d\Omega) = I $$
where $I$ denotes the $2\times 2$ identity matrix and
$$ P(\Omega) = P(\theta,\phi) = (I + \hat{n} \cdot \sigma)/2, $$
($\Omega$ is the spherical angle) with the notation $$ \hat{n} \cdot \sigma = \sum_{a=1}^3 n^a \sigma^a $$
and $\sigma^a$ are the Pauli matrices and $\hat{n}=(\sin(\theta)\cos(\phi),\sin(\theta)\sin(\phi),\cos(\theta))$ and $\mu(d\Omega)= \sin(\theta)d\theta d\phi /(2\pi)$. In other words
$$ \int_0^\pi \sin(\theta) d\theta \int_0^{2 \pi} d\phi P(\theta,\phi) = 2\pi I. $$
Integration of an operator on a finite dimensional space does not require any particular care. Simply integrate each component of the operator (matrix).
Added: If you want a more formal answer, in general you can define integration on a Banach space replacing absolute values with norms.
