Timeline for Quantum Field theory - integral notation
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Feb 6, 2015 at 19:47 | comment | added | Eric | Great answer! Thank you. For the identity we even don't need the functional calculus, we know that $Q^{\prime}$ as a selfadjoint operator admits a unique spectral measure $E$, thus by using the properties of a spectral measure we got $I=E(\sigma(Q^{\prime}))= \int_{\sigma(Q^{\prime})} 1 E(d\lambda)$, which can be of course written in a different notation which involves bra and kets. | |
| Feb 6, 2015 at 19:34 | comment | added | Igor Khavkine | A good rule of thumb in interpreting this notation is to make sure that your interpretation works when the Hilbert space is finite dimensional and all integrals are replaced by a sums. | |
| Feb 6, 2015 at 19:32 | comment | added | Igor Khavkine | Your first formula gives $Q'$ and not $I$, with $I$ given correctly by your second formula. The prototypical example of $Q'$ is the position operator in quantum mechanics. It has a simple continuous spectrum (in 1-dimension, that is) and that is why its "eigenvectors" are convenient for writing down a resolution of the identity. Essentially, you are representing identity using functional calculus $I = f(Q) = \int dq' f(q') |q'\rangle\langle q'|$, where $f(x) \equiv 1$. | |
| Feb 6, 2015 at 16:06 | comment | added | Eric | Thanks, I know that you have spectral theorem for any normal operator defined on a Hilbert space, however the notation for the resolution of identity in the example which I gave about was for the identity operator, which is in fact diagonal. One quick question so it supposed to be $$I= \int d q^{\prime} q^{\prime} \left|q^{\prime} \right> \left<q^{\prime}\right|$$ instead of $$I= \int d q^{\prime} \left|q^{\prime} \right> \left<q^{\prime}\right|$$? | |
| Feb 6, 2015 at 16:02 | vote | accept | Eric | ||
| Feb 6, 2015 at 13:13 | history | answered | Igor Khavkine | CC BY-SA 3.0 |