- #1
Alex Cros
- 28
- 1
Hi all,
I'm trying to find a mathematical way of showing that given a complete set $$\left |a_i\right \rangle_{i=1}^{i=dim(H)}∈H$$ together with the usual property of $$\left |\psi\right \rangle = ∑_i \left \langle a_i\right|\left |\psi\right \rangle\left |a_i\right \rangle ∀ \left |\psi\right \rangle∈H$$. Now, by letting the set $$\left | a_i \right \rangle_{i=1}^{i=dim(H)} → \left |a_i\right \rangle_{i=1}^{i=∞}$$ and $$\left |a_{i+1}\right \rangle = \left |a_i\right \rangle+\left |δ\right \rangle$$ as $$ δ→0$$ (in the sense of $$\left |a_{i+1}\right \rangle∈Neighborhood(\left |a_i\right \rangle)$$) we should obtain the familiar expression $$\left |\psi\right \rangle = ∫ da \left \langle a\right|\left |\psi\right \rangle\left |a\right \rangle ∀ \left |\psi\right \rangle∈H$$.
How could this be linked in a rigorous way without the usual "for the continuous case replace sum by integral".
Thanks in advance!
PD: Sorry for the latex form, writing in physics forums can be daunting without any packages...
I'm trying to find a mathematical way of showing that given a complete set $$\left |a_i\right \rangle_{i=1}^{i=dim(H)}∈H$$ together with the usual property of $$\left |\psi\right \rangle = ∑_i \left \langle a_i\right|\left |\psi\right \rangle\left |a_i\right \rangle ∀ \left |\psi\right \rangle∈H$$. Now, by letting the set $$\left | a_i \right \rangle_{i=1}^{i=dim(H)} → \left |a_i\right \rangle_{i=1}^{i=∞}$$ and $$\left |a_{i+1}\right \rangle = \left |a_i\right \rangle+\left |δ\right \rangle$$ as $$ δ→0$$ (in the sense of $$\left |a_{i+1}\right \rangle∈Neighborhood(\left |a_i\right \rangle)$$) we should obtain the familiar expression $$\left |\psi\right \rangle = ∫ da \left \langle a\right|\left |\psi\right \rangle\left |a\right \rangle ∀ \left |\psi\right \rangle∈H$$.
How could this be linked in a rigorous way without the usual "for the continuous case replace sum by integral".
Thanks in advance!
PD: Sorry for the latex form, writing in physics forums can be daunting without any packages...