Timeline for Explanation of Dirac's proof of arbitrary ket being expressible with eigenkets of observable
Current License: CC BY-SA 3.0
10 events
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| Jul 16, 2014 at 14:46 | vote | accept | user3237992 | ||
| Jul 16, 2014 at 11:40 | comment | added | By Symmetry | I think Dirac is trying to prove that no ket can appear in both the integral and in the sum and is doing this by proving that the normalisation of the 2 sets of kets cannot be done consistently, i.e $\delta(0)\ne 1$ and no amount of pulling out factors is going to help. His method of proving this is a little unusual. You're right he has written down the same equation twice with some dummy label, but all he then does with these equation substitute one into the other. Substituting an equation into itself is a perfectly legal manipulation, its just normally not very helpful. | |
| Jul 16, 2014 at 11:04 | comment | added | user3237992 | Using Symmetry's answer I was able to construct a more or less correct proof, but I can't follow Dirac's train of thought. He labels the same ket in the integral and in the sum differently (thus also the same bra is differently labeled). So when he constructs (32) & (33) from (31) by multiplying both sides first by $\langle\xi^lb|$ and second by $\langle\xi^la|$, he considers them as two different equations, however since only the labels are different but they are the same bra, there's no difference between the two equations, still, he considers them to be. Can somebody explain this? | |
| Jul 15, 2014 at 21:13 | comment | added | ACuriousMind♦ | @yuggib: Forgiven and duly noted. :) | |
| Jul 15, 2014 at 21:10 | comment | added | yuggib | @ACuriousMind Forgive my mathematical pedantry, but self-adjointness alone is still not a sufficient condition to have an orthonormal basis of eigenvectors...you need also that the operator is either compact or with compact resolvent. | |
| Jul 14, 2014 at 15:04 | comment | added | ACuriousMind♦ | @Danu: It's a side effect. But if the operator is not self-adjoint, you have no guarantee that it will be diagonalizable at all, even by states with complex eigenvalues. You really need self-adjointness to be sure that the eigenvectors form an orthonormal basis of the Hilbert space. | |
| Jul 14, 2014 at 14:55 | comment | added | Danu | @ACuriousMind Wasn't that due to the requirement that the eigenvalues be real? | |
| Jul 14, 2014 at 14:35 | history | edited | By Symmetry | CC BY-SA 3.0 |
At suggestion of @ACuriousMind added a point about operator for observables being Hermitian
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| Jul 14, 2014 at 13:47 | comment | added | ACuriousMind♦ | Good answer with physical intuition. One could, in more modern terms, mention that the completeness of the eigenstates is the reason why we demand observables to be self-adjoint. | |
| Jul 14, 2014 at 12:16 | history | answered | By Symmetry | CC BY-SA 3.0 |