Timeline for Inverse Quantum Operator
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Oct 29, 2015 at 4:44 | comment | added | Ali Moh | right, one basis is countable and one is not. I edited the answer to provide a practical demonstration rather than an abstract argument | |
| Oct 29, 2015 at 4:43 | history | edited | Ali Moh | CC BY-SA 3.0 |
added 804 characters in body
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| Oct 29, 2015 at 3:00 | comment | added | ACuriousMind♦ | I first thought this, too, but the state $\lvert x \rangle$ is not a valid state in the Hilbert space, i.e. it is not a proper eigenvector of $x$, but only lies inside the larger part of a rigged Hilbert space. It is not clear that the existence of a zero eigenvector in a larger space forbids the invertibility on the Hilbert space. | |
| Oct 29, 2015 at 2:43 | history | answered | Ali Moh | CC BY-SA 3.0 |