Timeline for Fock Space and fermionic annihilation & creation operators
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Feb 19, 2015 at 11:31 | comment | added | Mark Mitchison | @pindakaas No, there is no need to update the question. I think the question as it stands plus the accepted answer will be very clear to any future users. | |
| Feb 19, 2015 at 8:06 | comment | added | Kuhlambo | Thanks for the effort everyone. To get a complete summary of how the formalism is constructed starting at one particle states and going with tensor products all the way to the standard notation might be a bit much to ask here. This is why i am content with the marked answer. (although feel free to explain it all of course ^^ i would certainly appreciate it.) -- Should i update the question with the completed proof? | |
| Feb 19, 2015 at 7:58 | vote | accept | Kuhlambo | ||
| Feb 19, 2015 at 0:02 | answer | added | CR Drost | timeline score: 2 | |
| Feb 18, 2015 at 22:05 | comment | added | Sofia | @pindakaas The other operation, $C_b^{\dagger}C_a|\sim b,a⟩$, first of all leaves vacuum, $|\sim b,\sim a⟩$. You know for sure that the old fermion was removed. After that, you create a single fermion with spin-projection $b$, $|b, \sim a⟩$. | |
| Feb 18, 2015 at 22:05 | comment | added | Sofia | @pindakaas I don't know the complete answer, but you make a mistake. Let's deal only with two fermions, "old" and "new", identified by their places, the newly added fermion on the 1st position. The operation $C_b^{\dagger}|\sim b,a⟩$ introduces a new fermion, but one gets an antisymmetrical state, $(|b,a⟩ - |a,b⟩)/ \sqrt {2}$. Whether the old fermion remains with the value $a$ and the new one gets $b$ it is not known. Applying $C_a$ you get $(|b,\sim a⟩ - |\sim a,b⟩)/\sqrt {2}$, so, the old fermion or the new one was destroyed? You see, the position in the list is not fixed. (I continue) | |
| Feb 18, 2015 at 20:40 | answer | added | glS | timeline score: 2 | |
| Feb 18, 2015 at 19:14 | history | edited | Sofia | CC BY-SA 3.0 |
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| Feb 18, 2015 at 18:45 | history | edited | Sofia | CC BY-SA 3.0 |
deleted 8 characters in body
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| Feb 18, 2015 at 18:38 | history | edited | Sofia | CC BY-SA 3.0 |
clarification
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| Feb 18, 2015 at 18:32 | history | edited | Sofia | CC BY-SA 3.0 |
Minor grammar fixing
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| Feb 18, 2015 at 18:17 | comment | added | Kuhlambo | the answer presupposes what i was trying to prove, so thats not very satisfying. | |
| Feb 18, 2015 at 17:57 | history | edited | Kuhlambo | CC BY-SA 3.0 |
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| Feb 18, 2015 at 17:31 | comment | added | Qmechanic♦ | Possible duplicate: physics.stackexchange.com/q/62604/2451 | |
| Feb 18, 2015 at 16:51 | comment | added | Mark Mitchison | The action of the $C_a$ operators on the states $|a\ldots b\rangle$ includes a phase of $\pm 1$, which depends on an arbitrary choice of ordering the possible quantum numbers $a,b,\ldots$. It is simpler to take $\{C_a,C_b^\dagger\} = \delta_{ab}$ as a definition and then always write many-particle states like $C_1^\dagger C_2^\dagger \ldots C_m^\dagger \lvert 0 \rangle$ etc. | |
| Feb 18, 2015 at 16:34 | history | asked | Kuhlambo | CC BY-SA 3.0 |