Timeline for Fock Space and fermionic annihilation & creation operators
Current License: CC BY-SA 3.0
10 events
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| Apr 13, 2017 at 12:39 | history | edited | CommunityBot |
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| Feb 18, 2015 at 23:34 | history | edited | glS | CC BY-SA 3.0 |
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| Feb 18, 2015 at 22:59 | comment | added | glS | Your country is most probably mine too, so I understand the hour (there is no need for sinchronous conversations here though, so I don't see any problem with this factor). Anyway, I'm not contesting your voting choices. If you feel the answer is of low quality it is your right to vote whatever you like, I'm not offended by that. That said, I will try to make it clearer. | |
| Feb 18, 2015 at 22:50 | comment | added | Sofia | it also displeases me that Ballentine wrote a superficial proof, very questionable. So, let's talk tomorrow. Mark's comment is also not very useful. | |
| Feb 18, 2015 at 22:48 | comment | added | Sofia | I have no pleasure to give minus to someone. I just marked by that, the fact that this is not the answer. Let's see the situation tomorrow (now it's late in my country) and I'd be glad to remove the minus. | |
| Feb 18, 2015 at 22:43 | comment | added | glS | @Sofia indeed, that is not true, and I didn't say that. What I said is that the way you define your initial state is a manner of convention, and you can define it as (1) or as (2). Once you chose one or the other, you must stick with that convention. And of course we are dealing with fermionic operators so $$c_a^\dagger c_b^\dagger = - c_b^\dagger c_a^\dagger, \quad \text{for }a \neq b.$$ Concerning the question of the OP, you are right I didn't read that comment. I edited the question trying to more directly address the question. | |
| Feb 18, 2015 at 22:40 | history | edited | glS | CC BY-SA 3.0 |
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| Feb 18, 2015 at 22:39 | comment | added | Sofia | Your answer doesn't address the question posed by the user. He wants that somebody explain him what is wrong with the operations he does, on the initial state that he chose. I also looked in Ballentine and I disagree with the proof there. Besides that, I am not so sure that $C_a ^{\dagger} C_b ^{\dagger} = C_b ^{\dagger} C_a ^{\dagger}$ because the state created by the second variant is opposite in sign to the state created by the former variant. But let's discuss the issue tomorrow, maybe we'll clarify it. | |
| Feb 18, 2015 at 20:42 | comment | added | Sofia | did you see his comment to Mark? The user doesn't want to start from vacuum. | |
| Feb 18, 2015 at 20:40 | history | answered | glS | CC BY-SA 3.0 |