Wednesday, 14 January 2015

quantum mechanics - Expressing two particle state as a combination of single particle states



Say I have a Fock space H with basis K={|k|kN}. Then I consider the following single particle states:


|A=kKak|k, |B=kKbk|k.


I know that |k1k2=12(|k1|k2|k2|k1) is a valid fermionic two-particle state. I expected I could calculate the two particle state which contains particles A and B as


|AB?=12(|A|B|B|A).


But it turns out that


12(|A|B|B|A)=12k1,k2K(ak1bk2|k1|k2ak2bk1|k2|k1)=0.


So how do I write this two particle state |AB? It should be expressible as


|AB=k1,k2Kk1<k2ck1k2|k1k2,


but what is ck1k2? Is it ck1k2=ak1bk2? Why?



Answer




Why should


12k1,k2K(ak1bk2|k1|k2ak1bk2|k2|k1)=0


be true?


By switching the indices you get


12k1,k2K(ak1bk2|k1|k2ak1bk2|k2|k1)=12k1,k2K(ak1bk2ak2bk1)|k1|k2.


As you can see, the symmetric terms vanish, but the antisymmetric ones remain. From this equation, it is easy to see that


ck1,k2=12(ak1bk2ak2bk1).


Further information is given here.


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...