How do you derive Slater determinant from creation operator?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
Amentia
Messages
108
Reaction score
5
Hello,

Could someone provide me with a good proof or explain me here how we can derive Slater determinant for N fermions by starting with the vacuum state and the creation operators with anticommutation equations. I see that the idea of both these structures is similar but I cannot work it out rigorously.

Thank you.
 
Physics news on Phys.org
Amentia said:
Hello,

Could someone provide me with a good proof or explain me here how we can derive Slater determinant for N fermions by starting with the vacuum state and the creation operators with anticommutation equations. I see that the idea of both these structures is similar but I cannot work it out rigorously.

Thank you.
Gordon Baym's book Lectures on Quantum Mechanics covers this in the chapter on Second Quantization. His treatment of it is quite good, other than on occasion he does have a ## \sqrt{n!} ## that should be simply a ## \sqrt{n} ## or similar minor typo. I think the book is currently out of print but very good reading if you can get a copy of it.
 
Thank you, I will take a look at this chapter.