Past exam of the theoretical physics II course of the University of Cambridge / 2015 / Paper 1 / 5 / b

For identical fermions,

(b_{j}^{†})^{2} = 0

and

b_{2}^{†} b_{1}^{†} = - b_{1}^{†} b_{2}^{†}

. Therefore

a_{1}^{†} a_{2}^{†} ∣0 ⟩ = \frac{1}{2} (b_{1}^{†} + b_{2}^{†}) (b_{1}^{†} - b_{2}^{†}) ∣0 ⟩ = - b_{1}^{†} b_{2}^{†} ∣0 ⟩ .

(1)

The minus sign is a global phase for this output. The only occupancy is $(n_{1}, n_{2}) = (1, 1)$ , with probability one. Double occupancy of either specified single-particle mode is excluded; no extra internal states are present in this model.

 Ancestors (14)

5
Paper 1
2015
Past exam of the theoretical physics II course of the University of Cambridge
Theoretical Physics II course of the University of Cambridge
Theoretical Physics course of the University of Cambridge
Physics course of the year 3 of the natural sciences course of the University of Cambridge
Year 3 of the natural sciences course of the University of Cambridge
Year of the natural sciences course of the University of Cambridge
Natural sciences course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
List of universities
 Home

 View article source

 Discussion (0)

New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version

Past exam of the theoretical physics II course of the University of Cambridge / 2015 / Paper 1 / 5 / b

Solution

 Ancestors (14)

 Discussion (0)  Subscribe (1)

 Articles by others on the same topic (0)

 Discussion (0)