 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2018/1/a/solution

For independent bosonic modes, the canonical commutation relation is

[b_{i}, b_{j}^{†}] = δ_{ij}, [b_{i}, b_{j}] = [b_{i}^{†}, b_{j}^{†}] = 0.

For independent fermionic modes, the canonical anticommutation relations are

{c_{i}, c_{j}^{†}} = δ_{ij}, {c_{i}, c_{j}} = {c_{i}^{†}, c_{j}^{†}} = 0,

where

[A, B] = A B - B A

and

{A, B} = A B + B A

n_{i} = a_{i}^{†} a_{i},

with

a_{i} = b_{i}

c_{i}

as appropriate. It measures the mode's occupation number. Its possible eigenvalues are

n_{i} = 0, 1, 2, \dots for bosons; n_{i} = 0, 1 for fermions .

c_{i}^{2} = 0

implies

n_{i}^{2} = c_{i}^{†} (1 - c_{i}^{†} c_{i}) c_{i} = n_{i}

, so the number operator is a projection. This expresses the exclusion of double occupation of one fermionic mode. The total particle number is

N = \sum_{i} n_{i}

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