The wavefunctions of two identical particles in stated n and s are given by $${\phi _n}\left( {{r_1}} \right)$$ and $${\phi _s}\left( {{r_2}} \right)$$ , respectively. The particles obey Maxwell-Boltzmann statistics. The state of the combined two particles system is expressed as
A. $${\phi _n}\left( {{r_1}} \right) + {\phi _s}\left( {{r_2}} \right)$$
B. $$\frac{1}{{\sqrt 2 }}\left[ {{\phi _n}\left( {{r_1}} \right){\phi _s}\left( {{r_2}} \right) + {\phi _n}\left( {{r_2}} \right){\phi _s}\left( {{r_1}} \right)} \right]$$
C. $$\frac{1}{{\sqrt 2 }}\left[ {{\phi _n}\left( {{r_1}} \right){\phi _s}\left( {{r_2}} \right) - {\phi _n}\left( {{r_2}} \right){\phi _s}\left( {{r_1}} \right)} \right]$$
D. $${\phi _n}\left( {{r_1}} \right){\phi _s}\left( {{r_2}} \right)$$
Answer: Option D
A. $$\frac{1}{{1 + {e^{ - \varepsilon /{k_B}T}}}}$$
B. $$\frac{1}{{1 + 2{e^{\varepsilon /{k_B}T}}}}$$
C. $$\frac{1}{{2{e^{\varepsilon /{k_B}T}} + 4{e^{2\varepsilon /{k_B}T}}}}$$
D. $$\frac{1}{{2{e^{\varepsilon /{k_B}T}} - 4{e^{2\varepsilon /{k_B}T}}}}$$
A. $$\frac{1}{6}{E_F}$$
B. $$\frac{1}{5}{E_F}$$
C. $$\frac{2}{5}{E_F}$$
D. $$\frac{3}{5}{E_F}$$
A. clockwise
B. counter-clockwise
C. neither clockwise nor counter-clockwise
D. clockwise from X → Y and counter-clockwise from Y → X
A. $$\exp \left( {\frac{{\hbar \omega }}{{{k_B}T}}} \right) + 1$$
B. $$\exp \left( {\frac{{\hbar \omega }}{{{k_B}T}}} \right) - 1$$
C. $${\left[ {\exp \left( {\frac{{\hbar \omega }}{{{k_B}T}}} \right) + 1} \right]^{ - 1}}$$
D. $${\left[ {\exp \left( {\frac{{\hbar \omega }}{{{k_B}T}}} \right) - 1} \right]^{ - 1}}$$

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