We will study the Bogoliubov transformation in second-quantization by considering a bilinear Hamiltonian, ˆ α a u −v αˆ 1 u +v a † † † † H = E0 (a a + b b) + E1 (a b + ba ); † = (1.1) ↔ ˆ† ˆ† = 2 2 b† b v u v u − + u v − β β
Here, a, b ↔ a † , b † are bosonic annihilation ↔ creation operators, and we have the indicated linear † † transformation to αˆ , βˆ ↔ αˆ , βˆ . These diagonalize the Hamiltonian, as we will see. First: show that by inserting these expressions into the commutation relations for [a , b ] ↔ −[a † , b † ] that u 2 − v 2
=1.
For the proposed Bogoliubov transformation to be canonical, it must preserve the commutationrelations given that the original operators appropriately commute, † † † † † † † † † † [a, b ] = [a, b] = 0 = 0 = [a , b ] = −[a , b]; u2 −v2 =1 [αˆ , βˆ ] = [αˆ , βˆ ] = 0 = 0 = [αˆ , b ] = −[αˆ , βˆ ]; → (1.2) ???? † † ˆ ˆ [a, a † ] = 1 = [b, b† ] ˆ ˆ α α β β [ , ] 1 [ , ] = =
Preliminaries:
Is the transformation canonical? Take
commutators amidst the αˆ , βˆ (u 2
−v
2
↔ αˆ
†
the commutators on the LHS of (1.2) as given, and work out the
, βˆ , †
ˆ ] = [ au + vb† , a † v + bu ] = uv[a , a † ] + u 2 [a, b] + v 2 [b † , a † ] + vu [b † , b ] D.C . ˆ † † ) ⋅ [αˆ , β ˆ ] = 0† [ β , α =
uv + u
2
2
⋅ 0 + v ⋅ 0 − vu =
0↔
[αˆ , βˆ ] =
0 u 2 −v 2
↔ ˆ = 0 = −0 = −[ β , αˆ ]
=
0
ˆ† ] = −[αˆ , β †
(1.3)
ˆ † ] = [ au + vb† , av (u 2 − v2 ) ⋅ [αˆ , β av + b †u ] = uv[a , a ] + u 2 [ a, b † ] + v 2 [b † , a ] + vu [b † , b † ] 2
2
= uv ⋅ 0 + u ⋅ 0 − v ⋅ 0 + vu ⋅ 0 ↔
vb† , ua † (u 2 − v 2 ) ⋅ [αˆ , αˆ † ] = [ua + vb
(u − v ) ⋅ [ βˆ † , βˆ ] = [va + ub ub† , va † 2
2
+ vb] = u
2
+ ub ] = v
2
† [αˆ , βˆ ] =
2
u −v
† = 0 = −0 = − − [αˆ , βˆ ] †
0 2
(1.4)
[a , a † ] + uv[a , b ] + vu[b † , a † ] + v 2 [b † , b ] = +(u 2 − v 2 ) ↔ [αˆ , αˆ † ] = 1 †
†
†
2
†
[a , a ] + vu[a , b] + uv[b , a ] + u [b , b ] = −(u
2
−v
2
)↔
ˆ†] =1 [ βˆ , β
(1.5)
All other commutation relations follow by applying † to both sides or by using [ A, B ] = −[ B, A] (each, as twice demonstrated in (1.3) and (1.4)). Note, also, that the transformation being canonical just requires u 2 − v 2 Normalization:
≠
0.
let us compute the comutator [ a, a † ] ; from this comes the promised normalization,
vu − −[ βˆ † , βˆ ]v 2 = u 2 − v 2 [a, a † ] = 1 = [αˆ u − vβˆ † , αˆ †u − v βˆ ] = [αˆ , αˆ † ]u 2 − 0 ⋅ uv − 0 ⋅ vu Furthermore: demonstrate that the condition (u
2
+v
2
=1
(1.6)
a Hamiltonian without cross terms )E1 = 2uvE0 yields
ˆ† . ˆ ˆ and αˆ † β such as αβ † † Compute a † a , b†b , and a†b† ↔ ba , straight away (dirty work). Use (1.2) to favour occurrences of αˆ αˆ , βˆ βˆ , † † † 2 † 2 † † † 2 † 2 † † † a a = (uαˆ − v βˆ )(uαˆ − v βˆ ) = u αˆ αˆ + v βˆβˆ − uvαˆ βˆ − vu βˆαˆ = u αˆ αˆ + v (1 + βˆ βˆ ) − uv[αˆ βˆ + αˆβˆ ];
b†b = ( − vαˆ + u βˆ † )(−vαˆ † † †
ab
+ u βˆ ) =
ˆ − v βˆ )(−vαˆ + u βˆ = (uα †
†
v2αˆαˆ † + u 2 βˆ † βˆ − vuαˆβˆ − uvβˆ †αˆ †
† ˆ ˆ † + u 2αˆ † βˆ † + v 2 βα ˆˆ ) = −uvαˆ αˆ − vuββ
=
v2 (αˆ †αˆ + 1) + u2 βˆ † βˆ − uv[αˆ † βˆ† + αˆβˆ ];
↔ −uvαˆ
ˆ ˆ † + u 2 βα ˆ ˆ + v2αˆ † β ˆ † αˆ − vuββ
†
=
ba
Now, put (1.7) into the Hamiltonian of (1.1). Use the developed commutation relations summarized on the RHS of (1.2) to get operator-coefficients of αˆ †αˆ + βˆ † βˆ (diagonal) and αˆ † βˆ † + αˆβˆ (off-diagonal) so that you get,
(1.7)
H
=
2 † † 2 † † † † ˆ ˆ )) + E1((u 2 + v 2 )( αˆ † βˆ † + αβ ˆ ˆ ) − 2vu( αˆ † αˆ + βˆ † βˆ + 1)) E0 (u (αˆ αˆ + βˆ βˆ ) + v (2 + βˆ βˆ + αˆ αˆ ) − 2uv(αˆ βˆ + αβ
† 2 2 ˆ † βˆ † = ( E0 (u + v ) − 2vuE1 )(αˆ αˆ + βˆ βˆ ) − (2uvE0 − E1(u + v ))(α 2
2
†
In this form (1.8), it is clear that the condition (u
2
+v
2
2
†
†
2
(1.8)
)E1 = 2uvE0 eliminates the off-diagonal terms,
H = ( E0 (u + v ) − 2vuE1 )(αˆ αˆ + βˆ βˆ ) + E0 v − 2uvE1 2
ˆ ˆ ) + E0v 2 − 2uvE 1 + αβ
† † Oˆ ≡αˆ αˆ + βˆ βˆ
=
2 2 ˆ 2 ˆ +1] E0[( u + v ) O + v ] − 2 vuE1[ O
2
(1.9)
2
Use the parametrization (u , v) = (cosh t , sinh t ) to compute the eigenenergy λ = E0 − E 1 , implying that the † † diagonalized Hamiltonian takes the form H = λ (αˆ αˆ + βˆ βˆ ) + [const] .
Using the identities cosh x cosh y + sinh x sinh y = cosh( x + y ) and sinh x cosh y + cosh x sinh y = sinh( x + y ) , and also letting ∆ H
=
2
E0 sinh t − E1 sinh 2t , a constant shift in energy, we get, 2 H = E0 (Oˆ cosh 2t + sinh t ) − sinh 2tE1 (Oˆ + 1) = Oˆ ( E0 cosh 2t − E1 sinh 2t ) + ∆H
(1.10)
E
2 2 −1 E 10 . We use the trigNow notice the equivalence (u + v ) E1 = 2uvE0 ↔ E1 cosh 2t = E0 sinh 2t ↔ 2t = tanh
formulae cosh tanh −1 x =
1 1− x
2
=
1 H − ∆H = Oˆ E0 E 2 1 − ( E ) 1 0
Afterword:
sinh 2 x + 1 , and this Hamiltonian (1.10), sans the shift ∆ H , finally becomes,
− E1
2 = Oˆ E0 E 2 E 2 − E 2 1 − ( E ) 1 0 E 1 E 0
1 0
† † Oˆ ≡αˆ αˆ + βˆ βˆ
−
λ ≡ E − E ≡ λ (αˆ †αˆ + βˆ † βˆ ) (1.11) E0 2 − E 12 E 12
the Bogoliubov transformation for fermions is treated in exercise 1.7.
2
0
2 1
