102
4 Two Computational Schemes of χ (2)
4.5.2 Canonical Time Correlation Function
[Problem 4.2] Prove A; B = B; A for arbitrary operators A and B in
Eq. (4.24). This indicates that the operators in the canonical correlation function
are commutative like a classical one.
A; B is
A; B =
1
β
β
0
dλ exp(λH)A exp(−λH)B
(4.24)
=
1
β
β
0
dλ Tr
exp(−βH)
Q
exp(λH)A exp(−λH)B
,
where Q is the partition function. The variable λ in the above integral is transformed
to λ = β − λ. Consequently,
A; B =
−1
βQ
0
β
dλ
Tr
exp(−λ
H)A exp{(λ
− β)H}B
=
1
βQ
β
0
dλ
Tr
exp(−βH) exp(λ
H)B exp(−λ
H)A
=
1
β
β
0
dλ
exp(λ H)B exp(−λ H)A
= B; A .
Bibliography
1. Allen MP, Tildesley DJ (1987) Computer simulation of liquids. Clarendon Press, Oxford
2. Atkins PW, Friedman RS (2010) Molecular quantum mechanics, 5th edn. Oxford University
Press, Oxford
3. Bader JS, Berne BJ (1994) Quantum and classical relaxation rates from classical simulations.
J Chem Phys 100:8359–8366
4. Buch V, Tarbuck T, Richmond GL, Groenzin H, Li I, Shultz MJ (2007) Sum frequency
generation surface spectra of ice, water, and acid solution investigated by an exciton model.
J Chem Phys 127:204710
5. Egorov SA, Everitt KF, Skinner JL (1999) Quantum dynamics and vibrational relaxation. J
Phys Chem A 103:9494–9499
6. Frenkel D, Smit B (1996) Understanding molecular simulation. Academic, New York
7. Gan W, Wu B-H, Zhang Z, Guo Y, Wang H-F (2007) Vibrational spectra and molecular
orientation with experimental configuration analysis in surface sum frequency generation
(SFG). J Phys Chem C 111:8716–8725
4 Two Computational Schemes of χ (2)
4.5.2 Canonical Time Correlation Function
[Problem 4.2] Prove A; B = B; A for arbitrary operators A and B in
Eq. (4.24). This indicates that the operators in the canonical correlation function
are commutative like a classical one.
A; B is
A; B =
1
β
β
0
dλ exp(λH)A exp(−λH)B
(4.24)
=
1
β
β
0
dλ Tr
exp(−βH)
Q
exp(λH)A exp(−λH)B
,
where Q is the partition function. The variable λ in the above integral is transformed
to λ = β − λ. Consequently,
A; B =
−1
βQ
0
β
dλ
Tr
exp(−λ
H)A exp{(λ
− β)H}B
=
1
βQ
β
0
dλ
Tr
exp(−βH) exp(λ
H)B exp(−λ
H)A
=
1
β
β
0
dλ
exp(λ H)B exp(−λ H)A
= B; A .
Bibliography
1. Allen MP, Tildesley DJ (1987) Computer simulation of liquids. Clarendon Press, Oxford
2. Atkins PW, Friedman RS (2010) Molecular quantum mechanics, 5th edn. Oxford University
Press, Oxford
3. Bader JS, Berne BJ (1994) Quantum and classical relaxation rates from classical simulations.
J Chem Phys 100:8359–8366
4. Buch V, Tarbuck T, Richmond GL, Groenzin H, Li I, Shultz MJ (2007) Sum frequency
generation surface spectra of ice, water, and acid solution investigated by an exciton model.
J Chem Phys 127:204710
5. Egorov SA, Everitt KF, Skinner JL (1999) Quantum dynamics and vibrational relaxation. J
Phys Chem A 103:9494–9499
6. Frenkel D, Smit B (1996) Understanding molecular simulation. Academic, New York
7. Gan W, Wu B-H, Zhang Z, Guo Y, Wang H-F (2007) Vibrational spectra and molecular
orientation with experimental configuration analysis in surface sum frequency generation
(SFG). J Phys Chem C 111:8716–8725