Solution to Problem Set 1

A general time-dependent hamiltonian describing only single particle processes is,

H(t) = ∑_{i,j} V_{ij}(t)c_{i}^{†}c_{j}.  (1)

Since the hamiltonian is hermitian, the matrix V_{ij} is symmetric. The time evolution operator is defined as,

U(t, t′) = T exp [− ∫_{t′}^{t} H(t′′)dt′′] ,  (2)

where T stands for time ordering operator. The matrix element of the evolution operator in one particle space is defined as,

U_{ij}(t, t′) = ⟨ 0|c_{i}U(t, t′)c_{j}^{†}|0 ⟩.  (3)

The matrix B(t, t′) is defined in terms of the matrix V (t) as,

B(t, t′) = T exp [− ∫_{t′}^{t} V (t′′)dt′′] .  (4)

Each term in the above exponential is to be understood as multiplication of several matrices of the form V_{ij}(t), after appropriate time ordering.

1. First we examine the time evolution of B(t, t′), by comparing it with B(t + δt, t′). We have,

B(t + δt, t′) = T exp [− ∫_{t′}^{t} V (t′′)dt′′ − δtV (t+)]
= T ∑_{n} \frac{(−1)^{n}}{n!} [∫_{t′}^{t} V (t′′)dt′′ + δtV (t+)]^{n}
= B(t, t′) − T ∑_{n=1}^{∞} \frac{(−1)^{n−1}}{(n − 1)!} [∫_{t′}^{t} V (t′′)dt′′]^{n−1} δtV (t+) + O(δt)^{2}
= B(t, t′) − δtV (t+)B(t, t′).  (5)

Here t+ = t+ϵ, with ϵ → 0. Due to time ordering, in the last line of the above expression, V (t+) moves to the left of all other terms. Thus, B(t, t′) obeys the differential equation,

\frac{∂}{∂t}B_{ij}(t, t′) = −V_{ik}(t)B_{kj}(t, t′).  (6)

Similarly, it is easy to show that the time evolution operator satisfies the differential equation,

\frac{∂}{∂t}U(t, t′) = −H(t)U(t, t′)  (7)

Using equations (3) and (7) we get,

\frac{∂}{∂t}U_{ij}(t, t′) = −⟨ 0|c_{i}H(t)U(t, t′)c_{j}^{†}|0 ⟩
= − ∑_{k,l} ⟨ 0|c_{i}V_{kl}(t)c_{k}^{†}c_{l}U(t, t′)c_{j}^{†}|0 ⟩
= − ∑_{l} V_{il}(t)⟨ 0|c_{l}U(t, t′)c_{j}^{†}|0 ⟩
= −V_{il}(t)U_{lj}(t, t′).  (8)

Comparing equations (6) and (8) we find that both B_{ij}(t, t′) and U_{ij}(t, t′) satisfy the same differential equation. Moreover, at the initial condition, i.e. at t = t′, B_{ij}(t, t) = U_{ij}(t, t) = δ_{ij}. Thus,

B_{ij}(t, t′) = U_{ij}(t, t′).  (9)

This proves that the matrix B_{ij}(t, t′), as defined in equation (4), describes the evolution operator in the space of single particles.

2. We examine the time evolution of the Heisenberg operator (c_{H})_{i}, which is defined as,

(c_{H})_{i}(τ) = U(0, τ)c_{i}U(τ, 0).  (10)

Using equation (7) we get,

\frac{∂}{∂τ} (c_{H})_{i}(τ) = U(0, τ) [H(τ), c_{i}] U(τ, 0).

For the quadratic hamiltonian defined in equation (1) we get,

[H(τ), c_{i}] = − ∑_{l} V_{il}(τ)c_{l}.

Note that the above expression does not depend whether the particles are fermionic or bosonic. Using the above two expressions we get,

\frac{∂}{∂τ} (c_{H})_{i}(τ) = − ∑_{l} V_{il}(τ)(c_{H})_{l}(τ).  (11)

Thus, we find that the time evolution of the annihilation operators (and also of the creation operators) can be entirely described in the space of single particles. This is the essential simplification that is possible for a hamiltonian describing one particle processes. Such simplification is not possible if the hamiltonian includes multi-particle processes.

Next we define the operator,

A_{i}(τ) = ∑_{j} B_{ij}(τ, 0)c_{j}.  (12)

Using equation (6), its time evolution is given by,

\frac{∂}{∂τ} A_{i}(τ) = − ∑_{j} V_{ik}(τ)B_{kj}(τ, 0)c_{j}
= −V_{ik}(τ)A_{k}(τ).  (13)

Comparing equations (11) and (13) we find that both the operators (c_{H})_{i}(τ) and A_{i}(τ) satisfy the same differential equation. Moreover, at initial time, i.e. t = 0, (c_{H})_{i}(0) = A_{i}(0) = c_{i}. Thus, they are the same operators and we can write,

(c_{H})_{i}(τ) = ∑_{j} B_{ij}(τ, 0)c_{j}.  (14)

This completes the demonstration of what we have already noted, namely, the time evolution of the annihilation operators (and also of creation operators) is given by the matrix of the evolution operator U(t, t′) in the space of single particles.