Peskin & Schroeder equation (4.17) define the operator, U(t,t0) = ei(t−t0)H0e−i(t−t0)H
where
H = H0+Hint
is the full Hamiltonian and
H0 is the free Hamilton, both in the Schrodinger picture. In equation (4.26), Peskin and Schroeder state that the operator satisfies the following identity,
U(t1,t2)U(t2,t3) = U(t1,t3)
where
t1≥t2≥t3. Does this imply that the free Hamiltonian commutes with the interaction
[H0,Hint] = 0 ?
Here is my argument that it does.
In the condition t1≥t2≥t3 take t2=0. The identity is then, U(t1,0)U(0,t3)=U(t1,t3) .
Substitute the definition,
eit1H0e−it1He−it3H0eit3H=ei(t1−t3)H0e−i(t1−t3)H
and simplify to get,
e−it1He−it3H0=e−it3H0e−it1H
with
t1≥0≥t3 . Put
t1=t and
t3=−t.
e−itHeitH0=eitH0e−itH
Expanding to second order in
t,
(1−itH−t22HH)(1+itH0−t22H0H0)=(1+itH0−t22H0H0)(1−itH−t22HH)
results in,
HH0=H0H
so that
[H0,H]−=0. Now
H=H0+Hint so the free Hamiltonian must commute with the interaction.
[H0,Hint]−=0
In Peskin and Schroeder, the context for this material is the self-interacting scalar field with Hamiltonian,
H=∫d3x(12π(t,x)2+12∂ϕ∂xr∂ϕ∂xr+V(ϕ)) .
In the classical theory, the PB is,
[H0,Hint]PB=−∫d3xδH0δπδHintδϕ=−∫d3x πdVdϕ=−ddt∫d3x V(ϕ(t,x))
Going over to quantum theory,
[H0,Hint]−=−iddt∫d3x V(ϕ(t,x))
so that
[H0,Hint]−=0 implies the integral of
V(ϕ) is a conserved charge; is this also a correct result?
Ref. 1 writes the correct formula
U(t,t′) = eiH0(t−t0)e−iH(t−t′)e−iH0(t′−t0),t ≥ t′,
which satisfies
U(t1,t2)U(t2,t3) = U(t1,t3),t1 ≥ t2 ≥ t3.
Here t0 is an arbitrary but fixed fiducial initial instant where operators and states in the Schrödinger picture, the Heisenberg picture and the interaction picture all agree. For t≠t0, the three pictures are no longer the same, although they are still unitary equivalent.
For t′=t0, eq. (4.25) simplifies to
U(t,t0) = eiH0(t−t0)e−iH(t−t0).
It appears that OP mistakenly replaces t0 in eq. (4.17) with an arbitrary time t′≤t. The resulting equation
U(t,t′) = eiH0(t−t′)e−iH(t−t′).(←Wrong!)
is not correct.
References:
- M.E. Peskin & D.V. Schroeder, An Intro to QFT; Section 4.2.
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