Saturday, 2 April 2016

quantum field theory - Does Peskin & Schroeder Eq. (4.26), U(t1,t2)U(t2,t3)=U(t1,t3) imply [H0,Hint]=0?


Peskin & Schroeder equation (4.17) define the operator, U(t,t0) = ei(tt0)H0ei(tt0)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 t1t2t3. Does this imply that the free Hamiltonian commutes with the interaction [H0,Hint] = 0 ?
Here is my argument that it does.


In the condition t1t2t3 take t2=0. The identity is then, U(t1,0)U(0,t3)=U(t1,t3) .

Substitute the definition, eit1H0eit1Heit3H0eit3H=ei(t1t3)H0ei(t1t3)H
and simplify to get, eit1Heit3H0=eit3H0eit1H
with t10t3 . Put t1=t and t3=t. eitHeitH0=eitH0eitH
Expanding to second order in t, (1itHt22HH)(1+itH0t22H0H0)=(1+itH0t22H0H0)(1itHt22HH)
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+12ϕxrϕxr+V(ϕ)) .
In the classical theory, the PB is, [H0,Hint]PB=d3xδH0δπδHintδϕ=d3x πdVdϕ=ddtd3x V(ϕ(t,x))
Going over to quantum theory, [H0,Hint]=iddtd3x 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?



Answer



Ref. 1 writes the correct formula



U(t,t) = eiH0(tt0)eiH(tt)eiH0(tt0),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 tt0, 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(tt0)eiH(tt0).


It appears that OP mistakenly replaces t0 in eq. (4.17) with an arbitrary time tt. The resulting equation


U(t,t) = eiH0(tt)eiH(tt).(Wrong!)


is not correct.


References:




  1. M.E. Peskin & D.V. Schroeder, An Intro to QFT; Section 4.2.


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