A question about <The Quantum Theory of Fields> P.120

Sam Wong
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The Quantum Theory of Fields, Steven Weinberg, P.120

above 3.3.24, it says,

If the matrix elements of W between H0 - eigenstates are sufficiently smooth functions of energy, then matrix elements of W(t) between smooth super-positions of energy eigenstates vanish for t→±∞

I can't get this. Could anyone please show this explicitly?
 
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Well I'm not sure, but here's a stab.. Weinberg has W(t) ≡ eiH0t W e-iH0t

Take the matrix elements of both sides, between energy eigenstates Ψα and Ψβ.

α,W(t)Ψβ) = eiEαtα,WΨβ) e-iEβt = ei(Eα - Eβ)tα,WΨβ)

Now what he wants us to look at is matrix elements between smooth superpositions of energy eigenstates. So, um, take ∫∫f1(Eα) f2(Eβ) dEα dEβ of both sides, where f1, f2 are arbitrary smooth functions. The point now is that as t → ±∞ the RHS is basically the high-frequency limit of a Fourier transform. But the Fourier transform of a sufficiently smooth function goes to zero at ω = ±∞, so we have to conclude that the RHS is zero in the limit, and hence so is the LHS!
 
Thanks for your reply!

I was not sure about what he meant by smooth superposition.

I think your understanding make sense.
 
Bill_K said:
Well I'm not sure, but here's a stab.. Weinberg has W(t) ≡ eiH0t W e-iH0t

Take the matrix elements of both sides, between energy eigenstates Ψα and Ψβ.

α,W(t)Ψβ) = eiEαtα,WΨβ) e-iEβt = ei(Eα - Eβ)tα,WΨβ)

Now what he wants us to look at is matrix elements between smooth superpositions of energy eigenstates. So, um, take ∫∫f1(Eα) f2(Eβ) dEα dEβ of both sides, where f1, f2 are arbitrary smooth functions. The point now is that as t → ±∞ the RHS is basically the high-frequency limit of a Fourier transform. But the Fourier transform of a sufficiently smooth function goes to zero at ω = ±∞, so we have to conclude that the RHS is zero in the limit, and hence so is the LHS!


I read your explanation again and I think your explanation proves that the matrix element of W(t) is diagonal only. It is not guaranteed vanishing for α=β
 
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