Homework 2 Solution

# Either the commutators with exactly 1 creation

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: r in class. Sol. I’m not going to do every one of the commutation relations, but a few will suﬃce to demonstrate the relations for the creation and annihilation operators. For convenience, I’m going to set y = 0. Consider: ˆ ˆ φ(x), φ(0) d3 pd3 q (2π )6 = [π (x), π (0)] = ˆ ˆ − d3 pd3 q (2π )6 1 [ap , aq ]eip·x + [ˆp , aq ]e−ip·x + [ˆp , aq ]eip·x + [ˆp , aq ]e−ip·x ˆˆ a† ˆ † a ˆ† a† ˆ 4Ep Eq Ep Eq [ap , aq ]eip·x + [ˆp , aq ]e−ip·x − [ˆp , aq ]eip·x − [ˆp , aq ]e−ip·x ˆˆ a† ˆ † a ˆ† a† ˆ 4 Since both of these expressions are equal to zero, we can add or subtract them on a term-by term basis, and the two will cancel. either the commutators with exactly 1 creation operator or the ones with 2 creation or two annihilation operators are identically zero. Finally, consider: ˆ φ(x), π (0) ˆ = = i 2 iδ (x) d3 pd3 q [ap , aq ]eip·x − [ˆp , aq ]e−ip·x − [ˆp , aq ]eip·x + [ˆp , aq ]e−ip·x ˆˆ a† ˆ † a ˆ† a† ˆ (2π )6 and so: ˆ d3 x φ(x), π (0) ˆ =i = = i 2 i 2 d3 q [a0 , aq ] − [ˆ† , aq ] − [ˆ0 , aq ] + [ˆ† , aq ] ˆˆ a0 ˆ † a ˆ† a0 ˆ (2π )3 d3 q a ˆ† [a† , aq ] − [ˆ0 , aq ] ˆ0 ˆ (2π )3 which immediately gives the expected commutator for [ˆp , aq ]. Note that we used the relation: a† ˆ d3 xeip·x = (2π )3 δ (p) Likewise, we’ve already established that if these terms are non-zero, then the [ˆ, a] and [ˆ† , a† ] comaˆ aˆ mutators are zero. 3. Consider the following mode (in the real quantized scalar ﬁeld): |p = ap |...
View Full Document

Ask a homework question - tutors are online