Homework 2 Solution

As there is an odd number of operators pop 0 item

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Unformatted text preview: 0 ˆ† Compute the following: ˆ ˆ ˆ† ˆ (a) O = ap ap ap (in as simple a form as you can make it) Sol. I’ll freely admit that “simple” in this case is ambiguous. It would have been better, for, instance, to say, “in terms of a single operator plus whatever you need to add to it,” but even that leaves some ambiguity. Instead, I’ll show you what I really wanted. Recall that: ap , a† = (2π )3 δ (p − q ) ˆ ˆq and thus: ˆ = (2π )3 δ (0)ˆp + a† ap ap ′ a ˆp ˆ ˆ This is especially useful because the second term operatoring on an arbitrary state will almost certainly be zero. (b) p|ap ap |p ˆ† ˆ Sol. We can use the same relation above to expand this out: (2π )6 δ (0)2 0|0 + 2(2π )3 δ (0) 0|a† ap |0 + 0|a† ap a† ap |0 ˆp ˆ ˆp ˆ ˆp ˆ The latter two terms are identically zero, since: ap |0 = 0 ˆ so p|ap ap |p = (2π )6 δ (0)2 ˆ† ˆ which is simply the volume of the space, squared ˆ (c) p|O|p Sol. As there is an odd number of operators: ˆ p|O|p = 0 item Finally, consider the quantization of the complex scalar field. Previously, we showed that there is a conserved (classical) current: j µ = i(ψ ∗ ∂ µ ψ − ψ∂ µ ψ ∗ ) In QFT, of course, this current becomes an operator. (a) Write out (simplifying wherever possible) the “charge” operator, ˆ0 . You may find the operators j from problem 2 to be a useful guide. Sol. The quantized version of the operator is: ˙ˆ ˙ˆ ˆ0 = i(ψψ † − ψ † ψ ) j You can switch the order of both terms, but they need to be consistent. Recall: ˆ ψ (x) = d3 p (2π )3 ˆ ψ † (x) = d3 p (2π )3 and µ µ 1 c† e−ipµ x + ˆp eipµ x ˆp b 2Ep 1 ˆ† e−ipµ xµ + cp eipµ xµ bp ˆ 2Ep so ˆ0 j d3 pd3 q (2π )6 = = Ep 4Ep Eq µ c† e−ipµ x − ˆp eipµ x ˆp b µ µ − ˆ† e−ipµ x − cp eipµ x bp ˆ µ µ ˆ† e−iqµ xµ + cq eiqµ xµ bq ˆ c e−iqµ x + ˆq eiqµ x ˆq b µ Four of these terms identically cancel by the commutation relations (and an arbitrary switching of p and q ): c† ˆ† − ˆ† c† = 0 ˆp bq bp ˆq and similarly with the cˆ terms. This leaves: ˆb ˆ0 j = d3 pd3 q (2π )6 Ep 4Ep Eq µ c† cq e−i(pµ +qµ )x + cp c† ei(pµ +qµ )x ...
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