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  Vertex operator and normal ordering

+ 3 like - 0 dislike
1165 views

The two point function, or propagator for a free massless boson, ϕ in 2 dimensions is given by,

ϕ(z,ˉz)ϕ(ω,ˉω) = α2π{ln|zω2R|+ln|ˉzˉw2R|}

where R is an IR cutoff.

My question is:

How to prove that eikϕ(x) = :eikϕ(x):eαk22πln(a/2R),

where a is an UV cutoff, and :O: stands for normal ordering?

This post imported from StackExchange Physics at 2015-05-06 11:41 (UTC), posted by SE-user layman
asked Apr 29, 2015 in Theoretical Physics by layman (25 points) [ no revision ]
retagged May 6, 2015

This is a special case of the formula :eϕ(f):=eϕ(f)/eϕ(f) valid for any quasifree field ϕ and ϕ(f)=dμ(x)ϕ(x)f(x), where denotes vacuum expectation. It can be taken as the definition of normal ordering; functional differentiation then gives the usual rules.

1 Answer

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Here

ikϕ(x)ikϕ(x)=αk2πln(a/2R),

where a is an UV cutoff.

Now we can write (as all the ϕ's are located at x i.e. Time ordered {ϕn(x)}=ϕn(x) )

{ikϕ}n(x) = :{ikϕ}n(x):+all contractions= :{ikϕ}n(x):+ nC2(αk2πln(a/2R)):{ikϕ}n2(x):+nC2 n2C22(αk2πln(a/2R))2:{ikϕ}n4(x):+= :{ikϕ}n(x):+n(n1)(αk22πln(a/2R)):{ikϕ}n2(x):+n(n1)(n2)(n3)2!(αk22πln(a/2R))2:{ikϕ}n4(x):+

We expand the vertex operator,

eikϕ(x)=n=0(ikϕ)n(x)n!=n=0:{ikϕ}n(x):n!+(αk22πln(a/2R))n=2:{ikϕ}n2(x):(n2)!+12!(αk22πln(a/2R))2n=4:{ikϕ}n4(x):(n4)!+=n=0:{ikϕ}n(x):n![1+(αk22πln(a/2R))+12!(αk22πln(a/2R))2+]= :eikϕ(x):e(αk22πln(a/2R)).

Q.E.D.

This post imported from StackExchange Physics at 2015-05-06 11:41 (UTC), posted by SE-user layman
answered May 6, 2015 by layman (25 points) [ no revision ]

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