Friday, September 15, 2017

quantum field theory - How does the normalization of an n-particle state |nmathbfkrangle work?


You can expand the free, real scalar field in the following manner ϕ(x)=d3k(2π)312ωk[eiωkx0+ikxak+e+iωkx0ikxak]

where ωk=|k|2+m2 and the operators ak and ak obey the bosonic canonical commutation relations (CCRs): [ak,ap]=[ak,ap]=0       ,  [ak,ap]=δ(kp)


From here you build a symmetric Fock space as follows: You define a vacuum state |0 such that: ak|0=0

which is assumed to be normalized 0|0=1. Then a single-particle state can be defined as |1kak|0
which are orthonormal 1k|1p=δ(kp) on account of the CCRs. From here you can construct a state containing Ns=1ns particles: |(n1)k1(n2)k2(nN)kN(ak1)n1n1!(ak2)n2n2!(akN)nNnN!|0
where there are ns particles with the momentum ks.



I am confused about the way the following states are normalized: |nk=(ak)nn!|0

What is the normalization property of mp|nk? Or also np|nk? For example, I find that 2p|2k=120|apapakak|0=120|ap[akap+δ(pk)]ak|0=120|apakapak|0+12δ(pk)0|apak|0=120|apak[akap+δ(pk)]|0+12δ(pk)1p|1k=0+12δ(pk)1p|1k+12δ(pk)1p|1k=δ(pk)2


This makes no sense, as the square of the Dirac delta function is ill-defined. What am I doing wrong? How does the normalization actually work here?


EDIT: I have been using definitions for particle states that correspond to discrete labels - it is wrong to use them for continuous state labels. You cannot have more than one particle with exactly the same momentum. As usual Weinberg (Chapter 4) says it the clearest, he calls an N-particle state; |k1k2kN=ak1ak2akN|0

Which has the normalization condition p1p2pN|k1k2kM=δNMPNj=1δ(kjpP(j))
where P is the permutation over the integers 1,,N. This is symmetric under interchange of any of the labels and 0 if NM.



Answer



There is absolutely nothing wrong, you've simply abused the notation by setting different momenta exactly equal to each other. As a simple example, consider just one-particle states, where p|qδ(pq).

If you naively set the two momenta equal to each other, you would get p|pδ(0).
Then you might complain this is not mathematically correct, because δ(0) is not a valid function or even a functional; it's not even defined at all. But the general overlap is perfectly well-defined as a functional of two variables, i.e. δ(pq). Your mistake was simply to set those variables equal, when you should have kept them explicit to integrate over later.


Similarly, for two-particle states the normalization is p1p2|q1q2δ(p1q1)δ(p2q2)±δ(p1q2)δ(p2q1)

with the sign depending on the particle statistics. The right-hand side is a perfectly well-defined functional of four variables, which you must integrate over later. However, if you naively start setting these variables equal to each other you'll get, for bosons, pp|qqδ(pq)2
which is not well-defined.


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