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  How to derive canonical commutation relations between operators of massive field theory and corresponding massless theory?

+ 2 like - 0 dislike
3219 views

Let's have real massless scalar field theory:
L=12(μφ)2

Corresponding EOM reads
2φ=0

Let quantize it: for πφ=0φ
[ˆφ(x),ˆπφ(y)]=iδ(xy),[ˆφ(x),ˆφ(y)]=[ˆπφ(x),ˆφ(y)]=0,


or in terms of creation-destruction operators,
[ˆap,ˆak]=δ(pk),[ˆap,ˆak]=[ˆap,ˆak]=0

Suppose we turn on mass. Then we have EOM
(2+m2)φ=0

with relations
[ˆbp,ˆbk]=δ(pk),[ˆbp,ˆbk]=[ˆbp,ˆbk]=0


The question: how to derive (at least for zero mode) commutation relations between ap,bk,
[ˆap,ˆbk]=?

My attemption

My idea is that to rewrite creation operator in terms of ˆφ,ˆπφ:
ˆap=d3x(iˆπ0φ(x)+p00ˆφ0(x))eip0x,

ˆbp=d3x(iˆπMφ(x)+pM0ˆφM(x))eipMx,

where superscripts M,0 denote massive and massless theory correspondingly.


Then
[ˆak,ˆbp]=d3xd3yeipMxik0y[iπ0φ(x)+k00ˆφ0(x),iˆπMφ(y)+pM0ˆφM(y)]


So the task is "reduced" to definition of canonical commutation relations
[ˆφ0(x),ˆπMφ(y)],...

Is it true that
[ˆφ0(x),ˆπMφ(y)]=ieipM0t+ip00tδ(xy)

If yes, how to argue this statement?

asked Nov 12, 2015 in Theoretical Physics by NAME_XXX (1,060 points) [ revision history ]
edited Nov 12, 2015 by NAME_XXX

The representations of the creation/annihilation operators in the two Hilbert spaces are unitarily inequivalent, and hence not compatible; i.e., there is no simple way to define the bk in terms of the ak or conversely. Thus commutation relations between them do not make sense.

@ArnoldNeumaier :

it is related to my previous question, http://www.physicsoverflow.org/33896/generation-axion-particle-states-coherent-oscillations-field , and particularly to your comment "...VEVs are observable as particle masses. They are a property of the elementary excitations of the physical vacuum state...".
I need to prove that
ˆHMˆa0|0d3pEMpˆbpˆbpˆa0|0=mˆb0|0


For that I need to show that
[ˆbp,ˆa0]=δ(p),

or something like that.

@Arnold Neumaier :

I.e., I need to inentify zero mode states in initial massless theory with zero mode (but nonzero energy) states in corresponding massless theory. I want to do it in a language of commutation relations.

1 Answer

+ 2 like - 0 dislike

The representations of the creation/annihilation operators in the two Hilbert spaces are unitarily inequivalent, and hence not compatible; i.e., there is no simple way to define the bk in terms of the ak or conversely. Thus commutation relations between them do not make sense.


If you want to have a common Hilbert space for the massless and the massive case, you need to work in an approximation with a short distance (large momentum) cutoff, taken to infinity at the end. At finite cutoff, the two c/a operators are related by a unitary Bogoliubov transformation. (The latter diverge when the cutoff is removed, hence the exact theories have no common Hilbert space.)

answered Nov 12, 2015 by Arnold Neumaier (15,787 points) [ no revision ]

Thank you! But could you write the sketch of showing the relation of operators in massive and massless theories trough Bogoliubov transformation by providing finite cutoff, if you please? Unfortunately, I don't understand this principle.

I don't have the time for writing the full calculations, but here is the recipe:

A Bogoliubov transformation is a linear transformation U of the field, and transforms the correlations of the free field accordingly. You need to work out the details for the second-order correlations of a free field ϕ and some transform Φ=Uϕ ; then find out which transformation transforms the correlations for ϕ=ϕ0 into those for Φ=ϕm.

To keep things transparent, work in the momentum representation, and introduce the c/a operators for the two fields only after you know what U is.

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