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

+ 2 like - 0 dislike

Let's have real massless scalar field theory:
L = \frac{1}{2}(\partial_{\mu}\varphi )^{2}

Corresponding EOM reads
\partial^{2}\varphi = 0

Let quantize it: for $\pi_{\varphi} = \partial_{0}\varphi$
[\hat{\varphi}(\mathbf x), \hat{\pi}_{\varphi}(\mathbf y)] = i\delta (\mathbf x - \mathbf y), \quad [\hat{\varphi}(\mathbf x), \hat{\varphi}(\mathbf y)] = [\hat{\pi}_{\varphi}(\mathbf x), \hat{\varphi}(\mathbf y)] = 0,
or in terms of creation-destruction operators,
[\hat{a}_{\mathbf p}, \hat{a}_{\mathbf k}^{\dagger}] = \delta (\mathbf p - \mathbf k), \quad [\hat{a}_{\mathbf p}, \hat{a}_{\mathbf k}] = [\hat{a}_{\mathbf p}^{\dagger}, \hat{a}_{\mathbf k}^{\dagger}] = 0

Suppose we turn on mass. Then we have EOM
(\partial^{2} +m^{2})\varphi = 0

with relations
[\hat{b}_{\mathbf p}, \hat{b}^{\dagger}_{\mathbf k}] = \delta (\mathbf p - \mathbf k), \quad [\hat{b}^{\dagger}_{\mathbf p}, \hat{b}^{\dagger}_{\mathbf k}] = [\hat{b}_{\mathbf p}, \hat{b}_{\mathbf k}] = 0
The question: how to derive (at least for zero mode) commutation relations between $a_{\mathbf p}, b_{\mathbf k}$,
[\hat{a}_{\mathbf p}, \hat{b}_{\mathbf k}^{\dagger}] = ?

My attemption

My idea is that to rewrite creation operator in terms of $\hat{\varphi}, \hat{\pi}_{\varphi}$:
\hat{a}_{\mathbf p} = \int d^{3}\mathbf x (i\hat{\pi}^{0}_{\varphi}(x) + p^{0}_{0}\hat{\varphi}^{0}(x))e^{-ip^{0}x},

\hat{b}_{\mathbf p} = \int d^{3}\mathbf x (i\hat{\pi}^{M}_{\varphi}(x) + p_{0}^{M}\hat{\varphi}^{M}(x))e^{-ip^{M}x},

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

[\hat{a}_{\mathbf k}, \hat{b}^{\dagger}_{\mathbf p}] = \int d^{3}\mathbf x d^{3}\mathbf y e^{ip^{M}x - ik^{0}y}[i\pi^{0}_{\varphi}(x) + k^{0}_{0}\hat{\varphi}^{0}(x), -i\hat{\pi}^{M}_{\varphi}(y) + p^{M}_{0}\hat{\varphi}^{M}(y)]
So the task is "reduced" to definition of canonical commutation relations
[\hat{\varphi}^{0}(\mathbf x), \hat{\pi}^{M}_{\varphi}(\mathbf y)], ...
Is it true that
[\hat{\varphi}^{0}(\mathbf x), \hat{\pi}^{M}_{\varphi}(\mathbf y)] = ie^{-ip^{M}_{0}t + ip^{0}_{0}t}\delta (\mathbf x - \mathbf y)
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 $b_k$ in terms of the $a_k$ 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
\hat{H}^{M} \hat{a}_{0}^{\dagger}|0 \rangle \equiv \int d^{3}\mathbf p E^{M}_{\mathbf p}\hat{b}^{\dagger}_{\mathbf p}\hat{b}_{\mathbf p}\hat{a}_{0}^{\dagger}|0 \rangle = m \hat{b}^{\dagger}_{0}| 0\rangle
For that I need to show that
[\hat{b}_{\mathbf p}, \hat{a}^{\dagger}_{0}] = \delta (\mathbf 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 $b_k$ in terms of the $a_k$ 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 $\phi$ and some transform $\Phi=U\phi$ ; then find out which transformation transforms the correlations for $\phi=\phi_0$ into those for $\Phi=\phi_m$.

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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