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

+ 2 like - 0 dislike
1232 views

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

$$\partial^{2}\varphi = 0$$

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.

Then
$$[\hat{a}_{\mathbf k}, \hat{b}^{\dagger}_{\mathbf p}] = \int d^{3}\mathbf x d^{3}\mathbf y e^{ip^{M}x - ik^{0}y}[i\pi^{0}_{\varphi}(x) + k^{0}_{0}\hat{\varphi}^{0}(x), -i\hat{\pi}^{M}_{\varphi}(y) + p^{M}_{0}\hat{\varphi}^{M}(y)]$$
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?

edited Nov 12, 2015

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.

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.

+ 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 (15,468 points)

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.

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOv$\varnothing$rflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.