- #1

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In any case, does anyone know the answer for equal times, or unequal times?

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- Thread starter geoduck
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- #1

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In any case, does anyone know the answer for equal times, or unequal times?

- #2

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\begin{equation}

\langle \bar{\psi}_i \psi_i \rangle = c

\end{equation}

where c is some constant and we sum over flavour indeces. I don't see how standard commutation relations can hold with a non-zero VEV of the product of these fields.

- #3

tom.stoer

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A well-known example is the axial anomaly where the classically conserved charge Q

- #4

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A well-known example is the axial anomaly where the classically conserved charge Q_{5}fails to commute with the Hamiltonian after quantization & regularization, i.e. [H,Q_{5}] ≠ 0

Are you saying something like [f(X),P]=i (df/dx) might not hold, where f(X) is a complicated operator product of X's, if f(X) requires regularization?

- #5

tom.stoer

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In QFT you don't have X and P as operators but field operators like [itex]\phi(x)[/itex], [itex]\pi(y)[/itex] and commutators like

[tex][\phi(x),\pi(y)] = i\delta(x-y)[/tex]

Let's make an example: the axial charge is constructed from fermionic field operators as

[tex]Q_5 = \int_{\mathbf{R}^3}d^3r\,\bar{\psi}(x)\,\gamma^0\, \gamma_5 \,\psi(x)[/tex]

Applying the naive commutators before regularization one finds

[tex][H,Q_5] = 0[/tex]

which would correspond to a conserved axial current

[tex]\partial_\mu j^\mu_5 = 0[/tex]

But due to gauge-invariant regularization preserving

[tex]\partial_\mu j^\mu = 0[/tex]

one is forced to introduce a regulator violating the axial symmetry; therefore one finds

[tex][H,Q_5] = \mathcal{A}[/tex]

and therefore

[tex]\partial_\mu j^\mu_5 = \mathcal{a}[/tex]

where A represents the anomaly (which is a term depending on the spatial dimension and topology).

The commutators of the fermionic Fourier components for [itex]\psi(x)[/itex] remain valid, but the commutators of the Fourier coefficients of [itex]j^\mu_5(x)[/itex] are modified. So what happens is that due to quantization and regularization the structure of a composite operator may change. The conclusion is that at least for composite operators the naive replacement of Poisson brackets with commutators and the naive application of Fourier transformation of field operators need not always be valid.

- #6

DrDu

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Don't the commutators e.g. in QED vanish during renormalization due to Z becoming infinite?

- #7

tom.stoer

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Which commutators? For the axial charge?

Starting with the anomaly equation

[tex]\partial_\mu j^\mu_5 = \mathcal{a}[/tex]

one finds

[tex]i[H,Q_5] = \partial_0 Q_5 = \int_{\mathcal{M}^D} d^{D-1}x\,a_D(x)[/tex]

for D-dimensional space-time and spatial manifold D (assuming reasonable boundary conditions or compact M). This expression does not vanish in general

Starting with the anomaly equation

[tex]\partial_\mu j^\mu_5 = \mathcal{a}[/tex]

one finds

[tex]i[H,Q_5] = \partial_0 Q_5 = \int_{\mathcal{M}^D} d^{D-1}x\,a_D(x)[/tex]

for D-dimensional space-time and spatial manifold D (assuming reasonable boundary conditions or compact M). This expression does not vanish in general

Last edited:

- #8

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Which commutators? For the axial charge?

I think what is meant is that if the canonical commutation relations hold for the bare fields, then what about for the renormalized fields? The commutator should be 1/Z less, where Z is the field strength normalization, which goes to infinity.

This is worrisome because if you form bilinears out of the renormalized fields, then their commutators vanish, which ought to mean that there can be no microcausality???

(assuming reasonable boundary conditions or compact M).

I am always confused by this. If you have:

∫∫∫(div f)d

then by Stoke's theorem this is the flux of f over the surface, and if you mandate that fields vanish at the surface, then this term is zero.

But isn't it true for a scalar function g=g(x,y,z):

[tex]\int\int\int \partial_y g(x,y,z) dxdydz [/tex]

that the above expression also vanishes? So requiring the integrand to be a 3-divergence is a little too strong: if it's merely a partial derivative then wouldn't it vanish?

So not only does ∫∫∫(div f)d

[tex]\int\int\int \partial_x f_1(x,y,z) dxdydz=\int\int\int \partial_y f_2(x,y,z) dxdydz=\int\int\int \partial_z f_3(x,y,z) dxdydz=0 [/tex]

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