Understanding "Integrating Out": Examples and Explanations

[Lapidus]

Over and over again I read in physics text the words "integrating out". Sometimes it is clear or I have a vague feeling what is meant, sometimes not.

Could someone tell me what is meant by 'integrating out', perhaps with some easy examples.

if this is the wrong section forum, please redirect

thank you

 

[tiny-tim]

Hi Lapidus!

I think "integrating out" means exactly the same as "integrating" …

rather like "wake up" and "wake" or "meet with" and "meet".

Also, there's a subtext that if you have a function of x y and z, and you integrate wrt x, then the x disappears, and you have a function of only y and z …

so the x has been "integrated out"! ​

 

[Pengwuino]

I hear this phrase too lately and I don't know what it means either. I've been hearing it in the context of gravitational thermodynamics and I read things like "integrating out degrees of freedom" and I don't know what that means.

 

[chrispb]

The idea is that if one has a heavy particle/degree of freedom, one could do the path integral over all of those field configurations to remove it as a degree of freedom, leaving you with an effective field theory with light degrees of freedom. In practice, people don't actually do the path integral since we don't know how to, but instead, we write down all possible operators that one could imagine making with diagrams with the heavy particles and matching the coefficients of those operators onto the coefficients of diagrams involving the heavy particles.

For example, suppose one had a field theory with real scalars A and B, and one wrote down the following lagrangian: L = (dA)^2/2 + (dB)^2/2 - M^2 B^2/2 - gAB^2/2. Now, if you were working at energy scales below M, you couldn't create B particles out of collisions of As, so the low energy effective field theory would only contain A as a degree of freedom. If you did experiments, you would see something that behaved as an A^4 operator with some coefficient lambda/4!; in other words, you'd observe that two As could scatter. From the high energy point of view, what you have is a 4-A diagram with a B running in a box in the middle and various crosses thereof. If one wished to obtain lambda in terms of the high energy parameters g and M, one would evaluate the diagram in the low energy theory (-i*lambda) and in the high energy theory (evaluating all the box diagrams at one loop, or higher if you'd rather), and matching the two to extract lambda. If one wanted to properly match all behavior one could get out of the original lagrangian, you would need to include a whole tower of operators in your low energy effective theory; for example, (dA)^2 A^2.

Anyway, once one has a low energy effective description of As, you would say you've integrated out the B. Note that the low energy theory is nonrenormalizable; you can get A^6 type terms and whatnot. If you go ahead and compute the corrections, you'll see that they're suppressed by powers of M (so that the operator has dimension 4 still), and so observing higher dimension operators in a field theory could be indicative of new physics occurring at the scale that the operators are suppressed by. It is a hint that some new physics comes into make the theory renormalizable again. Hope this helps!

 

[Dale]

If you have some function of x and y then
$$\int_a^b f(x,y) \, dx = g(y)$$

You have "integrated out" the x.

 

[Lapidus]

thanks everybody

 

[khemist]

Is this like a contraction?

 

[tiny-tim]

hi khemist!

exactly like a contraction!

a contraction would be ∑_i f_ij

i and j are just labels, so we can also write it ∑_i f(i,j)

now make i x, and let there be infinitely many of them (and make j y) …

∑ becomes ∫, and ∑_i f(i,j) becomes ∫_x f(x,y),

which we usually write with a "dx", as ∫ f(x,y) dx

 

[khemist]

Is this a similar contraction when using tensors?

 

[tiny-tim]

yes

i showed only two indices, to keep it simple, but it would be similar for any number and type of tensor indices

 

[khemist]

Awesome thanks. When contracting it seems apparent we lose information about the original function. What would one call that information?

 

[tiny-tim]

merely corroborative detail, intended to give artistic verisimilitude to an otherwise bald and unconvincing narrative!

 

[Dickfore]

Integrating out in field theory jargon means the following:

Suppose we have a partition function given as a path integral (which I will discretize). This path integral is over two kinds of fields, let us call them $\phi_{i}$ and $\psi_{\alpha}$, where the index $i$ labels all the possible modes of the "phi"-field, and the index $\alpha$ labels the possible modes of the "psi"-field. Usually, these labels are the space-time coordinates, or after Fourier transforming, the 4-momentum components. Furthermore, we suppose we had performed a Wick rotation so that time is imaginary. The obtained "Euclidean" action:
$$S_{E}\left[\lbrace \phi_{i} \rbrace, \lbrace \psi_{\alpha}\rbrace\right] = \sum_{i, \alpha}{L_{E}(\lbrace \phi_{i} \rbrace, \lbrace \psi_{\alpha}\rbrace)}$$
involves a double summation over these indices and $L_{E}(\lbrace \phi_{i} \rbrace, \lbrace \psi_{\alpha}\rbrace)$ is a corresponding Lagrangian density.
The partition function is then:
$$Z = \int{\prod_{i, \alpha}{d\phi_{i} \, d\psi_{\alpha} \, e^{-S_{E}\left[\lbrace \phi_{i} \rbrace, \lbrace \psi_{\alpha}\rbrace\right]}}}$$
Now, "integrating out", as pointed out by others, simply means we perform the integral over the "psi"-fields only, i.e. we perform the following integral:
$$\int{\prod_{\alpha}{d\psi_{\alpha} \, e^{-S_{E}\left[\lbrace \phi_{i} \rbrace, \lbrace \psi_{\alpha}\rbrace\right]}}} \equiv e^{-S^{\mathrm{eff}}_{E}\left[\lbrace \phi_{i} \rbrace\right]}$$
We had emphasized the fact that the integral still depends parametrically on the "phi"-fields. The reason for the peculiar exponential form is that the partition function can still be written as:
$$Z = \int{\prod_{i}{d\phi_{i} \, e^{-S^{\mathrm{eff}}_{E}\left[\lbrace \phi_{i} \rbrace\right]}}}$$
In this regard, $S^{\mathrm{eff}}_{E}\left[\lbrace \phi_{i} \rbrace\right]$ may be treated as an "effective action" for the "phi"-fields after the "psi"-fields have been integrated out.

This procedure is far from trivial. One has to choose which fields modes are unimportant and have to be integrated out. Usually, we integrate high momentum modes and leave low momentum modes, but there are differences. Then, the "psi"-integral is also far from trivial and most be done perturbatively.