# Lie derivatives of Quantum Fields

1. Aug 13, 2011

### TomCurious

Hello, this question will essentially concern quantum field theory in curved spacetime, and it has two parts to it.

I have recently acquired DeWitt's treatment of the formalism, which immediately discusses the role of killing vectors in the theory. Specifically, given a killing vector field K^a (forgive me, I am still learning LaTex), we may form a 'generalized momentum' given by:

P = integral (Tab K^a dΣ^b) [1]

Where Tab is the stress energy tensor, and dΣ^b is the volume form corresponding to a cauchy hypersurface.

The book continues, explaining that this P has the following effect on the fields:

[Φ, P] = L_p (Φ) [2]

Where L_p denotes the Lie derivative with respect to P.

Now, the questions are as follows:

1. I can see a possible way for [2] to be true, following from the definition of a Lie derivative in General Relativity (i.e. given vector fields X and Y, [X, Y] = L_x(Y) ), but the Φ is not a vector field (in the GR sense), but a 'scalar' quantum operator on the Fock space. Moreover, P is scalar (perhaps it is also an operator?), and thus liable to the same concern.

2. Making the substitution [1] for P in [2], I cannot seem to extrude the right hand side of [2]. My main issue in this is that I want to discern precisely how important the killing vector field is. Or, is it possible to construct arbitrarily a scalar operator (e.g. Q), and use this on another scalar (as [Φ, Q] = L_q(Φ) )? Could I use it on a vector operator. etc? And, more importantly, how exactly do I go about calculating the Lie derivative of these quantum fields? There is a lovely formula (abstract, and index based) in General Relativity - what is it in QFT?

Thank you in advance.

2. Aug 13, 2011

### Finbar

What book are you reading?

I think that these operators P and $\phi$ must be vectors in some space. But obviously not space-time. Possibly they are vectors in Hilbert space??? In that case the Lie derivative is well defined.

3. Aug 13, 2011

### TomCurious

I am reading http://www.sciencedirect.com/science/article/pii/0370157375900514 (as I downloaded it via ebookee, I was inclined to think it was a book - my apologies). I have seen this elsewhere in the literature, however.

Undoubtedly, Φ is an operator in a Hilbert Space (the Fock space, to be precise) - so perhaps we are considering the Lie Derivative there. Nonetheless, the fact is that the original killing field, which is used to find P, is a spacetime vector.

Of course, as Tab, semi-classically, depends on the fields Φ's, then I can see that P does as well. Thus, we can take the commutator, expanding using the properties of the commutator and the canonical commutation relations from QFT, between P and Φ.

Nonetheless, I am uncertain as to how this would constitute a Lie Derivative (in fact, I had not considered Lie Derivatives in Hilbert Spaces). Explanation here would be appreciated.

4. Aug 13, 2011

### samalkhaiat

If the configuration variables are transformed according to
$$x^{a}\rightarrow x^{a} + g_{A}k^{aA}, \ \ (1)$$
with A is a multi-index taking values in the set $\{\emptyset , \{a\},\{ab\}, …\}$, local field operators will be subject to a unitary transformations
$$\bar{\Phi}(x) = U^{-1}(g) \Phi (x) U(g), \ \ (2)$$
If eq(1) leaves the action integral unchanged, Noether theorem provides us with time-independent “charge” of the form
$$Q^{A} = \int d \sigma^{a} T_{ab}k^{bA}$$
Using the fundamental Poisson brackets or the equal time commutation relations, one can prove the following properties for Q
1) $Q^{A}$ is covariant with respect to the index A.
2) $Q^{A}$ satisfies the Lie algebra of the group in question.
3) $Q^{A}$ generates the correct unitary transformation on local fields: If we write
$$U(g) = \exp (ig_{A}Q^{A}),$$
the infinitesimal form of eq(2) becomes
$$\bar{\Phi}(x) - \Phi (x) = g_{A}[iQ^{A}, \Phi (x)]$$
The left hand side of this equation is the change in the functional form of $\Phi (x)$ evaluated at two points $\bar{P}$ and $P$ having the same coordinate value $x^{a}$. This, essentially, is the Lie derivative
$$g_{A}\mathcal{L}^{A}\Phi (x) \equiv \delta \Phi (x) = g_{A}[iQ^{A}, \Phi (x)]$$

5. Aug 13, 2011

### TomCurious

Thank you for that gracious exposition, samalkhaiat, you were wonderfully clear.

However, I do have three questions:

1. Given two killing vectors (e.g. X and Y), their commutator is also a Killing Vector. Accordingly, the killing vectors with Lie bracket form a Lie subalgebra of vector fields on the manifold M. If M is complete, this corresponds to the Lie algebra of the isometry group on M. With this in mind, I was wondering if there was any relation between this isometry group and the symmetry group corresponding to transformations of the fields.

2. I am able to come up with a rather 'hand-waving' argument in which the form of the time-independant charge is arrived at (where Tab is calculated from functional derivatives of the Lagrangian), but as a reference I would appreciate a more complete derivation. Could you direct me to (or supply me with) one?

3. This is a topic which is quite fascinating to me, and I would like to know more about it. As per your statement regarding the proofs for the mentioned properties, I can again conceive of some rather 'hand-waving' arguments that they are correct, though higher levels of rigor escape me. Could you direct to some text/lecture notes that would elaborate on this subject?

6. Aug 14, 2011

### samalkhaiat

It is the same group. The local fields form a representation of the (space-time) symmetry group of M, i.e., they transform by finite dimensional matrix representation of symmetry group. That is to say that eq(2) should be understood as
$$U^{-1}(g)\Phi_{s}(x)U(g) = D_{s}{}^{r}\Phi_{r}(x^{a} – g_{A}k^{Aa})$$
If $k^{Aa}(x)=0$, the symmetry is called internal and the corresponding time-independent (and Lorentz scalar) charge is not related to the energy-momentum tensor.
Also, the symmetry group of (compactified version of) M need not be an isometry. Conformal symmetry, where the (conformal) killing fields generate non-linear coordinate transformation, is an example of space-time symmetry.

I think I covered this in
www.physicsforums.com/showthread.php?t=172461

I am not aware of any text/report which prove all the above three theorems, certainly not for a generic field theory. But, it can be done. It is lengthy though, perhaps this is why text books don’t do it. May be one day I do it for you.

regards

sam

Last edited: Aug 14, 2011
7. Aug 15, 2011

### femtofranco

Hi, I'm also interested in QFT in Curved Spacetime, but am very new at it. Reading this, I am further behind than I thought. I tried to read that post you link to, sam, but I am not very comfortable with it.

Looking at your first post here, you wrote:

$x^{a}\rightarrow x^{a} + g_{A}k^{aA}, \ \ (1)$

What precisely does $$g_{A}$$ denote? Is it a constant, or a matrix? Because you then form U(g) to operate on the field?

Also, let's say I have a vector field given by a general formula, e.g. say K is a conformal killing vector. How exactly do I perform the transformation denoted by (1)? Since K is a vector field it is a derivative, so how do I add it to x?

8. Aug 15, 2011

### samalkhaiat

Hi,
$g_{A}$ is a constant representing the group parameter(s). $k^{Ac}$ is a function of x. The transformation is generated by the vector field
$$K(x) = g_{A}k^{Ac}(x)\partial_{c}$$
You could also regard $g_{A}k^{Ac}$ itself as a vector field operating on x. Consider for example Lorentz transformations; here the index A is a double space-time index (ab), the parameter is $g_{A}\equiv \omega_{ab}$, so you have
$$g_{A}k^{Ac}= \frac{1}{2}\omega_{ab}(\eta^{bc}x^{a} - \eta^{ac}x^{b}) \ \ (3)$$
Using the fact $\eta^{ac} = \partial^{a}x^{c}$, we can introduce the orbital angular momentum operator into Lorentz transformations as follows
$$g_{A}k^{Ac} = - \frac{i}{2}\omega_{ab}L^{ab}x^{c},$$
where
$$L^{ab} = i (x^{a}\partial^{b} – x^{b}\partial^{a}).$$
We can also give a (spin-1) matrix representation of the Lorentz transformations (3) as follows
$$g_{A}k^{Ac} = - \frac{i}{2}\omega_{ab}(M^{ab})^{c}{}_{d}x^{d},$$
where M is the spin-1 matrix ( the cd element of M)
$$(M^{ab})^{c}{}_{d} = i( \eta^{bc}\delta^{a}_{d} - \eta^{ac} \delta^{b}_{d})$$

I hope that was helpful.

Regards

sam

9. Aug 15, 2011

### femtofranco

This is quite nifty, but let me make sure I understand this.

To find the transformed x, you simply add the components of the vector K(x) to x.

As such, the final form from your example would be:

$x^{c}\rightarrow (1 - \frac{i}{2}\omega_{ab}L^{ab})x^{c}$

We would then insert this into the variation of the action with respect to x, and adapt the fields accordingly, to see if the variation is zero (and, thus find a conserved quantity).

Now, suppose we are working with a Killing field, given by

$K = a \partial_{b}$

(Since the Killing equation can only specify K up to a constant factor)
As such, the new quantity is given by:

$x^{a}\rightarrow x^{a} + a$

This, as expected, looks just like a translation, and so the symmetry group is the translation subgroup of the Poincare group.

Am I mistaken anywhere?

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