I Complex scalar field commutation relations

1. Aug 9, 2017

TeethWhitener

I'm trying to derive the commutation relations of the raising and lowering operators for a complex scalar field and I had a question. Let's start with the commutation relations:
$$[\varphi(\mathbf{x},t),\varphi(\mathbf{x}',t)]=0$$
$$[\Pi(\mathbf{x},t),\Pi(\mathbf{x}',t)]=0$$
$$[\varphi(\mathbf{x},t),\Pi(\mathbf{x}',t)]=i\delta(\mathbf{x}-\mathbf{x}')$$
as well as the expression for the lowering operator:
$$a(\mathbf{k})=\int {d^3 x e^{-ikx}(i\Pi^{\dagger}(\mathbf{x},t)+\omega\varphi(\mathbf{x}',t))}$$
Now we calculate the commutator:
$$[a(\mathbf{k}),a(\mathbf{k}')]= \int{ d^3 xd^3x' e^{-i(kx+k'x')}\left(-[\Pi^{\dagger}(\mathbf{x},t),\Pi^{\dagger}(\mathbf{x}',t)]+\omega^2[\varphi(\mathbf{x},t),\varphi(\mathbf{x}',t)]+i\omega\left([\varphi(\mathbf{x},t),\Pi^{\dagger}(\mathbf{x}',t)] -[\varphi(\mathbf{x}',t),\Pi^{\dagger}(\mathbf{x},t)] \right) \right) }$$
The first two commutators clearly disappear. My question is about the other portion:
$$[\varphi(\mathbf{x},t),\Pi^{\dagger}(\mathbf{x}',t)] -[\varphi(\mathbf{x}',t),\Pi^{\dagger}(\mathbf{x},t)]$$
So I'm aware that this term also disappears, but I'm a little confused as to how. I understand that $\varphi$ and $\varphi^{\dagger}$ are supposed to be considered "independent" fields, but does that mean that $[\varphi,\Pi^{\dagger}]=0$? Is this derivable? What makes fields independent of one another?
The other option is that $[\varphi,\Pi^{\dagger}]\propto\delta(\mathbf{x}-\mathbf{x}')$ and the offending term disappears because the delta functions cancel.

2. Aug 9, 2017

Staff: Mentor

Maybe I shouldn't answer as I don't know enough of the background, but from the point of view of bilinearity of the commutator:

Am I right that it should be $a(\mathbf{k})=\int {d^3 x e^{-ikx}(i\Pi^{\dagger}(\mathbf{x},t)+\omega\varphi(\mathbf{x},t))}$ and $a(\mathbf{k}')=\int {d^3 x e^{-ikx}(i\Pi^{\dagger}(\mathbf{x}',t)+\omega\varphi(\mathbf{x}',t))}$, such that the mixed terms are (without the factors $i$ and $\omega$)
$$[\Pi^{\dagger}(\mathbf{x},t),\varphi(\mathbf{x}',t)] + [\varphi(\mathbf{x},t),\Pi^{\dagger}(\mathbf{x}',t)] \\ = - [\varphi(\mathbf{x}',t),\Pi^{\dagger}(\mathbf{x},t)] + [\varphi(\mathbf{x},t),\Pi^{\dagger}(\mathbf{x}',t)] \\ = - i\delta(\mathbf{x}'-\mathbf{x}) + i\delta(\mathbf{x}-\mathbf{x}')$$

Just a thought.

3. Aug 10, 2017

TeethWhitener

Yes this was my thinking when I mentioned the "other option" in the OP. But I've also read that $\varphi$ and $\varphi^{\dagger}$ should be regarded as independent fields, without a definition or an indication of what that entails, so I was left wondering if maybe independent fields commute.

4. Aug 10, 2017

king vitamin

Yes, the field and its complex conjugate can be treated as independent, so those commutators just vanish on their own. There's a nice discussion of this in Coleman's lecture notes, see page 56 here: https://arxiv.org/abs/1110.5013

5. Aug 10, 2017

TeethWhitener

This is a wonderful resource, thank you! So when we say two fields are independent, we mean that the variation of the action disappears independently for each of the fields?

6. Aug 10, 2017

vanhees71

Yes! Alternatively you can of course also rewrite everything in two real fields by setting
$$\phi=\frac{1}{\sqrt{2}} (\phi_1+\mathrm{i} \phi_2),$$
where the factor $1/\sqrt{2}$ is just for convenience.

With the complex fields the complete canonical quantization goes as follows. You start from the Lagrangian (for simplicity I consider free fields only):
$$\mathcal{L}=\partial_{\mu} \phi^* \partial^{\mu} \phi-m^2 \phi^* \phi.$$
The canonical field momenta are
$$\Pi=\frac{\partial \mathcal{L}}{\partial \dot{\phi}}=\partial_t \phi^*, \quad \Pi^*=\frac{\partial \mathcal{L}}{\partial \dot{\phi}^*}=\partial_t \phi.$$
When quantizing you make these to operators and assume the canonical bosonic equal-time commutators, where you treat $\hat{\phi}$ and $\hat{\phi}^{\dagger}$ as independent field-degrees of freedom. E.g., you have
$$[\hat{\phi}(t,\vec{x}),\hat{\Pi}(t,\vec{y})]=\mathrm{i} \delta^{(3)}(\vec{x}-\vec{y}), \quad [\hat{\phi}^{\dagger}(t,\vec{x}),\hat{\Pi}(t,\vec{y})]=0.$$

7. Aug 10, 2017

TeethWhitener

Ok thanks. One more question. Let's say we have a Lagrangian density $\mathcal{L}(\varphi,\psi)$ that is a function of two independent operator fields $\varphi$ and $\psi$ such that:
$$\delta S = 0 = \int{ d^4 x \left[\left(\frac{\partial\mathcal{L}}{\partial\varphi}\right)\delta\varphi +\left(\frac{\partial\mathcal{L}}{\partial\psi}\right)\delta\psi\right]}$$
I understand that $\partial_{\varphi}\mathcal{L} = \partial_{\psi}\mathcal{L} = 0$, but why does this imply that $\varphi$ and $\psi$ commute?

8. Aug 10, 2017

vanhees71

If your $\mathcal{L}$ doesn't contain time derivatives, there are no canonical momenta (or better they are identically 0). Then the fields commute at equal times by definition (canonical quantization).

9. Aug 10, 2017

TeethWhitener

Hmm, I used a bad example. My point was, let's say you have a Lagrangian density:
$$\mathcal{L}=\partial_{\mu} \varphi \partial^{\mu} \psi-m^2 \varphi \psi$$
$$\Pi_{\varphi}=\frac{\partial \mathcal{L}}{\partial \dot{\varphi}}=\partial_t \psi, \quad \Pi_{\psi}=\frac{\partial \mathcal{L}}{\partial \dot{\psi}}=\partial_t \varphi$$
such that $\varphi$ and $\psi$ are independent, i.e.,
$$\delta S = 0 = \int{ d^4 x \left[\left(\frac{\partial\mathcal{L}}{\partial\varphi}\right)\delta\varphi+ \left(\frac{\partial\mathcal{L}}{\partial\dot{\varphi}}\right)\delta\dot{\varphi} +\left(\frac{\partial\mathcal{L}}{\partial\psi}\right)\delta\psi+\left(\frac{\partial\mathcal{L}}{\partial\dot{\psi}}\right)\delta\dot{\psi}\right]}$$
where all the partial derivatives vanish. Can you derive from this that, e.g., $[\varphi,\Pi_{\psi}]=0$?

Edit: Clearly the partial derivatives won't vanish in this case, but let's say instead that each of the independent fields $\varphi$ and $\psi$ satisfies its own Euler-Lagrange equations.

10. Aug 10, 2017

TeethWhitener

I think the clearest way to phrase my question is thus: If two fields are independent, does that mean that their action variations vanish independently, or does that mean that they commute, or both? If both, can we derive one condition from the other, or do we simply impose both conditions on the fields?

11. Aug 10, 2017

king vitamin

The problem with answering the question as asked is that you are doing classical field theory manipulations, and then subsequently asking about "deriving" a quantum mechanical identity.

Perhaps it is easier to say that you can derive the following classical identity:

$$\left\{\varphi,\Pi_{\psi}\right\} = \frac{\partial \varphi}{\partial \varphi} \frac{\partial \Pi_{\psi}}{\partial \Pi_{\varphi}} - \frac{\partial \varphi}{\partial \psi} \frac{\partial \Pi_{\psi}}{\partial \Pi_{\psi}} = 0$$

directly from the classical field theory specified. Then you do the procedure called "canonical quantization," aka

$$\left\{ \cdot , \cdot \right\} \rightarrow \frac{1}{i \hbar} \left[ \cdot , \cdot \right]$$

at which point you have your desired result. This should come with all the requisite precautions about canonical quantization.

12. Aug 11, 2017

Staff: Mentor

Hmmmm. Here you have an issue. It may seem that say we have a spin half fields and a spin 1 photon field you can of course simply add the Lagrangian's and you have the combined system. Vary it and yes its the same as varying each separately. But you run into an issue related to gauge symmetry. You see both have global gauge symmetry - which is perfectly OK. But global - what about locality? SR more or less screams at you it should be local. So we check that - it isn't. But how to make it so jumps straight out at you - you have to add a certain interaction term - voila - you get EM.

That is the issue - combining more or less implies they must interact in many case due to symmetry considerations. It's in fact fundamental to the standard model and one of the most beautiful things man has discovered.

Its all explained in the following book, which I will not hide from you is not held in high esteem by some, maybe even most, on this form, but I really like:
http://physicsfromsymmetry.com/

BTW its not because its wrong - it isn't - it because it was written by a a then masters student and many felt, not without some justification, such important material really needed a professor and/or researcher with much more experience. This has led to typo's, minor errors etc textbooks should not really have. Personally I don't mind such things - but others are less forgiving so to speak.

Thanks
Bill

13. Aug 11, 2017

TeethWhitener

This helps. Clearly I get that quantizing one field and its conjugate momentum comes from the Poisson bracket. I don't know why it didn't occur to me that the commutation relations of two independent fields would also be inherited from the Poisson bracket. Chalk it up to lack of sleep.

One nitpick: Isn't the Poisson bracket in this case:
$$\{\varphi,\Pi_{\psi}\} = \frac{\partial\varphi}{\partial\varphi}\frac{\partial\Pi_{\psi}}{\partial\Pi_{\varphi}} - \frac{\partial\varphi}{\partial\Pi_{\varphi}}\frac{\partial\Pi_{\psi}}{\partial\varphi}+\frac{\partial\varphi}{\partial\psi}\frac{\partial\Pi_{\psi}}{\partial\Pi_{\psi}}-\frac{\partial\varphi}{\partial\Pi_{\psi}}\frac{\partial\Pi_{\psi}}{\partial\psi}$$

I was wondering about this: when I think about commuting operators, I tend to think about symmetries: If we have two commuting operators $AB = BA$ which we apply to, e.g., an eigenstate of $B$, we get $AB|b\rangle = Ab|b\rangle = bA|b\rangle = B(A|b\rangle)$ so that the operator $B$ is invariant under the transformation $A$, i.e., $A$ is a symmetry of $B$ (and vice versa, since they share an eigenspectrum). But I don't know how useful or accurate it is to think about independent fields as being "symmetries" of one another.

14. Aug 12, 2017

samalkhaiat

Nitpick properly, in field theory the Poisson manifold is infinite-dimensional product manifold $\mathcal{P} = \mathcal{B} \times \mathcal{B}^{*}$, with $\mathcal{B} = C^{\infty}(\mathbb{R}^{n})$ being a Banach space, and $\mathcal{B}^{*} = \mbox{den}(\mathbb{R}^{n})$, the space of densities defined via the pairing $\langle \vec{\varphi} , \vec{\pi} \rangle = \int d^{n}x \varphi^{a}(x) \pi_{a}(x)$. The Poisson bracket of two $C^{\infty}$ functions is then given by $$\{ G , H \} = \int d^{n}x \left( \frac{\delta G}{\delta \varphi^{a}} \frac{\delta H}{\delta \pi_{a}} - \frac{\delta G}{\delta \pi_{a}} \frac{\delta H}{\delta \varphi^{a}} \right) .$$

As for your original question, recall the definition of the conjugate momentum $$\pi_{a} = \frac{\partial}{\partial \dot{\varphi}^{a}} \mathcal{L}(\varphi , \nabla \varphi , \dot{\varphi}) .$$ Assuming that these set of equations can be solved for $\dot{\varphi}^{a}$. That is $$\dot{\varphi}^{a} = \dot{\varphi}^{a}(\varphi , \nabla \varphi , \pi) .$$ Now we can calculate the following Poisson brackets

$$\{ \dot{\varphi}^{a}(x) , \varphi^{b}(y) \}_{x^{0} = y^{0}} = - \frac{\delta \dot{\varphi}^{a}(x)}{\delta \pi_{b}(y)} = - \frac{\partial \dot{\varphi}^{a}(x)}{\partial \pi_{b}(x)} \delta^{3}(x-y) , \ \ \ (1)$$ $$\{ \dot{\varphi}^{a}(x) , \pi_{b}(y) \}_{x^{0} = y^{0}} =\frac{\delta \dot{\varphi}^{a}(x)}{\delta \varphi^{b}(y)} = \left( \frac{\partial \dot{\varphi}^{a}}{\partial \varphi^{b}} + \frac{\partial \dot{\varphi}^{a}}{\partial(\partial_{j}\varphi^{b})} \partial_{j} \right)\delta^{3}(x-y) .$$

In your case of complex scalar field, you have $$\dot{\phi} = \Pi_{\phi^{*}} , \ \ \Rightarrow \ \frac{\partial \dot{\phi}}{\partial \Pi_{\phi}} = 0 .$$ So in this case, Eq(1) gives you $$\{\dot{\phi}(x) , \phi (y) \}_{x^{0} = y^{0}} = - \frac{\partial \dot{\phi}}{\partial \Pi_{\phi}} \delta^{3}(x-y) = 0 .$$

15. Aug 13, 2017

TeethWhitener

Hi @samalkhaiat , thanks for your response. I enjoy your posts but they usually take me some time to digest. In this case, your post is quite a bit beyond my understanding, but I'd still like to try to understand as much as I can of it, so I have a question. (I migh

For the equation:
$$\{ G , H \} = \int d^{n}x \left( \frac{\delta G}{\delta \varphi^{a}} \frac{\delta H}{\delta \pi_{a}} - \frac{\delta G}{\delta \pi_{a}} \frac{\delta H}{\delta \varphi^{a}} \right)$$
I would naively assume that it's the continuous limit of the Poisson bracket I'm familiar with:
$$\{A,B\} =\sum_{i=1}^n \left(\frac{\partial A}{\partial q_i}\frac{\partial B}{\partial p_i} -\frac{\partial A}{\partial p_i}\frac{\partial B}{\partial q_i}\right)$$
With $a$ in the integral being an index analogous to $i$ in the sum. But this doesn't seem right to me: I'm under the impression that both sets of indices run from 1 to n, n being the dimension of the manifold. So what exactly is the $d^nx$ that we integrate over? Is it the space of smooth functions on the manifold?

16. Aug 16, 2017

samalkhaiat

1) In the continuous limit, you can think of the index $i$ on the coordinates $(q_{i},p_{i})$ as a multi-index $(a , \vec{x})$ taking values in the set $\{1,2, \cdots , d \} \times \mathbb{R}^{3}$. So, you can write $$(q_{i},p_{i})(t) = (q_{(a , \vec{x})} , p_{(a , \vec{x})}) (t) \equiv (Q^{a} , P_{a})(\vec{x},t) .$$ So, for every $a \in \{1 , 2 , \cdots , d \}$, the pair $(Q^{a},P_{a})$ are functions of the space-time point $(\vec{x},t) \in \mathbb{R}^{3} \times \mathbb{R} = \mathbb{R}^{(3,1)}$, i.e., you have fields. And, of course, the sum $\sum_{i = 1}^{\infty}$ get replaced by $\sum_{a = 1}^{d}\int_{\mathbb{R}^{3}} d^{3}\vec{x} .$

2) Have you heard the saying “Classical Mechanics is a $(0+1)$-dimensional field theory”? Indeed, one can think of the pairs $(q_{i},p_{i})(t)$ as functions of the “space-time” point $t \in \mathbb{R}^{(0,1)}$, i.e., you have a $(0 +1)$-dimensional field theory. Now, in the usual $(3 +1)$-dimensional field theory, every observable $A(\phi , \pi)$ defines the following (Hamiltonian) vector field $$X_{A} = \sum_{a = 1}^{d} \int_{\mathbb{R}^{3}} d^{3}x \left( \frac{\delta A}{\delta \phi^{a} (\vec{x},t)}\frac{\delta}{\delta \pi_{a}(\vec{x},t)} - \frac{\delta A}{\delta \pi_{a}(\vec{x},t)}\frac{\delta}{\delta \phi^{a}(\vec{x},t)}\right) ,$$ and the Poisson bracket is simply $X_{A}(B) = \{A , B \}$. So, for field theory on the “space-time” $\mathbb{R}^{(0,1)}$, the vector field associated with the observable $A(q,p)$ becomes $$X_{A(q,p)} = \sum_{i = 1}^{d} \int_{\mathbb{R}^{0}} d^{0}x \left( \frac{\delta A}{\delta q^{i}} \frac{\delta}{\delta p_{i}(t)} - \frac{\delta A}{\delta p_{i}(t)}\frac{\delta}{\delta q^{i}(t)}\right) = \sum_{i}^{d} \left( \frac{\partial A}{\partial q^{i}}\frac{\partial}{\partial p_{i}} - \frac{\partial A}{\partial p_{i}}\frac{\partial}{\partial q^{i}} \right) .$$ This vector field gives you the Poisson bracket that you are familiar with $X_{A}(B) = \{ A , B\}_{(q,p)}$.

3) To apply the canonical formalism to continuous system, it is more convenient to work with a denumerable infinity (of degree of freedom) instead of continuous infinity. To do this, you can divide the continuous (3-dimensional) spatial volume into very small cubical cells. Clearly, one can label these cells by discrete variable. So, one has infinite number of cells of volume $\delta V_{i}$. In order to apply the canonical formalism, one needs to associate a coordinate pair $(q_{i}(t), \dot{q}_{i}(t))$ to the $i$th cell. These can be identified with the average value of $\varphi (\vec{x} ,t)$ and $\dot{\varphi}(\vec{x} , t)$ over the $i$th cell: $$\left( q_{i}(t) , \dot{q}_{i}(t) \right) \equiv \frac{1}{\delta V_{i}} \int_{(\delta V_{i})} d^{3}\vec{x} \left( \varphi (\vec{x} , t) , \dot{\varphi}(\vec{x} , t)\right) .$$ This method, described in many textbooks, allows you to translate the usual canonical formalism of discrete systems to field theory. However, there is an alternative method that can help you understand the connection between the continuous and the discrete Poisson brackets quite easily. For simplicity, I will consider real scalar field $\varphi (\vec{x},t)$ described by the Lagrangian $$L = \int_{\mathbb{R}^{3}} d^{3}x \mathcal{L} \left(\varphi , \dot{\varphi} , \nabla \varphi \right)(\vec{x} ,t) . \ \ \ \ \ \ \ \ \ (3.1)$$ The field conjugate momentum is defined by $$\pi (\vec{x} , t) = \frac{\partial \mathcal{L}}{\partial \dot{\varphi}}(\vec{x},t) . \ \ \ \ \ \ \ \ \ \ \ (3.2)$$ Let $\{ u_{m}(\vec{x}) , \vec{x} \in \mathbb{R}^{3} , \ m = 1,2,3, \cdots \}$ be a complete set of real orthonormal functions $$\sum_{m=1}^{\infty} u_{m}(\vec{x}) u_{m}(\vec{y}) = \delta^{3} (\vec{x} - \vec{y}) , \ \ \ \ \ \ \ \ (3.3a)$$ $$\int_{\mathbb{R}^{3}} d^{3}x \ u_{m}(\vec{x}) u_{n}(\vec{x}) = \delta_{mn} . \ \ \ \ \ \ \ \ (3.3b)$$ Now we can write the following expansions $$\varphi (\vec{x} , t) = \sum_{m = 1}^{\infty} q_{m}(t) \ u_{m}(\vec{x}) , \ \ \ \ \ \ \ (3.4a)$$ $$\dot{\varphi} (\vec{x} , t) = \sum_{m = 1}^{\infty} \dot{q}_{m}(t) \ u_{m}(\vec{x}) . \ \ \ \ \ \ \ (3.4b)$$ Using Eq(3.3b), we obtain the inverted expansions

$$q_{n}(t) = \int_{\mathbb{R}^{3}} d^{3}x \ \varphi (\vec{x} ,t) \ u_{n}(\vec{x}) , \ \ \ \ \ \ (3.5a)$$

$$\dot{q}_{n}(t) = \int_{\mathbb{R}^{3}} d^{3}x \ \dot{\varphi} (\vec{x} ,t) \ u_{n}(\vec{x}) . \ \ \ \ \ \ (3.5b)$$ Next we interpret $(q_{n} , \dot{q}_{n})(t)$ as countable set of infinite coordinates and velocities, and apply the usual canonical formalism using the Lagrangian function $L(q , \dot{q})(t)$. Indeed, using (3.1) and the relations (3.4), it is easy to show that

$$\frac{\partial L}{\partial q_{n}} - \frac{d}{dt}(\frac{\partial L}{\partial \dot{q}_{n}}) = \int_{\mathbb{R}^{3}} d^{3}x \left( \frac{\partial \mathcal{L}}{\partial \varphi} - \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\varphi)} \right) \right) u_{n}(\vec{x}) .$$ If the left hand side is zero, then the completeness of the set $\{u_{n}\}$ leads to the field equation $$\frac{\partial \mathcal{L}}{\partial \varphi} - \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\varphi)} \right) = 0 . \ \ \ \ \ \ (3.6)$$ Conversely, if the field equations (3.6) are satisfied then $$\frac{\partial L}{\partial q_{n}} - \frac{d}{dt}(\frac{\partial L}{\partial \dot{q}_{n}}) = 0 , \ n = 1,2,3, \cdots . \ \ \ \ (3.7)$$ Thus, at every $(\vec{x} , t)$ the field equation (3.6) is equivalent to the infinite set of equations (3.7) for $(q_{n},\dot{q})(t)$, and vice versa.

Now that we have Lagrangian formalism, we can pass to the Hamiltonian formalism and address the issue of Poisson brackets. To do this we define momenta $p_{n}(t)$ canonically conjugate to the coordinates $q_{n}(t)$ by the “usual” relation $$p_{n}(t) \equiv \frac{\partial L}{\partial \dot{q}_{n}(t)} = \int_{\mathbb{R}^{3}} d^{3}x \frac{\partial \mathcal{L}}{\partial \dot{\varphi}(\vec{x},t)} \frac{\partial \dot{\varphi}(\vec{x},t)}{\partial \dot{q}_{n}(t)} .$$ Then, from the definition (3.2) of $\pi (\vec{x},t)$ and the expansion (3.4b), we obtain $$p_{n}(t) = \int_{\mathbb{R}^{3}} d^{3}x \ \pi (\vec{x},t) \ u_{n}(\vec{x}) . \ \ \ \ \ \ \ (3.8a)$$ This can be inverted to give $$\pi (\vec{x},t) = \sum_{n}^{\infty} \ p_{n}(t) \ u_{n}(\vec{x}) . \ \ \ \ \ \ \ \ (3.8b)$$ We now have enough relations to describe any observable $A$ as a function of the pair $(q_{n} , p_{n})(t)$ or as a functional $\mathcal{A}$ of $(\varphi , \pi)(\vec{x},t)$. Indeed, $$A(q , p)(t) = \mathcal{A}(\varphi , \pi )(\vec{x},t) .$$ So, we can calculate $$\frac{\partial A}{\partial q_{n}(t)} = \int_{\mathbb{R}^{3}} d^{3}x \ \frac{\delta \mathcal{A}}{\delta \varphi (\vec{x},t)} \frac{\partial \varphi (\vec{x},t)}{\partial q_{n}(t)} ,$$ which is $$\frac{\partial A}{\partial q_{n}(t)} = \int_{\mathbb{R}^{3}} d^{3}x \ \frac{\delta \mathcal{A}}{\delta \varphi (\vec{x},t)} \ u_{n}(\vec{x}) .$$ And, similarly for the observable $B$, we find $$\frac{\partial B}{\partial p_{n}(t)} = \int_{\mathbb{R}^{3}} d^{3}y \ \frac{\delta \mathcal{B}}{\delta \pi (\vec{y},t)} \ u_{n}(\vec{y}) .$$ Thus $$\sum_{n}^{\infty} \frac{\partial A}{\partial q_{n}} \frac{\partial B}{\partial p_{n}} = \int_{\mathbb{R}^{6}} d^{3}x d^{3}y \ \frac{\delta \mathcal{A}}{\delta \varphi (\vec{x},t)} \frac{\delta \mathcal{B}}{\delta \pi (\vec{y},t)} \sum_{n} u_{n}(\vec{x}) u_{n}(\vec{y}) .$$ Then, the completeness relation (3.3a) leads to $$\sum_{n}^{\infty} \ \frac{\partial A}{\partial q_{n}} \frac{\partial B}{\partial p_{n}} = \int_{\mathbb{R}^{3}} d^{3}x \ \frac{\delta \mathcal{A}}{\delta \varphi (\vec{x},t)} \frac{\delta \mathcal{B}}{\delta \pi (\vec{x},t)} .$$ This gives us the following Poisson bracket on the infinite-dimensional phase-space of our field theory

$$\sum_{n}^{\infty} \left( \frac{\partial A}{\partial q_{n}} \frac{\partial B}{\partial p_{n}} - \frac{\partial A}{\partial p_{n}} \frac{\partial B}{\partial q_{n}}\right) = \int_{\mathbb{R}^{3}} d^{3}x \left( \frac{\delta \mathcal{A}}{\delta \varphi} \frac{\delta \mathcal{B}}{\delta \pi} - \frac{\delta \mathcal{A}}{\delta \pi} \frac{\delta \mathcal{B}}{\delta \varphi} \right) .$$ The same method can be applied for field theory with finite-component field $\varphi^{a}(\vec{x},t)$. In the Poisson bracket, this requires you to sum over the fields index $a$.

Last edited: Aug 16, 2017
17. Aug 16, 2017

TeethWhitener

I hadn't, but the ensuing explanation is exceptionally clear and helpful. I'm still working my way through point 3), so I might have some questions later, but your posts have been very edifying. Thank you!