What is “momentum density” and why it important to QFT?

I am reading Quantum Field Theory for the Gifted Amateur. On page 98, they provide a summary of a basic canonical quantization procedure:

• Step I: Write down a classical Lagrangian density in terms of the field. This is the creative part because there are lots of possible Lagrangians. After this step, everything else is automatic.
• Step II: Calculate the momentum density and work out the Hamiltonian density in terms of fields.
• Step III: Now treat the fields and momentum density as operators. Impose commutation relations on them to make them quantum mechanical.
• Step IV: Expand the field in terms of creation/annihilation operators. This will allow us to use occupation numbers and stay sane.
• Step V: That's it. Congratulations, you are now the proud owner of a working quantum field theory, provided you remember the normal ordering interpretation.

I don't understand what momentum density is or why it comes up at this point in the quantization process. If by momentum, they mean like the operator $\hat{p}$, what about the position operator $\hat{x}$? Why isn't there a position density operator needed too? Everything else in the procedure makes sense to me except Step II. I assume Hamiltonian density is the Hamiltonian counterpart to Lagrangian density.

Can someone explain what momentum density is and why it's needed at this step in the procedure?

• Are you familiar with how to go from the Lagrangian formulation to the Hamiltonian formulation in the case of mechanics? – Robin Ekman Feb 6 '15 at 1:53
• No. I haven't gone through that carefully. If that's the term from classical mechanics, then I know where to find info to answer this question and it all makes sense now. – Stan Shunpike Feb 6 '15 at 1:57
• Part of my hesitance in learning the Hamiltonian was the complexity of symplectic manifolds. I haven't been able to grasp tensor products, and therefore wedge products. So I hadn't really paid much attention to it. But I suppose that's not really relevant to the question I am asking so knowing how to switch from Lagrangian to Hamiltonian would be useful and important to know. – Stan Shunpike Feb 6 '15 at 2:00
• @StanShunpike - One cannot understand canonical quantization without understanding the Hamiltonian formulation of classical mechanics. Further, knowledge of symplectic manifolds, etc. are not required to have the basic knowledge of the Hamiltonian formalism that is required to understand QM. – Prahar Feb 6 '15 at 2:38
• Oh really? Super! I was so...for lack of a better word....turned off by that because I just don't have the knowledge yet to understand flows on manifolds in the physics context yet. But that's super. It shouldn't take long to become familiar with it then. Much less time then I anticipated. – Stan Shunpike Feb 6 '15 at 3:14

1. A field $\phi^{\alpha}:[t_i,t_f]\times \mathbb{R}^3\to \mathbb{R}$ is the field-theoretic version of a (generalized) position variable $q^i:[t_i,t_f]\to \mathbb{R}$ in point mechanics. Note that the physical position space $\mathbb{R}^3$ typically plays very different roles in field theory and in point mechanics.$^1$

2. Momentum density $\pi_{\alpha}$ is the natural field-theoretic generalization of momentum variable $p_i$ from point mechanics. The (Lagrangian) momentum densities are $$\tag{1}\pi_{\alpha}~:=~\frac{\partial{\cal L}}{\partial\dot{\phi}^{\alpha}}$$ in analogy with $$\tag{2}p_i~:=~\frac{\partial L}{\partial\dot{q}^i}$$ in point mechanics.

3. Note that there is another notion of momentum $P^i=T^{0i}$ coming from the stress-energy-momentum tensor $T^{\mu\nu}$, cf. e.g. this Phys.SE post.

4. One should perform a Legendre transformation $$\tag{3}\dot{\phi}^{\alpha} \quad\longleftrightarrow\quad \pi_{\alpha}$$ to get to the Hamiltonian formulation. Note in particular that (Hamiltonian) momentum densities $\pi_{\alpha}$ are independent variables.

5. The Hamiltonian formulation is needed$^2$ in order to impose the canonical commutation relations (CCRs) necessary for quantization.

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$^1$ The notion of spacetime, position and field can more generally be defined with the help of differential geometry and the notion of a manifold.

$^2$ Here we only discuss the traditional approach. For manifestly covariant Hamiltonian formulation, see also e.g. this and this Phys.SE posts.

• Regarding point 5, I read here physics.stackexchange.com/q/21866/66165 that we don't necessarily need the Hamiltonian formulation. Am I misunderstanding that post? – Stan Shunpike Feb 6 '15 at 19:10
• It depends on what you want to do. See also e.g. this Phys.SE post. – Qmechanic Feb 6 '15 at 19:26