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Take a free relativistic QFT, say for a real scalar field $\phi$ with the Lagrangian density $$ \mathscr{L} = \frac{ \partial_{\mu} \phi \partial^\mu \phi - m^2 \phi^2}{2} \ . $$ After quantization we have quantum operators $\hat{\phi}(t,\mathbf{x})$ and $\hat{\pi} (t,\mathbf{x}) := \partial_0 \hat{\phi}(t,\mathbf{x})$ which obey the standard commutation relations.

Considering the standard field expansion in terms of creation and annihilation operators (where $E_p = \sqrt{p^2 + m^2}$) $$ \hat{\phi}(t,\mathbf{x}) = \int \frac{d^3 p}{\sqrt{(2\pi)^3 2 E_p}} \left( e^{- i E_p t + i \mathbf{p} \cdot \mathbf{x}} \hat{a}_{\mathbf{p}} + e^{+ i E_p t - i \mathbf{p} \cdot \mathbf{x}} \hat{a}^{\dagger}_{\mathbf{p}} \right) \ , $$ is it true that the bosonic Fock space $\mathcal{F}$ can be decomposed as a tensor product of $\mathcal{F}_{\mathbf{p}}$ for each momentum $\mathbf{p}$? Something like $$ \mathcal{F} = \bigotimes_{\mathbf{p} \in \mathbb{R}^{3}}\mathcal{F}_{\mathbf{p}} ? $$ I think this is true, but I don't understand how one can prove this. How can one see that this is true, if it is?

(Or is it rather a direct sum $\mathcal{F} = \bigoplus_{\mathbf{p}} \mathcal{F}_{\mathbf{p}}$?)

DanielC
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QuantumEyedea
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1 Answers1

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Once you impose the canonical commutation relations, the situation is very similar to the quantum harmonic oscillator. From the expression for $\phi(x, t)$ and $\pi(x, t)$ that you wrote, you can show that the Hamiltonian is $$H = \int \frac{d^3 p}{(2\pi)^3 2E_p} E_p a^\dagger_{\textbf{p}} a_{\textbf{p}}$$ up to a constant shift which should be renormalized away. From this, it follows that $$ [H, a^\dagger_{\textbf{p}}] = E_p a^\dagger_{\textbf{p}}, \quad [H, a_{\textbf{p}}] = -E_p a_{\textbf{p}} $$ which means that $a_{\textbf{p}}$ applied to an energy eigenstate will lower its energy by $\sqrt{p^2 + m^2}$. The physical requirement that energy be bounded from below then tells us that there is a state annihilated by all $a_{\textbf{p}}$. This is the vacuum and we can consider states of the form $$ (a^\dagger_{\textbf{p}_1})^{n_1} \left | 0 \right >, \quad (a^\dagger_{\textbf{p}_2})^{n_2} \left | 0 \right > $$ which are built on top of it. The left collection is a basis for $\mathcal{F}_{\textbf{p}_1}$ and the right one is a basis for $\mathcal{F}_{\textbf{p}_2}$. In order for $$ (a^\dagger_{\textbf{p}_1})^{n_1} (a^\dagger_{\textbf{p}_2})^{n_2} \left | 0 \right > $$ to be a basis for $\mathcal{F}_{\textbf{p}_1} \otimes \mathcal{F}_{\textbf{p}_2}$, we need to show that they are linearly independent. But we can do better than this and show that they are orthogonal because the canonical commutation relations enforce $$ \left < 0 \right | (a_{\textbf{p}_2})^{m_2} (a_{\textbf{p}_1})^{m_1} (a^\dagger_{\textbf{p}_1})^{n_1} (a^\dagger_{\textbf{p}_2})^{n_2} \left | 0 \right > \propto \delta_{m_1, n_1} \delta_{m_2, n_2}. $$ Knowing this, we can iterate and show that we get a basis for $\mathcal{F}_{\textbf{p}_1} \otimes \mathcal{F}_{\textbf{p}_2} \otimes \mathcal{F}_{\textbf{p}_3} \otimes \dots$ with enough modes. Therefore the Fock space (the span of all states you get by creating particles from the vacuum) is a tensor product.

Of course I don't claim that this settles the question of what the admissible distributions are when you want to take a linear combination of energy eigenstates weighted by some $f(\textbf{p})$. I also didn't show that the vacuum is unique. But even when there are many vaccua, the Fock space for each one is a tensor product. To see that it's not a direct sum, consider the simplest case where each creation operator squares to zero. A 3 qubit system in this case has $$ (a^\dagger_1)^{n_1}(a^\dagger_2)^{n_2}(a^\dagger_3)^{n_3} \left | 0 \right > $$ as a basis. This is 8 dimensional because each $n_i$ is either 0 or 1. Not 6 dimensional as we would expect from a direct sum.

Connor Behan
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