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What is the difference between the momentum $p$ in $e^{i\mathbf{p}\cdot{\mathbf{x}}}$ in the Fourier transform of a scalar field and the corresponding conjugate momenta $\pi(x)$ of the scalar field?

tmwilson26
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Quantum Fields
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1 Answers1

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The $p$ in the Fourier transform of the (free real scalar) field $$ \phi(x) = \int \left(a(\vec p)\exp(-\mathrm{i}px) + a^\dagger(\vec p)\exp(\mathrm{i}px)\right)\frac{\mathrm{d}^3p}{2p^0} $$ is a number, it is essentially a change in coordinates on $\mathbb{R}^n$ like every Fourier transform.

The canonically conjugate momentum $\pi(x) := \frac{\partial L}{\partial \dot{\phi}}$ is another field, together with $\phi(x)$ it forms a set of coordinates for the classical (field) phase space (which is infinite-dimensional for fields).

In particular, the $p$ from the Fourier transform does not become an operator in the course of canonical quantization, it is different from the momentum operator $$ P^\mu = \int p^\mu a^\dagger(\vec p)a(p) \mathrm{\vec d}^3 p$$ which, also, is not the operator associated to the canonically conjugate momentum.

ACuriousMind
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  • thanks for your answer, but P in the exponential gives the value of momentum P when we insert it inside the Klein-Gordon equation, we get a mass shell condition $m^{2}+P^{2}=E^{2}$ so it has a physical meaning not just a coordinate change – Quantum Fields Jun 15 '15 at 01:26
  • @QuantumFields: That's because we Fourier transform only the spatial part $\vec x$ of $x = (x^0,\vec x)$, and get $p^0 = E_p$ precisely from the constraint $p^2 = -m^2$. Note the weird-looking integration measure, which is discussed e.g. in this post. – ACuriousMind Jun 15 '15 at 15:26