Skip to main content
Physics

Return to Answer

added 30 characters in body
Source Link
yuggib
yuggib
  • 12.2k
  • 1
  • 24
  • 48

You can utilize the construction of the $Q$ space, as described in Reed and Simon vol.2, page 228-230.

Oversimplifying, you can make the analogy $\lvert \phi\rangle \sim \lvert x\rangle$, but the associated momentum is not $p$$\hat{p}$, but $\pi$$\hat{\pi}$ (the canonical conjugate momentum of the field $\phi$$\hat{\phi}$).

With slightly more precision: the Fock space is isomorphic to an $L^2$ space where $\phi$$\hat{\phi}$ acts as the multiplication by the function $x$ (is a "variable" of the $L^2$ space), and $\pi$$\hat{\pi}$ as the (functional) derivative $-i\frac{\delta}{\delta x}$; and in this context you can define the "eigenfunctions" (they do not belong to the $L^2$ obviously) $\lvert\phi\rangle$ and $\lvert\pi\rangle$ with the usual meaning as (infinite dimensional) position and momentum eigenfunctions. The precise construction is detailed in the reference above.

You can utilize the construction of the $Q$ space, as described in Reed and Simon vol.2, page 228-230.

Oversimplifying, you can make the analogy $\lvert \phi\rangle \sim \lvert x\rangle$, but the associated momentum is not $p$, but $\pi$ (the canonical conjugate momentum of the field $\phi$).

With slightly more precision: the Fock space is isomorphic to an $L^2$ space where $\phi$ acts as the multiplication by the function $x$ (is a "variable" of the $L^2$ space), and $\pi$ as the (functional) derivative $-i\frac{\delta}{\delta x}$; and in this context you can define the "eigenfunctions" (they do not belong to the $L^2$ obviously) $\lvert\phi\rangle$ and $\lvert\pi\rangle$ with the usual meaning as (infinite dimensional) position and momentum eigenfunctions. The precise construction is detailed in the reference above.

You can utilize the construction of the $Q$ space, as described in Reed and Simon vol.2, page 228-230.

Oversimplifying, you can make the analogy $\lvert \phi\rangle \sim \lvert x\rangle$, but the associated momentum is not $\hat{p}$, but $\hat{\pi}$ (the canonical conjugate momentum of the field $\hat{\phi}$).

With slightly more precision: the Fock space is isomorphic to an $L^2$ space where $\hat{\phi}$ acts as the multiplication by the function $x$ (is a "variable" of the $L^2$ space), and $\hat{\pi}$ as the (functional) derivative $-i\frac{\delta}{\delta x}$; and in this context you can define the "eigenfunctions" (they do not belong to the $L^2$ obviously) $\lvert\phi\rangle$ and $\lvert\pi\rangle$ with the usual meaning as (infinite dimensional) position and momentum eigenfunctions. The precise construction is detailed in the reference above.

Source Link
yuggib
yuggib
  • 12.2k
  • 1
  • 24
  • 48

You can utilize the construction of the $Q$ space, as described in Reed and Simon vol.2, page 228-230.

Oversimplifying, you can make the analogy $\lvert \phi\rangle \sim \lvert x\rangle$, but the associated momentum is not $p$, but $\pi$ (the canonical conjugate momentum of the field $\phi$).

With slightly more precision: the Fock space is isomorphic to an $L^2$ space where $\phi$ acts as the multiplication by the function $x$ (is a "variable" of the $L^2$ space), and $\pi$ as the (functional) derivative $-i\frac{\delta}{\delta x}$; and in this context you can define the "eigenfunctions" (they do not belong to the $L^2$ obviously) $\lvert\phi\rangle$ and $\lvert\pi\rangle$ with the usual meaning as (infinite dimensional) position and momentum eigenfunctions. The precise construction is detailed in the reference above.