- #1
schieghoven
- 85
- 1
Hello,
What is a quantum state? Put generalised functions/Schwartz distributions to one side, because a) they're not a Hilbert space, and b) they can't be multiplied, so it's hopeless to even begin to think about Feynman diagrams.
One-particle quantum states seem to be fairly well understood. The state of the system is a function [itex] \psi: \mathbb{R}^3 \rightarrow \mathbb{C} [/itex], and [tex]|\psi(x)|^2[/tex] gives the probability density of finding the particle near the space point x. Let's denote by [itex] \Omega_1 [/itex] the space of one-particle states. [itex] \Omega_1 [/itex] is a Hilbert space with inner product
Building on this, an n-particle state [tex] \psi [/tex] is presumably a function of n space points (x1, x2, ... xn). Assume Bose symmetry, so [tex] \psi [/tex] is totally symmetric with respect to x1, x2, ... xn. So in this case [itex] \psi: (\mathbb{R}^3)^n \rightarrow \mathbb{C} [/itex] and [tex]|\psi(x_1, x_2, \ldots, x_n)|^2[/tex] gives the probability density of finding the n particles near the space points x1, x2, ..., xn. The set [itex] \Omega_n [/itex] of all n-particle states has a canonical inner product
Cheers,
Dave
What is a quantum state? Put generalised functions/Schwartz distributions to one side, because a) they're not a Hilbert space, and b) they can't be multiplied, so it's hopeless to even begin to think about Feynman diagrams.
One-particle quantum states seem to be fairly well understood. The state of the system is a function [itex] \psi: \mathbb{R}^3 \rightarrow \mathbb{C} [/itex], and [tex]|\psi(x)|^2[/tex] gives the probability density of finding the particle near the space point x. Let's denote by [itex] \Omega_1 [/itex] the space of one-particle states. [itex] \Omega_1 [/itex] is a Hilbert space with inner product
[tex]
\langle \psi, \phi \rangle = \int d^3x \; \psi^*(x) \phi(x) \qquad \qquad (1)
[/tex]
This Hilbert space is known as [itex]L^2(\mathbb{R}^3)[/itex]. The states [tex]\psi, \phi \in \Omega_1[/tex] evolve in time according to an equation of motion, and the the inner product (1) is constant in time. Equivalently, the system evolves by a unitary transformation on [itex] \Omega_1 [/itex].\langle \psi, \phi \rangle = \int d^3x \; \psi^*(x) \phi(x) \qquad \qquad (1)
[/tex]
Building on this, an n-particle state [tex] \psi [/tex] is presumably a function of n space points (x1, x2, ... xn). Assume Bose symmetry, so [tex] \psi [/tex] is totally symmetric with respect to x1, x2, ... xn. So in this case [itex] \psi: (\mathbb{R}^3)^n \rightarrow \mathbb{C} [/itex] and [tex]|\psi(x_1, x_2, \ldots, x_n)|^2[/tex] gives the probability density of finding the n particles near the space points x1, x2, ..., xn. The set [itex] \Omega_n [/itex] of all n-particle states has a canonical inner product
[tex]
\langle \psi, \phi \rangle = \frac{1}{n!} \int d^3x_1 \ldots d^3x_n \;
\psi^*(x_1, \ldots, x_n) \phi(x_1, \ldots, x_n)
[/tex]
and is a Hilbert space. So far so good. This is not just rigorous - L^2 spaces are stock concepts in pure math - but it's readily understandable as well. States have a direct physical interpretation at all times, not just at asymptotic [itex] t \rightarrow \pm \infty [/tex]. So is it possible to formulate field theory from this standpoint? In field theory, particle number changes with time, so let's suppose the set of all states is\langle \psi, \phi \rangle = \frac{1}{n!} \int d^3x_1 \ldots d^3x_n \;
\psi^*(x_1, \ldots, x_n) \phi(x_1, \ldots, x_n)
[/tex]
[tex]
\Omega = \mathbb{C} \oplus \Omega_1 \oplus \Omega_2 \oplus \ldots
\oplus \Omega_n \oplus \ldots
[/tex]
(The C is for the vacuum.) Is it possible to define the dynamics of the system in terms of an equation of motion for the n-particle 'wavefunctions'? Would this be another route towards constructing the Feynman series?\Omega = \mathbb{C} \oplus \Omega_1 \oplus \Omega_2 \oplus \ldots
\oplus \Omega_n \oplus \ldots
[/tex]
Cheers,
Dave